मंथिकाManthika
Menu
IN REVIEW

Free while we finish the books. A book in review is free to read and use, and so is its PDF. Once the book is finished, its PDF is ₹299.

A grid of numbers, treated as one object

Two rows of a class mark-sheet. Drop the names and subject headings, keep the six numbers in place, and the grid is one matrix, A.

Your class teacher keeps a mark-sheet. The students’ names run down the left side. The subjects run across the top. Each box holds one student’s marks in one subject. Read along a row and you see one student. Read down a column and you see one subject.

Many sheets work like this. A shop’s price list is a grid of items and prices. A bill lists each item with its quantity, rate and amount. The register from Chapter 1 is a grid too. In each one, a number means something only because of its place. Move it to another box and it means something else.

A grid with one name

Now treat the whole grid as one thing. Give it a single name, such as $A$, and do arithmetic with it. A grid of numbers used this way is a MATRIX.

Kanada
A SCHOLAR INDIA REMEMBERS

Kanada lived in north India between the 6th and the 2nd century BCE. He is named as the founder of the Vaisheshika, the school of particulars (from viśeṣa, the particular thing that makes one atom differ from another). He taught that matter is built from tiny parts that join in twos and threes. Ask of a matrix what he asked of matter. What kind of thing is it? It is built from numbers, but it is not a number. So test each number habit before you use it.

Kanada · the sparrow
Adding two mark-sheets works the way adding numbers does. Multiplying can break a habit: two matrices with no zero entries multiply to all zeros.

Here is the question for the whole chapter. You know how to add, scale and multiply numbers. If a whole grid is one thing, which of that arithmetic still works? And what changes? By the last page you will know which habits from numbers still hold, and which two break.

Check yourself
  1. What does a MATRIX make of a whole grid of numbers?
    1. a list of rows only, with columns handled separately
    2. one packaged object, with its own arithmetic
    3. a single number once every entry is combined
    4. several independent numbers, each handled on its own
    Check your answer
    1. a list of rows only, with columns handled separately — Nothing in this definition separates rows from columns as different kinds of object, the WHOLE grid, rows and columns together, is what gets packaged.
    2. ✓ one packaged object, with its own arithmetic — (B) The definition is explicit, a matrix bundles the whole grid into ONE object that gets added, scaled, and multiplied as a single unit.
    3. a single number once every entry is combined — Packaging a grid into one object does not shrink it down to a single number, the grid stays a grid, with every entry intact.
    4. several independent numbers, each handled on its own — A matrix is not a collection of independent numbers, the whole point here is that the grid is packaged as ONE object, handled as a unit.
  2. Adding and scaling work for matrices the way they do for numbers. Which ONE of these rules fails for matrices?
    1. equality, which no longer requires matching orders
    2. addition, which stops being entry by entry for larger matrices, where entries must combine across rows
    3. scalar multiplication, which stops distributing over addition
    4. multiplication: two matrices with no zero entry can multiply to the zero matrix
    Check your answer
    1. equality, which no longer requires matching orders — Equality still needs matching orders throughout, that requirement is never dropped.
    2. addition, which stops being entry by entry for larger matrices, where entries must combine across rows — Addition is entrywise for matrices of any order, nothing here singles out larger matrices as an exception.
    3. scalar multiplication, which stops distributing over addition — Scalar multiplication distributing over addition is one of the rules that DOES survive matrix multiplication, it is not the broken one.
    4. ✓ multiplication: two matrices with no zero entry can multiply to the zero matrix — (D) Of these four, only multiplication breaks a habit from numbers. Two matrices with no zero entry can multiply to the zero matrix, $O$.
  3. In the class mark-sheet, Roll 1’s Maths mark is 42. What makes 42 a Maths mark for Roll 1, and not some other mark?
    1. its place, in Roll 1’s row and the Maths column
    2. its size compared with the other marks on the sheet
    3. the total of all six marks on the sheet
    4. the numbers written next to it in the same row
    Check your answer
    1. ✓ its place, in Roll 1’s row and the Maths column — (A) A number in a grid means something only because of its place. Move 42 to another box and it becomes another student’s mark, or a mark in another subject.
    2. its size compared with the other marks on the sheet — The size of a mark does not say whose it is. The same 42 in Roll 2’s Physics box would be Roll 2’s Physics mark.
    3. the total of all six marks on the sheet — The total of the sheet is one number for all six boxes. It cannot say what one box means.
    4. the numbers written next to it in the same row — The numbers next to 42 are Roll 1’s other marks. They do not tell you that 42 is the Maths one. Its column does.

↑ Back to top

Before you start

You met every one of these facts once already. Try each check below.

If any of these felt new, read the page named before going on.

↑ Back to top

Order, entries, and named matrix shapes

●KEY-TERM
This matrix has 2 rows and 3 columns, so its order is 2 × 3. Rows are always said first.

Look again at the mark-sheet. It has 2 students and 3 subjects. So it has 2 rows and 3 columns. To find a mark, you find the student first and then the subject. A matrix is read the same way, row first and then column.

The order of a matrix

So a matrix with 2 rows and 3 columns has ORDER 2 by 3. It is written $2 \times 3$, and it has $2 \times 3 = 6$ entries. A matrix with $m$ rows and $n$ columns has order $m \times n$. Rows always come first.

Row 2, column 3 is the second student’s Chemistry mark, 31. Every entry has an address: its row first, then its column.

Each number in a matrix is an ENTRY, also called an element. In the mark-sheet, the entry in row 2, column 3 is the second student’s mark in the third subject. If the matrix is called $A$, that entry is $a_{2 3}$. Keep the space between 2 and 3. They are two labels, the row and the column. They are not the number twenty-three. The entry in row $i$ and column $j$ is $a_{i j}$.

Equal means same order and same entries. The second pair share six numbers but not the same order; the third pair differ in one entry.

Two matrices are EQUAL when two things are true. They have the same order. And each entry equals the entry in the same place in the other: $a_{i j} = b_{i j}$ for every $i$ and $j$. Two mark-sheets are the same only if every box matches.

Finding unknown entries

Let us use that. Find $x$ and $y$ when

$\begin{pmatrix} x & 2 \\ 3 & y \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ 3 & -1 \end{pmatrix}$.

*Both are $2 \times 2$,* so set each entry equal to the entry in the same place. The top-left entries give $x = 5$, and the bottom-right entries give $y = -1$.

So one matrix equation gives you as many plain equations as it has entries.

Try one where the entries combine. Find $x$ and $y$ when

$\begin{pmatrix} x+y & 2 \\ 3 & x-y \end{pmatrix} = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}$.

*The off-diagonal entries, $2$ and $3$, already match,* so the new information is in the other two: $x + y = 7$ and $x - y = 1$. Adding the two equations gives $2x = 8$, so $x = 4$; then $y = 3$.

Your turn: a matrix has 2 rows and 5 columns. What is its order? How many entries does it have?

Check yourself
  1. What is the order of a matrix with $m$ rows and $n$ columns?
    1. $m \times m$, since order always reports a square count
    2. $n \times m$, columns counted first
    3. $m \times n$, rows first, then columns
    4. $m + n$, the total count of rows and columns
    Check your answer
    1. $m \times m$, since order always reports a square count — Order names $m$ and $n$ exactly as they are, nothing forces them to be equal, and most matrices are not square.
    2. $n \times m$, columns counted first — Order always reports rows first, swapping to columns-first describes a different, unrelated convention.
    3. ✓ $m \times n$, rows first, then columns — (C) Order reports rows first, then columns, $m \times n$, never the other way round.
    4. $m + n$, the total count of rows and columns — Order is two separate numbers, rows and columns, not their sum, $m + n$ throws away which count is which.
  2. Exactly when are two matrices $A$ and $B$ EQUAL?
    1. every entry matches, even if the two orders differ
    2. they share the same order, regardless of what the entries are
    3. they share the same order and every corresponding entry matches
    4. they have the same number of entries in total, however arranged
    Check your answer
    1. every entry matches, even if the two orders differ — Matching entries mean nothing unless the two matrices even have the same order to begin with, a $2 \times 3$ cannot equal a $3 \times 2$, whatever the numbers are.
    2. they share the same order, regardless of what the entries are — Same order says nothing about whether the actual entries agree, the entry test still has to run.
    3. ✓ they share the same order and every corresponding entry matches — (C) Equality needs BOTH tests together, same order AND matching entries, position for position.
    4. they have the same number of entries in total, however arranged — A bare count of entries, however arranged, checks neither shape nor position, two very different matrices can share a total entry count.
  3. Why does the entry $a_{i j}$ notation matter enough that $i$ and $j$ are always read as two separate labels, never merged?
    1. because separate labels let a matrix skip the column count
    2. because $a_{i j}$ names one exact position: row $i$, column $j$
    3. because $i$ and $j$ must always be equal for the notation to work
    4. because merging them would change a matrix’s order
    Check your answer
    1. because separate labels let a matrix skip the column count — Naming a position still needs BOTH a row number and a column number, separate labels do not let anyone skip the column.
    2. ✓ because $a_{i j}$ names one exact position: row $i$, column $j$ — (B) Row-then-column is what pins down ONE exact position, merge the labels and the position a claim is talking about goes ambiguous.
    3. because $i$ and $j$ must always be equal for the notation to work — $i$ and $j$ are independent labels; an entry like $a_{1 3}$ has $i \neq j$, and that is the ordinary case, not an exception.
    4. because merging them would change a matrix’s order — A matrix’s order comes from $m$ and $n$, not from how its entry labels are written, merging $i$ and $j$ would not change the order at all.
●KEY-TERM
One student’s marks make a row matrix, 1 × 3. One subject’s marks make a column matrix, 2 × 1.

Take one row of the mark-sheet on its own. It holds one student’s marks in every subject. A matrix with just one row is a ROW MATRIX. Now take one column on its own. It holds every student’s marks in one subject. A matrix with just one column is a COLUMN MATRIX. A row matrix has order $1 \times n$, and a column matrix has order $m \times 1$.

Square, diagonal, scalar, identity: each is the one before with one more condition. In the last three, every entry off the diagonal is 0.

The other names are definitions, so here they are plainly. A SQUARE MATRIX has as many rows as columns, so its order is $n \times n$. Its MAIN DIAGONAL runs from the top-left corner to the bottom-right corner. It holds the entries $a_{1 1}$, $a_{2 2}$ and so on.

Every square matrix sits in the outer ring. Each ring inward adds one condition: diagonal, then scalar, then only the identity I at the centre.
Kanada
A SCHOLAR INDIA REMEMBERS

Do any two of these names clash? Check it on the identity $I$ of order $2 \times 2$. It is square. Its entries off the diagonal are $0$, so it is diagonal. Its diagonal entries match, so it is scalar. They are all $1$, so it is the identity. That is four names, and none clash, because each name sits inside the one before. A matrix that fails a step stops there.

Kanada · the sparrow

Diagonal, scalar and identity matrices

Inside the square matrices sit three more names. Each one adds one condition to the one before.

A zero matrix can have any order. Both of these are O: every entry 0, one 2 × 3 and one 2 × 2.

A ZERO MATRIX has every entry $0$. It is written $O$, and it can have any order at all.

Telling the types apart

Try one. Take $\begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}$. It is square, and both entries off the diagonal are $0$. So it is diagonal. Its diagonal entries, 2 and 5, are not the same. So it is not scalar. Each name tells you something about the whole matrix at a glance. The rules later in the chapter use them.

Your turn: is the $2 \times 2$ zero matrix $O$ a diagonal matrix? Is it a scalar matrix?

Check yourself
  1. What is a SCALAR matrix?
    1. any square matrix, whatever its diagonal entries are
    2. diagonal, with every diagonal entry equal
    3. a diagonal matrix whose diagonal entries are all zero
    4. any matrix with exactly one row
    Check your answer
    1. any square matrix, whatever its diagonal entries are — A general square matrix need not be diagonal at all, off-diagonal entries can be anything. Scalar keeps diagonal’s zero-off-diagonal condition and adds one more.
    2. ✓ diagonal, with every diagonal entry equal — (B) Scalar means diagonal, PLUS every diagonal entry is the same single number, not necessarily $1$, not necessarily $0$.
    3. a diagonal matrix whose diagonal entries are all zero — All-zero diagonal entries describe the ZERO matrix specifically, scalar only requires the diagonal entries to be EQUAL, not zero.
    4. any matrix with exactly one row — A one-row matrix is a ROW matrix. Scalar has nothing to do with the number of rows.
  2. Which named shape can have ANY order at all, not just a square one?
    1. the zero matrix
    2. a diagonal matrix
    3. a scalar matrix
    4. the identity matrix
    Check your answer
    1. ✓ the zero matrix — (A) Every other named shape here assumes square; the zero matrix alone can be $m \times n$ for any $m$, $n$, every entry $0$.
    2. a diagonal matrix — A diagonal matrix names its OFF-diagonal entries, which only makes sense once rows and columns match , square is required.
    3. a scalar matrix — A scalar matrix is a diagonal matrix by definition, and diagonal itself requires a square order.
    4. the identity matrix — The identity matrix is the scalar matrix with diagonal entries $1$, it needs a square order like every scalar matrix does.
  3. Diagonal, scalar, and identity matrices nest inside one another. Why does every scalar matrix count as diagonal too?
    1. scalar keeps diagonal’s off-diagonal-zero condition, and just adds one more
    2. scalar and diagonal are two unrelated conditions that happen to overlap
    3. scalar is a weaker condition than diagonal, so diagonal implies scalar instead
    4. scalar only applies to $2 \times 2$ matrices, where diagonal always holds anyway
    Check your answer
    1. ✓ scalar keeps diagonal’s off-diagonal-zero condition, and just adds one more — (A) Scalar is diagonal PLUS “all diagonal entries equal”, every condition diagonal demands is still required, with one more layered on top.
    2. scalar and diagonal are two unrelated conditions that happen to overlap — The two conditions are not unrelated, scalar is built directly on top of diagonal, sharing its off-diagonal-zero requirement exactly.
    3. scalar is a weaker condition than diagonal, so diagonal implies scalar instead — Diagonal is the weaker, more general condition; scalar narrows it further. The implication runs scalar-to-diagonal, never the reverse.
    4. scalar only applies to $2 \times 2$ matrices, where diagonal always holds anyway — Nothing about the nesting depends on matrix size, a $5 \times 5$ scalar matrix is diagonal for the same reason a $2 \times 2$ one is.

↑ Back to top

Adding matrices and scaling by a number

▲CONCEPT
Term 1 plus Term 2, box by box. Roll 1’s Maths: 42 and 45 make 87. Every other entry adds the same way.

At the end of the year the school adds two mark-sheets. Term 1 marks are on one sheet and Term 2 marks are on the other. For the total, you add the two marks in the same box. That is the same student in the same subject. You never add one student’s Maths to another student’s English.

A 2 × 3 and a 3 × 2 cannot be added: some boxes in one have no partner in the other.
Two 2 by 3 matrices fill the same six places, so they add. A 2 by 3 and 3 by 2 share only four.

That works only if both sheets have the same students and the same subjects, in the same places. If one sheet has an extra subject, some boxes have no partner. Then there is no total for them.

Addition of matrices

Matrices add in the same way. You can add two matrices only when they have the same order. Then you add entry by entry. The entry in row $i$, column $j$ of $A + B$ is $a_{i j} + b_{i j}$. A $2 \times 3$ matrix and a $3 \times 2$ matrix cannot be added at all. Subtraction works the same way, entry by entry. Take the same pair: $A - B = \begin{pmatrix} -4 & 3 \\ 3 & 2 \end{pmatrix}$, top left, $1 - 5 = -4$.

Doubling a Term 1 sheet marked out of 50 gives marks out of 100. Every entry doubles: Roll 2’s Physics, 44, becomes 88.

Now suppose a test was marked out of 50, and the school wants every mark out of 100. You double every mark on the sheet. Multiplying a matrix by a number works like that. For a number $k$, each entry of $k A$ is $k$ times the entry of $A$ in the same place. This is SCALAR MULTIPLICATION, and $k A$ has the same order as $A$.

A check with small matrices

Let us try both on small matrices. Take

$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & 0 \\ 1 & 2 \end{pmatrix}$.

Add them entry by entry:

$A + B = \begin{pmatrix} 1 + 5 & 2 + 0 \\ 3 + 1 & 4 + 2 \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 4 & 6 \end{pmatrix}$.

Then multiply every entry of $A$ by 3:

$3 A = \begin{pmatrix} 3 \cdot 1 & 3 \cdot 2 \\ 3 \cdot 3 & 3 \cdot 4 \end{pmatrix} = \begin{pmatrix} 3 & 6 \\ 9 & 12 \end{pmatrix}$.

The number rules still hold: A + B and B + A give the same matrix, so you can add in either order.

Adding and scaling keep the habits you have from numbers.

Check it: with $k = 2$, $k(A + B) = \begin{pmatrix} 12 & 2 \\ 6 & 12 \end{pmatrix}$ and $k A + k B = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix} + \begin{pmatrix} 10 & -2 \\ 0 & 4 \end{pmatrix} = \begin{pmatrix} 12 & 2 \\ 6 & 12 \end{pmatrix}$. They agree.

So with additions and scalings you can rearrange freely, just as with numbers.

Your turn: for the same $A$ and $B$, find $2B - A$. Then find $2(B - A)$. Are they the same matrix?

Check yourself
  1. For a scalar $k$, what does entry $(i,j)$ of $k A$ equal?
    1. $k$, regardless of the entry it replaces
    2. $a_{i j}$ divided by $k$
    3. $k \cdot a_{i j}$
    4. $k + a_{i j}$
    Check your answer
    1. $k$, regardless of the entry it replaces — Scaling keeps the original entry and multiplies it, it never discards $a_{i j}$ and replaces it outright.
    2. $a_{i j}$ divided by $k$ — Scalar multiplication multiplies each entry by $k$; dividing describes a different, inverse operation.
    3. ✓ $k \cdot a_{i j}$ — (C) Scalar multiplication scales EVERY entry by $k$, entry $(i,j)$ of $k A$ is $k \cdot a_{i j}$.
    4. $k + a_{i j}$ — Adding $k$ to an entry describes entrywise addition with a constant matrix, not scalar multiplication.
  2. Why can a $2 \times 3$ matrix never be added to a $3 \times 2$ matrix?
    1. addition is entry by entry, and some positions have no partner
    2. because the entries would need to be scaled first
    3. because one of the two matrices must be the zero matrix, so that adding it leaves the result unchanged
    4. because addition only works on square matrices
    Check your answer
    1. ✓ addition is entry by entry, and some positions have no partner — (A) Addition matches entry $(i,j)$ of $A$ to entry $(i,j)$ of $B$. With different orders, some positions in one matrix have no partner in the other.
    2. because the entries would need to be scaled first — Scaling entries first changes nothing about whether two orders line up, a mismatched order stays mismatched either way.
    3. because one of the two matrices must be the zero matrix, so that adding it leaves the result unchanged — Neither matrix has to be the zero matrix; any two matrices of the same order add, whatever their entries are.
    4. because addition only works on square matrices — Addition needs matching order, not squareness, two $2 \times 5$ matrices add just as validly as two $3 \times 3$ ones.
  3. Addition and scalar multiplication both leave a matrix’s order unchanged. What does this fact rule out?
    1. that $A + B$ or $k A$ could come out with a different order from $A$
    2. that a matrix can be scaled by a negative number
    3. that $A + B$ could ever equal $B + A$
    4. that two matrices of the same order can ever be added, since combining two matrices should give a new order
    Check your answer
    1. ✓ that $A + B$ or $k A$ could come out with a different order from $A$ — (A) Adding and scaling keep the order, so the result never has a new one. A product or a transpose can change it.
    2. that a matrix can be scaled by a negative number — Scaling by a negative number is perfectly ordinary; the shape fact says nothing about the sign of $k$.
    3. that $A + B$ could ever equal $B + A$ — $A + B = B + A$ is commutativity, a separate guarantee stated directly above, the order-preserving fact is about SHAPE, not about swapping the two matrices.
    4. that two matrices of the same order can ever be added, since combining two matrices should give a new order — This fact confirms same-order matrices add freely, it rules nothing like that out.
Double A, then take away B entry by entry. Row 1, column 2: twice 2 is 4, and 4 minus −1 is 5.
Worked example

Add and scale two matrices: $2A - B$

  1. $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, $B = \begin{pmatrix} 5 & -1 \\ 0 & 2 \end{pmatrix}$
    You are given $A$ and $B$. Both are $2 \times 2$, so you can add them.
  2. $A + B = \begin{pmatrix} 1 + 5 & 2 + (-1) \\ 3 + 0 & 4 + 2 \end{pmatrix} = \begin{pmatrix} 6 & 1 \\ 3 & 6 \end{pmatrix}$
    Add the entries in the same place. No row mixes with another row.
  3. $2A = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix}$
    Scale first. Every entry of $A$ is doubled.
  4. $2A - B = \begin{pmatrix} 2 - 5 & 4 - (-1) \\ 6 - 0 & 8 - 2 \end{pmatrix} = \begin{pmatrix} -3 & 5 \\ 6 & 6 \end{pmatrix}$
    Now subtract $B$ from $2A$, entry by entry, as with addition.
  5. $2 \cdot 2 - (-1) = 5$ and $2 \cdot 3 - 0 = 6$
    Check two entries on their own. Row 1, column 2 and row 2, column 1 match the answer.
Worked example

The same idea, on a firm’s production table

  1. Table 3.1 (NCERT Introductory Microeconomics, Class 12) gives row $L=1$ as $A = \begin{pmatrix} 0 & 1 & 3 & 7 \end{pmatrix}$ for capital $K=0,1,2,3$
    The table gives a firm’s output for each amount of labour and capital it uses. Reading four entries straight off it, labour held fixed at $1$ unit.
  2. Row $L=2$ gives $B = \begin{pmatrix} 0 & 3 & 10 & 18 \end{pmatrix}$ for the same four capital values
    The next row of the table, labour held fixed at $2$ units instead.
  3. $A + B = \begin{pmatrix} 0 & 4 & 13 & 25 \end{pmatrix}$
    Matrix addition adds each matching entry: the total output of the two labour levels at each capital value.
  4. $2A = \begin{pmatrix} 0 & 2 & 6 & 14 \end{pmatrix}$
    Scalar multiplication doubles every entry the same way, as if labour $1$’s whole output row were scaled up.
Check yourself
  1. For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 5 & -1 \\ 0 & 2 \end{pmatrix}$, the worked example finds $A + B = \begin{pmatrix} 6 & 1 \\ 3 & 6 \end{pmatrix}$. What is $B + A$?
    1. $\begin{pmatrix} 6 & 1 \\ 3 & 6 \end{pmatrix}$ transposed
    2. a different matrix, since matrix addition is not commutative
    3. $\begin{pmatrix} 6 & 1 \\ 3 & 6 \end{pmatrix}$, identical to $A + B$
    4. undefined, since the order of addition matters
    Check your answer
    1. $\begin{pmatrix} 6 & 1 \\ 3 & 6 \end{pmatrix}$ transposed — Transposing swaps rows and columns, an unrelated operation to simply reversing the order of a sum.
    2. a different matrix, since matrix addition is not commutative — Addition IS commutative for matrices; the non-commutativity that breaks is specific to MULTIPLICATION, never carried over to addition.
    3. ✓ $\begin{pmatrix} 6 & 1 \\ 3 & 6 \end{pmatrix}$, identical to $A + B$ — (C) Addition IS commutative, $B + A$ lands on exactly the same matrix as $A + B$, entry for entry.
    4. undefined, since the order of addition matters — $A$ and $B$ share the same order, so $B + A$ is exactly as defined as $A + B$, nothing about order-of-addition is ever undefined here.
  2. Still using $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, $B = \begin{pmatrix} 5 & -1 \\ 0 & 2 \end{pmatrix}$, the worked example computes $2A - B = \begin{pmatrix} -3 & 5 \\ 6 & 6 \end{pmatrix}$. What is entry $(1,1)$ of $2A - B$ built from?
    1. $2 \cdot 1 - 2 \cdot 5$
    2. $2 \cdot 1 - 5$
    3. $1 - 5$
    4. $1 \cdot 1 - 5 \cdot 5$
    Check your answer
    1. $2 \cdot 1 - 2 \cdot 5$ — Only $A$ is scaled by $2$ in $2A - B$; $B$ enters the subtraction at its own, unscaled entry.
    2. ✓ $2 \cdot 1 - 5$ — (B) Entry $(1,1)$ of $2A$ is $2 \cdot 1 = 2$, and entry $(1,1)$ of $B$ is $5$; the subtraction is $2 \cdot 1 - 5 = -3$.
    3. $1 - 5$ — $2A$’s entry $(1,1)$ is $2 \cdot 1$, not the bare entry $1$, the scalar has to be applied before subtracting.
    4. $1 \cdot 1 - 5 \cdot 5$ — The formula calls for scaling then subtracting, not multiplying the two matrices’ entries together , that would be a different operation entirely.
  3. The worked example notes ‘no row mixes with another row’ while computing $A + B$ and $2A - B$. Which chapter idea does this illustrate?
    1. the symmetric-plus-skew-symmetric decomposition
    2. the compatibility condition that gates matrix multiplication, needed whenever two matrices are combined
    3. addition and scaling acting entry by entry, position by position
    4. the uniqueness of a matrix inverse
    Check your answer
    1. the symmetric-plus-skew-symmetric decomposition — The symmetric-plus-skew-symmetric decomposition is a later, separate idea, not what this addition-and-scaling example illustrates.
    2. the compatibility condition that gates matrix multiplication, needed whenever two matrices are combined — Compatibility is the condition gating whether $A B$ exists, this worked example only adds and scales, no multiplication involved.
    3. ✓ addition and scaling acting entry by entry, position by position — (C) Addition and scalar multiplication touch one position at a time, nothing from row 1 ever feeds into row 2’s result.
    4. the uniqueness of a matrix inverse — Inverse uniqueness concerns a completely different concept, invertibility, not addition or scaling.

↑ Back to top

Row times column

▲CONCEPT
The bill as row times column: each quantity times its price, then the two products added. 20 + 120 = 140 rupees.
The same purchase, drawn as a shop counter.

You buy 2 pens and 3 notebooks. A pen costs 10 rupees and a notebook costs 40 rupees. The shopkeeper multiplies each quantity by its price and adds. Two times 10 is 20, three times 40 is 120, and the bill is 140 rupees.

Writing the bill as matrices

Write what you bought as a row, 2 and 3. Write the prices as a column, 10 and 40. The bill is the row times the column. You multiply first by first and second by second, then add. That one step, row times column, is the whole of matrix multiplication.

Two customers, two shops. Each bill is one row of quantities times one column of prices: your friend at shop 1 pays 90 rupees.

Now a friend buys 5 pens and 1 notebook. Put the two of you in two rows. A second shop sells pens at 12 rupees and notebooks at 35 rupees. Put the two shops’ prices in two columns. Each customer at each shop gets one bill, and the bills form a grid.

$\begin{pmatrix} 2 & 3 \\ 5 & 1 \end{pmatrix} \begin{pmatrix} 10 & 12 \\ 40 & 35 \end{pmatrix} = \begin{pmatrix} 140 & 129 \\ 90 & 95 \end{pmatrix}$

The entry in row 2, column 1 is your friend at the first shop. It is $5 \times 10 + 1 \times 40 = 90$ rupees.

Any entry of a product

So here is the rule. The entry in row $i$, column $j$ of $A B$ is row $i$ of $A$ times column $j$ of $B$. You multiply matching entries and add the products.

The middle numbers must match, and the outer ones give the answer’s order: 2 × 3 by 3 × 2 gives 2 × 2.
Left: inner counts 3 and 3 share one edge, and the outer 2 and 2 give the order. Right: 3 and 2 cannot meet.

This needs the row and the column to be the same length. Every item bought needs exactly one price. So the number of columns of $A$ must equal the number of rows of $B$. If $A$ is $m \times n$ and $B$ is $n \times p$, then $A B$ is defined, and its order is $m \times p$. If the two inner numbers differ, $A B$ is not defined at all.

The order of a product

The outer numbers give the size of the answer. A $2 \times 3$ matrix times a $3 \times 2$ matrix is $2 \times 2$, as the figure above shows. One product works out every bill at once, for any number of customers and shops.

Your turn: take a different pair, with no entries given. $A$ has order $3 \times 2$ and $B$ has order $2 \times 4$. Is $A B$ defined, and what is its order? Is $B A$ defined?

Blue row 2 of A meets purple column 1 of B. The products 28, 45 and 66 add to 139, the orange entry.
Check yourself
  1. What is entry $(i,j)$ of the matrix product $A B$?
    1. the dot product of row $i$ of $A$ with column $j$ of $B$
    2. the sum of every entry in $A$’s row $i$
    3. $a_{i j} + b_{i j}$, the two matching entries added
    4. $a_{i j} \cdot b_{i j}$, the two matching entries multiplied
    Check your answer
    1. ✓ the dot product of row $i$ of $A$ with column $j$ of $B$ — (A) Matrix multiplication is row-times-column, multiply corresponding entries of row $i$ and column $j$, then add.
    2. the sum of every entry in $A$’s row $i$ — Every entry of $A B$ needs BOTH a row of $A$ and a column of $B$, summing $A$’s row alone leaves $B$ out entirely.
    3. $a_{i j} + b_{i j}$, the two matching entries added — Adding matching entries is what matrix ADDITION does, multiplication builds each entry from a dot product instead.
    4. $a_{i j} \cdot b_{i j}$, the two matching entries multiplied — Multiplying matching entries describes an ENTRYWISE product, a different operation that matrix multiplication explicitly is NOT.
  2. For $A B$ to even exist, what must be true of $A$’s and $B$’s orders?
    1. the number of rows of $A$ must equal the number of columns of $B$
    2. $A$ and $B$ must share the exact same order
    3. the number of columns of $A$ must equal the number of rows of $B$
    4. both $A$ and $B$ must be square
    Check your answer
    1. the number of rows of $A$ must equal the number of columns of $B$ — The rule matches $A$’s COLUMNS to $B$’s ROWS, never $A$’s rows to $B$’s columns, swapping them states the wrong pair.
    2. $A$ and $B$ must share the exact same order — Same order is what ADDITION needs, multiplication only needs the inner two counts to match, a far looser requirement.
    3. ✓ the number of columns of $A$ must equal the number of rows of $B$ — (C) That shared inner count is what actually gets summed over in the dot product, without it, row $i$ of $A$ and column $j$ of $B$ do not even have matching lengths.
    4. both $A$ and $B$ must be square — Neither $A$ nor $B$ needs to be square, only the shared inner count decides whether $A B$ exists.
  3. If $A$ is $m \times n$ and $B$ is $n \times p$, what is the order of the product $A B$?
    1. $m \times n$
    2. $m \times p$
    3. $p \times m$
    4. $n \times n$
    Check your answer
    1. $m \times n$ — The product’s order is not simply $A$’s order carried over, $B$’s column count $p$ replaces $A$’s column count $n$.
    2. ✓ $m \times p$ — (B) The outer two numbers give the product’s shape, $m \times p$, once the inner two, both $n$, have matched and cancelled out of the description.
    3. $p \times m$ — $m \times p$, not $p \times m$, the outer two numbers keep the order they started in, $A$’s rows first.
    4. $n \times n$ — The inner count $n$ matches and cancels out of the description, it never shows up in the product’s own order.
  4. $A$ is $2 \times 3$ and $B$ is $3 \times 2$. Why can $A B$ exist while $B A$, computed the other way round, could still fail in general for a differently-shaped pair?
    1. compatibility checks a different inner pair each way
    2. because $B A$ never exists once $A B$ does
    3. because only square matrices can ever be multiplied
    4. because $A B$ and $B A$ always share the same compatibility condition
    Check your answer
    1. ✓ compatibility checks a different inner pair each way — (A) Compatibility for $A B$ checks $A$’s columns against $B$’s rows; for $B A$ it checks $B$’s columns against $A$’s rows, a different pair of numbers, so one order can pass while the other fails.
    2. because $B A$ never exists once $A B$ does — One product existing settles nothing about the other, here BOTH happen to exist, but a differently-shaped pair could easily have only one defined.
    3. because only square matrices can ever be multiplied — Neither product needs a square matrix, compatibility is about the inner counts lining up, not about being square.
    4. because $A B$ and $B A$ always share the same compatibility condition — $A B$ checks $A$’s columns against $B$’s rows; $B A$ checks $B$’s columns against $A$’s rows, genuinely different pairs of numbers, not one shared condition.
  5. If the inner numbers of $A$ and $B$ do not match, what happens to $A B$?
    1. $A B$ is computed anyway, padded with zero entries
    2. $A B$ is computed using only the smaller of the two inner counts
    3. $A B$ is simply undefined, with no fallback rule
    4. $A B$ becomes the zero matrix automatically
    Check your answer
    1. $A B$ is computed anyway, padded with zero entries — There is no padding convention, a mismatched inner count leaves nothing to pad, the product is simply not defined.
    2. $A B$ is computed using only the smaller of the two inner counts — There is no fallback to the smaller count, compatibility is exact, not approximate.
    3. ✓ $A B$ is simply undefined, with no fallback rule — (C) A mismatched inner count leaves no shared length to run the dot product over, $A B$ is not defined at all, whatever $B A$ does.
    4. $A B$ becomes the zero matrix automatically — A mismatched product does not default to zero; it is UNDEFINED, a different status from “computed and equal to zero.”
Worked example

Multiply a $2 \times 3$ matrix by a $3 \times 2$ matrix, entry by entry

  1. $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$ (order $2 \times 3$), $B = \begin{pmatrix} 7 & 8 \\ 9 & 10 \\ 11 & 12 \end{pmatrix}$ (order $3 \times 2$)
    First check the sizes. $A$ is $2 \times 3$ and $B$ is $3 \times 2$.
  2. $3 = 3$: the column count of $A$ matches the row count of $B$
    The inner numbers match, so $A B$ is defined. The outer numbers say it is $2 \times 2$.
  3. Entry $(1,1) = 1 \cdot 7 + 2 \cdot 9 + 3 \cdot 11 = 7+18+33 = 58$
    Row 1 of $A$ times column 1 of $B$. Multiply in pairs, then add.
  4. Entry $(1,2) = 1 \cdot 8 + 2 \cdot 10 + 3 \cdot 12 = 8+20+36 = 64$
    Row 1 of $A$ times column 2 of $B$.
  5. Entry $(2,1) = 4 \cdot 7 + 5 \cdot 9 + 6 \cdot 11 = 28+45+66 = 139$
    Row 2 of $A$ times column 1 of $B$.
  6. Entry $(2,2) = 4 \cdot 8 + 5 \cdot 10 + 6 \cdot 12 = 32+50+72 = 154$
    Row 2 of $A$ times column 2 of $B$.
  7. *$A B = \begin{pmatrix} 58 & 64 \\ 139 & 154 \end{pmatrix}$*
    Put the four entries in their places. $B A$ would be $3 \times 3$, a different size altogether.
  8. $58 + 64 = 122$, and $1 \cdot 15 + 2 \cdot 19 + 3 \cdot 23 = 15 + 38 + 69 = 122$
    Check row 1 another way. The rows of $B$ add up to 15, 19 and 23. Row 1 of $A$ times those totals must give the sum of row 1 of $A B$.
Check yourself
  1. In the worked example, $A$ is $2 \times 3$ and $B$ is $3 \times 2$. Entry $(1,1)$ of $A B$ is computed as $1 \cdot 7 + 2 \cdot 9 + 3 \cdot 11$. Which two things does this sum use?
    1. the full matrices $A$ and $B$, entry by entry
    2. row 1 of $A$ and row 1 of $B$
    3. row 1 of $A$ and column 1 of $B$
    4. column 1 of $A$ and row 1 of $B$
    Check your answer
    1. the full matrices $A$ and $B$, entry by entry — Only ONE row of $A$ and ONE column of $B$ feed a single entry, the rest of both matrices is irrelevant to that one position.
    2. row 1 of $A$ and row 1 of $B$ — Row 1 of $A$ paired with row 1 of $B$ is not a rule matrix multiplication uses, one factor always supplies a ROW, the other a COLUMN.
    3. ✓ row 1 of $A$ and column 1 of $B$ — (C) Entry $(1,1)$ pairs row 1 of $A$ with column 1 of $B$, exactly the row-times-column rule.
    4. column 1 of $A$ and row 1 of $B$ — Entry $(1,1)$ specifically pairs $A$’s row with $B$’s column, not the reverse.
  2. The worked example finds $A B = \begin{pmatrix} 58 & 64 \\ 139 & 154 \end{pmatrix}$. What order does $B A$ have, if it were formed?
    1. $2 \times 3$, matching $A$’s own order
    2. $2 \times 2$, the same as $A B$
    3. undefined, since $B A$ can never be formed here
    4. $3 \times 3$, a different order
    Check your answer
    1. $2 \times 3$, matching $A$’s own order — $B A$’s order is built from $B$’s rows and $A$’s columns, not from $A$’s order carried straight over.
    2. $2 \times 2$, the same as $A B$ — $A B$’s order comes from $A$’s rows and $B$’s columns; $B A$’s order comes from $B$’s rows and $A$’s columns, the outer numbers are not the same pair.
    3. undefined, since $B A$ can never be formed here — $B$ is $3 \times 2$ and $A$ is $2 \times 3$: the inner counts are both $2$, so $B A$ IS defined, just at a different shape from $A B$.
    4. ✓ $3 \times 3$, a different order — (D) $B$ is $3 \times 2$ and $A$ is $2 \times 3$, the inner counts (both $2$) match, so $B A$ IS defined, but at order $3 \times 3$, not $2 \times 2$.
  3. The sums in a product are short. What skill does the worked example’s working practise?
    1. the symmetric-plus-skew-symmetric decomposition
    2. the uniqueness of a matrix inverse
    3. scalar multiplication scaling every entry equally
    4. row-times-column bookkeeping
    Check your answer
    1. the symmetric-plus-skew-symmetric decomposition — The symmetric-plus-skew-symmetric decomposition appears later and is not about row/column bookkeeping.
    2. the uniqueness of a matrix inverse — Inverse uniqueness is a fact about a different concept entirely, unrelated to tracking rows and columns during a product computation.
    3. scalar multiplication scaling every entry equally — Scalar multiplication touches one entry at a time, with no row-versus-column pairing to track at all.
    4. ✓ row-times-column bookkeeping — (D) Row-times-column bookkeeping is exactly where tracking errors creep in, the sums are simple, keeping the right row and column paired is the actual skill.

↑ Back to top

What survives, what doesn’t

▲CONCEPT
Quantities times prices is your bill, one number. Prices times quantities is a 2 × 2 table that nobody pays. Order matters.

Go back to your bill. Your row of quantities times the column of prices gave one number, 140 rupees. Now multiply the other way round, with the prices first. The column 10, 40 times the row 2, 3 pairs every price with every quantity. You get a table of four numbers, and it is not your bill. Nobody pays it. So for matrices the order of a product is not a choice.

The product in both orders

Here is that product:

$\begin{pmatrix} 10 \\ 40 \end{pmatrix} \begin{pmatrix} 2 & 3 \end{pmatrix} = \begin{pmatrix} 20 & 30 \\ 80 & 120 \end{pmatrix}$.

Your bill was the other order:

$\begin{pmatrix} 2 & 3 \end{pmatrix} \begin{pmatrix} 10 \\ 40 \end{pmatrix} = \begin{pmatrix} 140 \end{pmatrix}$.

The same two matrices give a $1 \times 1$ matrix one way and a $2 \times 2$ matrix the other way.

The same A and B multiplied both ways round. A B is 2 × 2, four numbers; B A is 3 × 3, nine numbers.
A new pair: here M N exists and is 2 × 3, but N M cannot be formed at all: 3 columns meet only 2 rows.

*So $A B$ and $B A$* can differ. Sometimes only one of them is defined. Even when both are defined and the same size, they can still differ, as the next section, “A B is not B A”, shows.

Grouping does not matter. Multiply A and B first, or B and C first: both routes end at the same matrix.
Multiplication spreads over addition. A times (B + C) and A B + A C give the same four entries.

Two other habits from numbers do survive. Each one needs the sizes to fit.

Using the surviving habits

Let us put the two together. Grouping lets you work out a long product from either end. Spreading lets you open brackets. Neither one lets you swap two factors. So hold the order of every product fixed, through every step.

Check yourself
  1. Which pair of rules survives matrix multiplication intact?
    1. associativity and distributivity
    2. commutativity and associativity
    3. commutativity and distributivity
    4. associativity and equality
    Check your answer
    1. ✓ associativity and distributivity — (A) $(A B) C = A (B C)$ and $A(B+C) = A B + A C$ both hold, associativity and distributivity survive; commutativity is the one that does not.
    2. commutativity and associativity — Commutativity is exactly the rule that does NOT survive, it cannot appear in the surviving pair.
    3. commutativity and distributivity — Distributivity does survive, but pairing it with commutativity still smuggles in the one rule that fails.
    4. associativity and equality — Equality is a relation between two matrices, not an algebraic manipulation rule like associativity or distributivity, it is not what is being contrasted here.
  2. $A(B+C)$ and $(A+B)C$ are both distributive rules, but they are treated as two separate facts rather than one. Why?
    1. because $B + C$ and $A + B$ never have compatible orders
    2. because only one of the two involves a sum at all
    3. the outside factor keeps its side
    4. because one of the two rules is actually false in general
    Check your answer
    1. because $B + C$ and $A + B$ never have compatible orders — Order compatibility governs whether a given product exists at all, not which of the two distributive rules applies, both rules assume compatible orders throughout.
    2. because only one of the two involves a sum at all — Both rules involve a sum, $(B+C)$ in one, $(A+B)$ in the other, so “only one involves a sum” is simply false.
    3. ✓ the outside factor keeps its side — (C) Since multiplication does not commute, distributing from the left ($A(B+C)$) and from the right ($(A+B)C$) are genuinely different claims, neither can substitute for the other.
    4. because one of the two rules is actually false in general — Both $A(B+C) = A B + A C$ and $(A+B)C = A C + B C$ are true statements, neither rule is false.
  3. A student regroups a product mid-calculation, swapping $A B$ to $B A$ to make the algebra tidier. Is this ever safe?
    1. no: order never swaps, only grouping can
    2. yes, since associativity permits regrouping factors freely
    3. no, because associativity itself is false for matrices
    4. yes, but only for $2 \times 2$ matrices
    Check your answer
    1. ✓ no: order never swaps, only grouping can — (A) Associativity lets you regroup WHICH factors multiply first, it never licenses swapping their ORDER. That stays fixed unless $A B = B A$ is separately verified.
    2. yes, since associativity permits regrouping factors freely — Associativity permits changing WHICH factors multiply first, never changing their ORDER, swapping $A B$ to $B A$ is a different, unlicensed move.
    3. no, because associativity itself is false for matrices — Associativity is one of the two rules that DOES survive matrix multiplication, it is not false.
    4. yes, but only for $2 \times 2$ matrices — Nothing about regrouping versus reordering depends on matrix size, the distinction holds for $2 \times 2$ matrices exactly as it does for any other size.

↑ Back to top

A B is not B A

✕MISCONCEPTION
First try

For $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$, I worked out $A B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$, then wrote $B A$ as the same matrix without multiplying the other way round.

Second look

$B A$ actually comes out $\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$, a different matrix, because row-times-column multiplication reads each product in the order the rows and columns are written.

$A B = B A$ only sometimes, never by default. Compute each product on its own before claiming the two are equal.

3
→
× 5
→
15
5
→
× 3
→
15
With numbers the order never matters: three fives and five threes are both 15.

With numbers, the order of the factors in a product never matters. Three fives and five threes are both fifteen. You have used this since primary school, often without noticing. The habit comes with you to matrices, and there it fails.

Two edits on one photo, in the two orders. The letter ends up facing two different ways, so F R and R F differ.

The photo editor on your phone shows it. Rotate a photo a quarter turn, then flip it left to right. Now start again with the same photo. Flip it first, then rotate it. The two pictures come out different. Each edit can be written as a matrix, and doing one edit after another is a product. So the order of the edits changes the answer.

How does a matrix move a picture? Write a point as a column, and multiply. A $2 \times 2$ matrix times the column $\begin{pmatrix} x \\ y \end{pmatrix}$ is a new column, a new point. The figures that follow show what a matrix does to the UNIT SQUARE, the square with corners $(0, 0)$, $(1, 0)$, $(0, 1)$ and $(1, 1)$. Its corner $(1, 0)$ lands on the matrix’s first column, and its corner $(0, 1)$ lands on the second column.

Swap the order and two entries change: A B has 2 in the top left, B A has 1. Same size, different matrices.
The unit square under A B has its far corner at (3, 2). Under B A the far corner lands at (2, 3).
Kanada
A SCHOLAR INDIA REMEMBERS

You call both of these multiplication. Are they the same kind of thing? With numbers, swapping the two factors changes nothing. With matrices, $A B$ pairs the rows of $A$ with the columns of $B$. In $B A$, the rows of $B$ meet the columns of $A$. Different rows meet different columns, so different entries come out. So $A B$ and $B A$ are two products. Work out both before you say they agree.

Kanada · the sparrow

Here is a pair of matrices whose two products differ:

$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$.

A working to check

FIRST TRY: writing $(A + B)^2$ as $(A + B)(A + B)$ spreads out to $A^2 + A B + B A + B^2$, and the middle terms combine to $A^2 + 2 A B + B^2$.

SECOND LOOK: $A B + B A$ is $2 A B$ only when $A B = B A$. The worked example below works out $A B$ and $B A$ for the pair above, entry by entry, and the two differ, so the middle terms do not combine that way. Work out both products before combining them; never assume $A B = B A$.

Some pairs do give the same product both ways, such as $A$ and $I$.

Your turn: take $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Find $A B$ and $B A$. Do they match?

Check yourself
  1. TRAP: ‘matrix multiplication should commute like ordinary numbers, so $A B$ and $B A$ must be the same matrix.’ Is the trap correct?
    1. yes: matrices always multiply the same way numbers do
    2. no: order can change the result
    3. no, because matrix multiplication is not even associative
    4. yes, but only for matrices with all-positive entries
    Check your answer
    1. yes: matrices always multiply the same way numbers do — This restates the trap the student is testing, exactly as though it were correct, the counterexample shows it is not.
    2. ✓ no: order can change the result — (B) The trap is false, order can change the result entirely, or leave one of the two products undefined altogether.
    3. no, because matrix multiplication is not even associative — Associativity ($(A B)C = A(B C)$) is a different rule from commutativity ($A B = B A$), associativity holds for matrices even though commutativity does not.
    4. yes, but only for matrices with all-positive entries — Whether entries are positive has no bearing on whether $A B = B A$, the counterexample uses only $0$s and $1$s and still disagrees.
  2. For $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$, it is given that $A B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ and $B A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$. What does this pair confirm?
    1. that $A$ and $B$ cannot both be invertible
    2. that two square matrices of the same order can give $A B \neq B A$
    3. that one of the two products must be a computational error
    4. that $A B$ and $B A$ always share the same diagonal entries, since swapping order should not change it
    Check your answer
    1. that $A$ and $B$ cannot both be invertible — This example is about commutativity, not invertibility, neither matrix’s inverse is discussed here.
    2. ✓ that two square matrices of the same order can give $A B \neq B A$ — (B) Both products exist, both come out $2 \times 2$, and they still disagree entry by entry, commutativity fails even in this friendly-looking case.
    3. that one of the two products must be a computational error — Both $A B$ and $B A$ are computed correctly here, the disagreement is the whole point, not a mistake to explain away.
    4. that $A B$ and $B A$ always share the same diagonal entries, since swapping order should not change it — The two matrices’ diagonals are $2,1$ and $1,2$, they differ, the opposite of what this option claims.
  3. Take $A = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. Before claiming $A B = B A$, what should you do?
    1. only check whether $B$ has a zero entry
    2. nothing: same-order square matrices always commute
    3. only check whether $A$ is diagonal
    4. compute both products and compare them directly
    Check your answer
    1. only check whether $B$ has a zero entry — A single zero entry in $B$ is not a general test for anything about commuting, the actual test is to compute and compare.
    2. nothing: same-order square matrices always commute — Same-order square matrices do not automatically commute, the counterexample above is exactly two such matrices that fail to commute.
    3. only check whether $A$ is diagonal — Whether $A$ is diagonal says nothing about the specific $B$ paired with it, commuting depends on the PAIR, not on one matrix alone.
    4. ✓ compute both products and compare them directly — (D) $A B = B A$ is a claim to test for a given pair, never one to assume, compute both sides and compare.
  4. A classmate claims: ‘$A B = B A$ whenever $A$ and $B$ are both square and the same size.’ Using $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ from the counterexample above, is the claim defensible?
    1. yes, provided both matrices are also invertible
    2. yes: same size and square is exactly the condition needed
    3. no, because $A$ and $B$ are not actually the same size
    4. no: both are $2 \times 2$ square, yet unequal
    Check your answer
    1. yes, provided both matrices are also invertible — Both $A$ and $B$ are in fact invertible here, and they still fail to commute, invertibility is unrelated to commuting.
    2. yes: same size and square is exactly the condition needed — Same size and square is exactly the condition $A$ and $B$ satisfy here, and they still do not commute , the claim is refuted, not confirmed.
    3. no, because $A$ and $B$ are not actually the same size — $A$ and $B$ are both $2 \times 2$, the exact same order, there is no size mismatch to point to.
    4. ✓ no: both are $2 \times 2$ square, yet unequal — (D) $A$ and $B$ are both $2 \times 2$, both square, and still $A B \neq B A$, same size and square is not enough to force commuting.
Worked example

Verify $A B \neq B A$ for $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$

  1. $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$
    This is the pair from the section above. Both products are defined, so work out both.
  2. $A B$ entry $(1,1) = 1 \cdot 1 + 1 \cdot 1 = 2$, entry $(1,2) = 1 \cdot 0 + 1 \cdot 1 = 1$
    Row 1 of $A$ times each column of $B$.
  3. $A B$ entry $(2,1) = 0 \cdot 1 + 1 \cdot 1 = 1$, entry $(2,2) = 0 \cdot 0 + 1 \cdot 1 = 1$
    Row 2 of $A$ times each column of $B$. So $A B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$.
  4. $B A$ entry $(1,1) = 1 \cdot 1 + 0 \cdot 0 = 1$, entry $(1,2) = 1 \cdot 1 + 0 \cdot 1 = 1$
    Now the other order. Row 1 of $B$ times each column of $A$.
  5. $B A$ entry $(2,1) = 1 \cdot 1 + 1 \cdot 0 = 1$, entry $(2,2) = 1 \cdot 1 + 1 \cdot 1 = 2$
    Row 2 of $B$ times each column of $A$. So $B A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$.
  6. *$A B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \neq \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = B A$*
    Both products are $2 \times 2$, and still they differ.
  7. $(A B)_{1 1} = 2$, but $(B A)_{1 1} = 1$
    Check one entry. A single entry that differs is enough to show $A B \neq B A$.
Check yourself
  1. The worked example computes $A B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ for $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$. What is entry $(1,1)$ of $A B$ built from?
    1. row 1 of $B$ dotted with column 1 of $A$
    2. entry $(1,1)$ of $A$ multiplied by entry $(1,1)$ of $B$ alone
    3. the sum of every entry in $A$
    4. row 1 of $A$ dotted with column 1 of $B$
    Check your answer
    1. row 1 of $B$ dotted with column 1 of $A$ — Row of $B$ dotted with column of $A$ describes $B A$’s rule, the reverse order from the $A B$ entry being asked about here.
    2. entry $(1,1)$ of $A$ multiplied by entry $(1,1)$ of $B$ alone — A single entry-pair multiplied together is only ONE term of the dot product, not the whole sum that builds the entry.
    3. the sum of every entry in $A$ — Summing every entry of $A$ alone leaves $B$ out completely, every product entry needs both matrices.
    4. ✓ row 1 of $A$ dotted with column 1 of $B$ — (D) $A B$’s entry $(1,1)$ is row 1 of $A$, $(1,1)$, dotted with column 1 of $B$ , $(1,1)$, giving $1 \cdot 1 + 1 \cdot 1 = 2$.
  2. The worked example finds $B A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$. Entry $(2,2)$ of $B A$ equals $1 \cdot 1 + 1 \cdot 1 = 2$. Which row and column produced it?
    1. row 1 of $A$, column 1 of $B$
    2. row 2 of $A$, column 2 of $B$
    3. row 2 of $B$, column 2 of $A$
    4. row 1 of $B$, column 2 of $A$
    Check your answer
    1. row 1 of $A$, column 1 of $B$ — Row 1, column 1 gives entry $(1,1)$, not entry $(2,2)$, the position asked about needs row 2 and column 2.
    2. row 2 of $A$, column 2 of $B$ — Row of $A$, column of $B$ is exactly $A B$’s rule, $B A$ needs the roles switched, $B$’s row and $A$’s column.
    3. ✓ row 2 of $B$, column 2 of $A$ — (C) $B A$’s entry $(2,2)$ pairs row 2 of $B$ with column 2 of $A$, the roles of $A$ and $B$ swap compared to $A B$’s rule.
    4. row 1 of $B$, column 2 of $A$ — Row 1 of $B$ pairs with column 1 of $A$, the rule for entry $(1,1)$ of $B A$, not entry $(2,2)$.
  3. The worked example confirms $A B \neq B A$ ‘even though both products are defined and both come out $2 \times 2$.’ Which earlier chapter fact does this directly demonstrate?
    1. the trap that multiplication should commute
    2. the uniqueness of a matrix inverse
    3. the symmetric-plus-skew-symmetric decomposition
    4. the zero-divisor behaviour of matrix multiplication
    Check your answer
    1. ✓ the trap that multiplication should commute — (A) This worked check is the concrete counterexample behind the ‘multiplication should commute’ trap, same order, both defined, still unequal.
    2. the uniqueness of a matrix inverse — Inverse uniqueness is proved with a different argument entirely, not tied to this specific $A$, $B$ pair.
    3. the symmetric-plus-skew-symmetric decomposition — The decomposition uses a different matrix and a different construction, unrelated to this pair.
    4. the zero-divisor behaviour of matrix multiplication — Zero-divisor behaviour is demonstrated with a different pair of matrices later on, not this counterexample.

↑ Back to top

Transpose, symmetric, and skew-symmetric matrices

●KEY-TERM
The same Term 1 marks with subjects down the side and students across. Row 1 on the left is column 1 on the right.

Suppose the office wants the mark-sheet the other way round. Now the subjects run down the side and the students run across the top. Every mark is still there. The second student’s mark in the third subject now sits in row 3, column 2. The rows have become columns, and the columns have become rows.

A 2 by 3 matrix and its 3 by 2 transpose. Blue row 1 becomes column 1, and violet row 2 becomes column 2.

Doing this to a matrix is called taking its TRANSPOSE. The transpose of $A$ is written $A^T$. NCERT and board papers also write it as A′, and both marks mean the same thing. Row 1 of $A$ becomes column 1 of $A^T$, row 2 becomes column 2, and so on. So the entry $a_{i j}$ moves to row $j$, column $i$. If $A$ is $m \times n$, then $A^T$ is $n \times m$.

Let us check one. Take

$A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$, which is $2 \times 3$.

Its first row holds 1, 2 and 3. That row becomes the first column of $A^T$, and row 2 becomes column 2:

$A^T = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix}$, which is $3 \times 2$.

Road distances in km, rounded. Delhi to Jaipur is about 280 and Jaipur to Delhi is the same: the table matches its own mirror.

Now think of the table of road distances printed in an atlas. Each city has a row and a column. The distance from Delhi to Jaipur is the same as the distance from Jaipur to Delhi. So the table reads the same across as down. Transpose it and nothing changes.

Symmetric and skew-symmetric matrices

A matrix like that is SYMMETRIC, which means $A^T = A$. It has to be square, because $A$ and $A^T$ must have the same order. Entry by entry, $a_{i j} = a_{j i}$. Each entry matches its mirror image across the main diagonal. For example, $\begin{pmatrix} 1 & 7 \\ 7 & 4 \end{pmatrix}$ is symmetric.

Symmetric: each entry equals its mirror, 7 and 7. Skew-symmetric: each entry is minus its mirror, 3 and −3, and the diagonal is 0.
A 1 on the diagonal rules this one out: a diagonal entry equal to its own negative has to be 0.

A SKEW-SYMMETRIC matrix is a square matrix with $A^T = -A$. Entry by entry, that says

$a_{i j} = -a_{j i}$ for every $i$ and $j$.

*Put $i = j$* to look at a diagonal entry:

$a_{i i} = -a_{i i}$.

*Add $a_{i i}$ to both sides,* then halve:

$2 a_{i i} = 0$, so $a_{i i} = 0$.

So every diagonal entry is $0$. For example, $\begin{pmatrix} 0 & 3 \\ -3 & 0 \end{pmatrix}$ is skew-symmetric.

KindRuleEntry by entry
Symmetric$A^T = A$$a_{i j} = a_{j i}$
Skew-symmetric$A^T = -A$$a_{i j} = -a_{j i}$, so the diagonal is $0$
Transpose the mark-sheet twice and you have the first sheet again: rows become columns, then columns become rows.
Add, then turn; or turn each, then add. Both give the same matrix: the transpose of a sum is the sum of the transposes.

Properties of the transpose

Four rules hold for every transpose. Board papers ask for them, the last one most of all.

The transpose of A B is Bᵀ times Aᵀ, with the order flipped. Keeping the order, Aᵀ times Bᵀ, gives a different matrix.

Why does the order flip? Say $A$ is $2 \times 3$ and $B$ is $3 \times 4$. Then $A B$ is $2 \times 4$, so $(A B)^T$ is $4 \times 2$. Now $B^T A^T$ is a $4 \times 3$ times a $3 \times 2$, which is $4 \times 2$ and fits. *But $A^T B^T$* is a $3 \times 2$ times a $4 \times 3$, which is not even defined.

Testing the flip

Check it with numbers. Take two small matrices and work out both sides, one line at a time.

Worked example

Check $(A B)^T = B^T A^T$ on two matrices

  1. $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 3 & 1 \end{pmatrix}$
    Take two $2 \times 2$ matrices, so every product below is defined.
  2. $A B = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}$
    Row times column: row 1 of $A$ with each column of $B$, then row 2.
  3. $(A B)^T = \begin{pmatrix} 7 & 3 \\ 2 & 1 \end{pmatrix}$
    Transpose $A B$: its rows become columns.
  4. $B^T = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$, $A^T = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$
    Transpose $B$ and $A$, each on its own.
  5. *$B^T A^T = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 7 & 3 \\ 2 & 1 \end{pmatrix}$*
    Multiply in the flipped order: the answer matches $(A B)^T$.
  6. Check: $A^T B^T = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 2 & 7 \end{pmatrix}$
    Keep the order instead, and the answer is a different matrix.

Your turn: write down $A^T$ for $A = \begin{pmatrix} 2 & -1 \\ 5 & 0 \end{pmatrix}$. Is $A$ symmetric, skew-symmetric, or neither?

Check yourself
  1. If $A$ has order $m \times n$, what is the order of its transpose $A^T$?
    1. $m \times m$, transpose keeps only the row count
    2. $n \times m$, rows and columns swap
    3. $n \times n$, transpose always produces a square matrix
    4. $m \times n$, transpose never changes the order
    Check your answer
    1. $m \times m$, transpose keeps only the row count — The column count matters just as much as the row count, dropping it loses half of what transpose actually does.
    2. ✓ $n \times m$, rows and columns swap — (B) Transpose swaps rows and columns, so the order flips too, $m \times n$ becomes $n \times m$.
    3. $n \times n$, transpose always produces a square matrix — Transpose only produces a square result when $A$ was already square; a non-square $A$ transposes to a different, still non-square, order.
    4. $m \times n$, transpose never changes the order — Transpose always swaps rows and columns, the order changes unless $m = n$ to begin with.
  2. Exactly when is a square matrix $A$ SKEW-SYMMETRIC?
    1. $A^T = -A$
    2. every entry of $A$ is negative
    3. $A$ has no diagonal entries at all
    4. $A^T = A$
    Check your answer
    1. ✓ $A^T = -A$ — (A) Skew-symmetric means the transpose equals the NEGATIVE of $A$, forcing $a_{i j} = -a_{j i}$ throughout.
    2. every entry of $A$ is negative — Nothing about skew-symmetry requires every entry to be negative, off-diagonal entries can be positive too, as long as $a_{i j} = -a_{j i}$ holds.
    3. $A$ has no diagonal entries at all — Every square matrix, skew-symmetric ones included, has a full diagonal, skew-symmetry forces those diagonal entries to be zero, not absent.
    4. $A^T = A$ — $A^T = A$ is SYMMETRIC. Skew-symmetric flips the sign instead: $A^T = -A$.
  3. Why must every diagonal entry of a skew-symmetric matrix equal $0$?
    1. because $i = j$ forces $a_{i i} = -a_{i i}$
    2. because transpose deletes the diagonal entirely
    3. because skew-symmetric matrices are never square
    4. because the definition states it as a separate rule
    Check your answer
    1. ✓ because $i = j$ forces $a_{i i} = -a_{i i}$ — (A) Setting $i = j$ in the skew-symmetric condition forces $a_{i i} = -a_{i i}$, which only $0$ can satisfy.
    2. because transpose deletes the diagonal entirely — Transpose moves entry $(i,j)$ to position $(j,i)$; it never removes an entry, diagonal or otherwise.
    3. because skew-symmetric matrices are never square — Skew-symmetric matrices ARE square, but that is the starting condition the definition assumes, not the reason the diagonal ends up zero.
    4. because the definition states it as a separate rule — The zero-diagonal fact is DERIVED by setting $i = j$ in the definition, it is a consequence, not an extra rule bolted on.
  4. One glance at a matrix’s diagonal can rule out skew-symmetry entirely. Why?
    1. a nonzero diagonal entry means the matrix cannot be square
    2. one nonzero diagonal entry breaks $a_{i i} = -a_{i i}$
    3. the diagonal has nothing to do with symmetric or skew-symmetric matrices at all
    4. skew-symmetric matrices must have every entry, diagonal or not, equal to zero
    Check your answer
    1. a nonzero diagonal entry means the matrix cannot be square — Squareness is a separate property from the diagonal entries, a non-square matrix has no diagonal test to run in the first place, but a nonzero diagonal entry does not by itself say anything about squareness.
    2. ✓ one nonzero diagonal entry breaks $a_{i i} = -a_{i i}$ — (B) Every diagonal entry of a skew-symmetric matrix is forced to $0$, spotting one nonzero entry settles the question immediately.
    3. the diagonal has nothing to do with symmetric or skew-symmetric matrices at all — The diagonal is precisely the shortcut here, every skew-symmetric matrix has a zero diagonal, so checking it is not irrelevant, it is decisive.
    4. skew-symmetric matrices must have every entry, diagonal or not, equal to zero — Skew-symmetry only forces the DIAGONAL to be zero, off-diagonal entries are free to be anything satisfying $a_{i j} = -a_{j i}$.

↑ Back to top

Symmetric plus skew-symmetric

▲CONCEPT
A splits into symmetric P and skew-symmetric Q. The pair 3 and 1 becomes 2 and 2 in P, and 1 and −1 in Q.

Every square matrix is the sum of a symmetric matrix and a skew-symmetric matrix, in exactly one way. For a square matrix $A$, the two parts are these.

A 2 × 3 matrix and its 3 × 2 transpose cannot be added, so A + Aᵀ is not defined. A must be square.

*$A$ must be square.* Otherwise $A$ and $A^T$ have different orders and cannot be added.

Before halving: A + Aᵀ is symmetric and A − Aᵀ is skew-symmetric. Halve each and they add back to A.

The two parts add back to $A$. First put both halves under one $(1/2)$:

$P + Q = (1/2)(A + A^T + A - A^T)$.

*The $A^T$ terms cancel,* and $A + A = 2 A$:

$P + Q = (1/2)(2 A) = A$.

The shape of each part

*Why is $P$ symmetric?* Take the transpose of $P$ and use the transpose rules, one line at a time.

Why it works

$P$ is symmetric and $Q$ is skew-symmetric

  1. $P^T = ((1/2)(A + A^T))^T$
    Take the transpose of both sides of $P = (1/2)(A + A^T)$.
  2. $P^T = (1/2)(A + A^T)^T$
    A number comes out of a transpose: $(k A)^T = k A^T$.
  3. $P^T = (1/2)(A^T + (A^T)^T)$
    The transpose of a sum is the sum of the transposes: $(A + B)^T = A^T + B^T$.
  4. *$P^T = (1/2)(A^T + A) = P$*
    Transposing twice gives back $A$, and the order of a sum does not matter.
  5. $Q^T = (1/2)(A^T - (A^T)^T) = (1/2)(A^T - A)$
    Do the same three steps to $Q = (1/2)(A - A^T)$.
  6. *$Q^T = -(1/2)(A - A^T) = -Q$*
    Take out a minus sign, so $Q$ is skew-symmetric.
A equals P plus Q. Across the dashed diagonal, P’s mirror pairs match and Q’s differ only in sign, with zeros down its diagonal.
Kanada
A SCHOLAR INDIA REMEMBERS

How can you write a square matrix as a sum of two matrices? Write it as $P$ plus $Q$: $P$ mirrors itself across the diagonal, and $Q$ mirrors itself with every sign changed, so its diagonal is all $0$. Is there another way to write it like that? No. The proof below shows that any such pair must be $P$ and $Q$. So the sum is fixed by $A$ itself, not by the person splitting it.

Kanada · the sparrow

Uniqueness of the split

Why only one way? Suppose $A = R + S$, with $R$ symmetric and $S$ skew-symmetric. Here is why $R$ must be $P$ and $S$ must be $Q$:

Why it works

Any such split is $P$ plus $Q$

  1. $A = R + S$
    Start from the split, with $R^T = R$ and $S^T = -S$.
  2. $A^T = R^T + S^T = R - S$
    Transpose both sides, then use $R^T = R$ and $S^T = -S$.
  3. $A + A^T = 2 R$
    Add the two equations: the $S$ terms cancel.
  4. $R = (1/2)(A + A^T) = P$
    Halve both sides: $R$ is the symmetric part.
  5. $A - A^T = 2 S$, so $S = (1/2)(A - A^T) = Q$
    Subtract the two equations instead, then halve.
  6. *$R = P$ and $S = Q$*
    So the split can only be $P$ plus $Q$, and no other split exists.

The split lets you study any square matrix as two simpler pieces. One piece mirrors itself across the diagonal. The other mirrors itself with every sign changed. The worked example below splits a $2 \times 2$ matrix with numbers.

Check yourself
  1. For a square matrix $A$, define $P = (1/2)(A + A^T)$. What property does $P$ have?
    1. $P$ is symmetric
    2. $P$ is always the zero matrix
    3. $P$ equals $A$ exactly
    4. $P$ is skew-symmetric
    Check your answer
    1. ✓ $P$ is symmetric — (A) $P^T = (1/2)(A^T + A) = P$, $P$ is symmetric by construction, since transpose distributes over the sum and $(A^T)^T = A$.
    2. $P$ is always the zero matrix — $P$ is the zero matrix only in the special case where $A$ is already skew-symmetric, not as a general rule.
    3. $P$ equals $A$ exactly — $P = A$ only when $A$ is already symmetric; for a general $A$, $P$ is a genuinely different matrix from $A$.
    4. $P$ is skew-symmetric — $Q$, not $P$, is the piece built to be skew-symmetric, $P$ satisfies the opposite condition, $P^T = P$.
  2. Why does $Q = (1/2)(A - A^T)$ come out skew-symmetric?
    1. because $Q$ is defined to be diagonal
    2. because subtraction always produces a skew-symmetric result
    3. because $Q^T = (1/2)(A^T - A) = -Q$
    4. because $A$ itself must already be skew-symmetric
    Check your answer
    1. because $Q$ is defined to be diagonal — $Q$’s definition says nothing about being diagonal; $Q$ can have nonzero off-diagonal entries, as the worked example shows.
    2. because subtraction always produces a skew-symmetric result — Subtracting two arbitrary matrices does not generally give a skew-symmetric result, the specific pairing $A$ and $A^T$ is what makes $Q$ skew-symmetric.
    3. ✓ because $Q^T = (1/2)(A^T - A) = -Q$ — (C) Transposing $Q$ flips the sign of the subtraction, giving exactly $-Q$, the skew-symmetric condition, derived directly from the definition.
    4. because $A$ itself must already be skew-symmetric — $A$ can be any square matrix at all, the decomposition works whether or not $A$ starts out skew-symmetric.
  3. Every square matrix decomposes into a symmetric part and a skew-symmetric part ‘in exactly one way.’ What does this rule out?
    1. that $A$ itself could already be symmetric
    2. that $P$ could have a nonzero diagonal entry
    3. that $Q$ could ever be the zero matrix
    4. another pair could split $A$ this way
    Check your answer
    1. that $A$ itself could already be symmetric — A symmetric starting $A$ still decomposes uniquely, it just happens to land on $Q = O$, which is fine.
    2. that $P$ could have a nonzero diagonal entry — $P$ is symmetric, and a symmetric matrix can carry any numbers on its diagonal. Only $Q$, the skew-symmetric part, must have a zero diagonal.
    3. that $Q$ could ever be the zero matrix — $Q = O$ happens exactly when $A$ is already symmetric, a normal, expected case, not something the uniqueness claim forbids.
    4. ✓ another pair could split $A$ this way — (D) “Exactly one way” means the specific $P$ and $Q$ built from $A$ and $A^T$ are the ONLY such pair, no alternative symmetric/skew-symmetric split exists.
  4. If a square matrix $A$ is already symmetric, what happens to its decomposition $A = P + Q$?
    1. both $P$ and $Q$ become the zero matrix
    2. the decomposition becomes impossible to define
    3. $Q$ becomes the zero matrix, and $P = A$
    4. $P$ becomes the zero matrix, and $Q = A$
    Check your answer
    1. both $P$ and $Q$ become the zero matrix — Only $Q$ vanishes here; $P$ does not, $P = (1/2)(A+A) = A$, the full matrix, not zero.
    2. the decomposition becomes impossible to define — The decomposition stays perfectly well-defined for a symmetric $A$, it simply lands on $Q = O$, not on “undefined.”
    3. ✓ $Q$ becomes the zero matrix, and $P = A$ — (C) $A^T = A$ makes $Q = (1/2)(A - A^T) = O$, the whole matrix lands in the symmetric part, $P = A$.
    4. $P$ becomes the zero matrix, and $Q = A$ — $A^T = A$ makes $Q = (1/2)(A - A^T) = O$ vanish, not $P$, $P$ instead becomes the whole of $A$.
Worked example

Split $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ into its symmetric and skew-symmetric parts

  1. $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$, so $A^T = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$
    Write down $A^T$ first. Its rows are the columns of $A$.
  2. $P = (1/2)(A + A^T) = (1/2)\begin{pmatrix} 4 & 4 \\ 4 & 8 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix}$
    Add $A$ and $A^T$ entry by entry, then halve.
  3. Check $P^T = \begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix} = P$
    $P$ reads the same across as down, so it is symmetric.
  4. $Q = (1/2)(A - A^T) = (1/2)\begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$
    Subtract $A^T$ from $A$ entry by entry, then halve.
  5. Check $Q^T = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = -Q$, and both diagonal entries of $Q$ are $0$
    $Q$ is skew-symmetric, and its diagonal is all zeros, as it must be.
  6. $P + Q = \begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} = A$
    Check that the two parts add back to $A$. They do, so the split is right.
Check yourself
  1. For $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$, the worked example finds $P = \begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix}$. What check confirms $P$ is symmetric?
    1. checking that every entry of $P$ is positive
    2. checking that $P^T = P$
    3. checking that $P + P = A$
    4. checking that $P$ has no zero entries
    Check your answer
    1. checking that every entry of $P$ is positive — Positive entries have nothing to do with symmetry, a symmetric matrix can have negative, zero, or positive entries.
    2. ✓ checking that $P^T = P$ — (B) Symmetric means $P^T = P$, swap rows and columns of $\begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix}$ and it reads back identically.
    3. checking that $P + P = A$ — $P + P = A$ would mean $2P = A$, a different and unrelated statement from the symmetry test $P^T = P$.
    4. checking that $P$ has no zero entries — Whether $P$ has zero entries is unrelated to symmetry, $P^T = P$ is the actual test.
  2. The worked example computes $Q = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ for $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$. Both diagonal entries of $Q$ are $0$. Why is this expected?
    1. because $Q$ was built by simply deleting $A$’s diagonal
    2. because $A$ itself has a zero diagonal
    3. it is a coincidence specific to this particular $A$
    4. every skew-symmetric matrix has a zero diagonal
    Check your answer
    1. because $Q$ was built by simply deleting $A$’s diagonal — $Q$ is built from $A - A^T$, entry by entry, not by deleting $A$’s diagonal, the zero diagonal is a CONSEQUENCE of the subtraction, not a construction step.
    2. because $A$ itself has a zero diagonal — $A$’s diagonal is $2, 4$, nonzero. It is $Q$, not $A$, whose diagonal is forced to zero.
    3. it is a coincidence specific to this particular $A$ — This is not a coincidence of this one example, every skew-symmetric matrix has an all-zero diagonal, by the same derivation as before.
    4. ✓ every skew-symmetric matrix has a zero diagonal — (D) $Q$ is skew-symmetric by construction, and every skew-symmetric matrix forces $a_{i i} = -a_{i i} = 0$ on its diagonal, not a coincidence of this example.
  3. The worked example confirms $P + Q = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} = A$ at the end. What does this final check verify?
    1. the split is exact, nothing lost or added
    2. that $A$ is already symmetric
    3. that the compatibility condition for multiplication holds
    4. that $A$ is invertible
    Check your answer
    1. ✓ the split is exact, nothing lost or added — (A) Adding $P$ and $Q$ back together and landing exactly on $A$ confirms the split loses nothing, the two pieces reconstruct the original matrix precisely.
    2. that $A$ is already symmetric — $A$ here is $\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$, which is NOT symmetric ($A^T \neq A$), that is exactly why both $P$ and $Q$ turn out nonzero.
    3. that the compatibility condition for multiplication holds — Compatibility gates whether a PRODUCT is defined; $P + Q$ is a sum of two same-order matrices, always defined.
    4. that $A$ is invertible — Invertibility is an unrelated question, this final step checks reconstruction, not whether $A$ has an inverse.

↑ Back to top

Invertible matrices

▲CONCEPT
Multiplying by I changes nothing, from either side, the way multiplying a number by 1 does.
5
→
× 4
→
× 1/4
undoes × 4
→
5
Chapter 1’s undo with numbers: times 4 turns 5 into 20, and times one quarter turns 20 into 5 again.

*The identity matrix $I$* is the matrix that changes nothing. Multiplying by $I$ is like multiplying by 1, so $A I = I A = A$. A square matrix $A$ is INVERTIBLE when some matrix $B$ of the same order undoes it, so that $A B = B A = I$. That $B$ is the INVERSE of $A$, and it is written $A^{-1}$. It is the same undo as the inverse of a function in Chapter 1.

B undoes A from both sides: A B and B A are both I. The definition asks for both products.
A moves the unit square: (1, 0) stays, (0, 1) goes to (1, 1). B moves it back. In either order the square returns.

A matrix moves a point: write the point as a column and multiply, the way the photo-editing matrices later in this chapter move a picture. Read the figure with the columns. $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ sends $(1, 0)$ to its first column, $(1, 0)$, so that corner stays. It sends $(0, 1)$ to its second column, $(1, 1)$, so the square leans right. $B = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$ sends $(1, 1)$ back to $(0, 1)$. Chapter 4 uses this picture again.

The definition asks for both products, $A B$ and $B A$, not just one. And only a square matrix can be invertible. If $A$ is $m \times n$ and $B$ is $n \times m$, then $A B$ is $m \times m$ and $B A$ is $n \times n$. Both can equal the same $I$ only when $m = n$.

Checking an inverse

Check with numbers. Take

$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$.

Multiply them, row times column:

$A B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$.

Now the other order:

$B A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$.

*Both products are $I$,* so $B$ undoes $A$, and $A^{-1} = B$.

This matrix has no inverse. Whatever it multiplies, the two rows of the answer come out equal, so I never appears.

Not every square matrix has an inverse. The zero matrix $O$ has none, because $O B = O$ for every $B$, never $I$. Chapter 4 calls a matrix like that SINGULAR. It also gives a number, the determinant, that tells you which matrices have an inverse.

B
→
× I
→
I = A C
C is the inverse of A
→
regroup
grouping rule
→
(B A) C
(B A) C
→
B A = I
B is the inverse of A
→
drop I
→
C
Why there is only one inverse: each step rewrites the matrix without changing it, so the B you started with equals C.

An invertible matrix has only one inverse. Let us see why. Suppose $B$ and $C$ are both inverses of $A$, so $A C = I$ and $B A = I$. Start from $B$ and change it one step at a time:

Why it works

Two inverses of $A$ are one matrix

  1. $B = B I$
    Multiplying by $I$ changes nothing.
  2. $B I = B (A C)$
    Write $I$ as $A C$, because $C$ is the inverse of $A$.
  3. $B (A C) = (B A) C$
    Regroup the product: grouping holds for matrices.
  4. $(B A) C = I C$
    Write $B A$ as $I$, because $B$ is the inverse of $A$.
  5. *$I C = C$, so $B = C$*
    Multiplying by $I$ changes nothing, so the chain that began at $B$ ends at $C$.

When one inverse is found

That helps in practice. If you find a $B$ with $A B = B A = I$, you have the inverse, and there is no other one to look for. Working out $A^{-1}$ for a given matrix is the next chapter’s job.

Your turn: is $B = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$ the inverse of $A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$? Check both products.

The inverse of a product

A product has an inverse too. Take invertible matrices $A$ and $B$ of the same order. Then $A B$ is invertible, and its inverse takes the two inverses in the opposite order: $(A B)^{-1} = B^{-1} A^{-1}$. To see why, multiply $A B$ by $B^{-1} A^{-1}$, once from each side.

Why it works

Why $(A B)^{-1} = B^{-1} A^{-1}$

  1. $(A B)(B^{-1} A^{-1}) = A (B B^{-1}) A^{-1} = A I A^{-1} = I$
    Regroup the brackets; the middle $B B^{-1}$ is $I$, and $I$ changes nothing.
  2. *$(B^{-1} A^{-1})(A B) = B^{-1} (A^{-1} A) B = B^{-1} I B = I$*
    The other order gives $I$ in the same way. Both products are $I$, so $B^{-1} A^{-1}$ is the inverse of $A B$.
Check yourself
  1. When is a square matrix $A$ called INVERTIBLE?
    1. some matrix $B$ satisfies $A B = I$ alone
    2. some $B$ satisfies $A B = B A = I$
    3. $A$ equals its own transpose
    4. $A$ has no zero entries anywhere
    Check your answer
    1. some matrix $B$ satisfies $A B = I$ alone — This definition requires BOTH $A B = I$ and $B A = I$, one-sided equality alone is not the stated condition.
    2. ✓ some $B$ satisfies $A B = B A = I$ — (B) Invertibility needs BOTH products to give $I$, $A B = I$ alone is not the definition used here.
    3. $A$ equals its own transpose — $A = A^T$ is the definition of symmetric, an unrelated property from having an inverse.
    4. $A$ has no zero entries anywhere — Having zero entries or not says nothing about invertibility, plenty of invertible matrices have zero entries, and plenty without any are singular.
  2. If $B$ and $C$ both satisfy $A B = B A = I$ and $A C = C A = I$, it follows that $B = B(A C) = (B A)C = C$. What does this show?
    1. that every square matrix has exactly two inverses
    2. that $B$ and $C$ must both equal the identity matrix
    3. that $A$ must equal both $B$ and $C$
    4. that an inverse, when it exists, is unique
    Check your answer
    1. that every square matrix has exactly two inverses — The derivation shows there can be AT MOST one such matrix, not that there are always exactly two , “exactly two” is the opposite conclusion.
    2. that $B$ and $C$ must both equal the identity matrix — $B$ and $C$ end up equal to EACH OTHER through the chain, nothing forces either one to equal $I$ itself.
    3. that $A$ must equal both $B$ and $C$ — The chain concludes $B = C$, saying nothing about $A$ equaling either inverse.
    4. ✓ that an inverse, when it exists, is unique — (D) The chain of equalities forces $B = C$, whenever an inverse exists, it is the ONLY matrix satisfying the two-sided condition.
  3. A matrix that is not invertible is called SINGULAR. What is true of a singular matrix’s inverse?
    1. it exists, but only using the adjoint method from Chapter 4
    2. it exists but is simply hard to compute
    3. it exists only after row-reducing the matrix
    4. it does not exist at all
    Check your answer
    1. it exists, but only using the adjoint method from Chapter 4 — The whole point of “singular” is that NO method, adjoint included, can produce an inverse that does not exist.
    2. it exists but is simply hard to compute — Singular does not mean “hard to compute”, it means no inverse exists at all, however hard one looked.
    3. it exists only after row-reducing the matrix — Nothing about row-reduction is mentioned in this chapter’s definition of singular.
    4. ✓ it does not exist at all — (D) Singular means no inverse exists, period, not “exists but is hard to find.”
  4. This chapter states only the definition of invertibility and the uniqueness fact. What is explicitly left for Chapter 4?
    1. the fact that an inverse, when it exists, is unique
    2. computing $A^{-1}$ via the adjoint
    3. the definition of matrix multiplication itself
    4. the definition of what a square matrix is
    Check your answer
    1. the fact that an inverse, when it exists, is unique — Uniqueness IS proved here, in this chapter, using the $B = B(A C) = (B A)C = C$ chain, nothing about it is deferred.
    2. ✓ computing $A^{-1}$ via the adjoint — (B) This chapter only asks WHETHER an inverse exists; the adjoint method for actually BUILDING one waits for Chapter 4.
    3. the definition of matrix multiplication itself — Matrix multiplication is one of the very first ideas this chapter builds, it is not deferred to Chapter 4.
    4. the definition of what a square matrix is — “Square matrix” is defined earlier in this same chapter, nothing about that definition is pushed to Chapter 4.

↑ Back to top

A B = O does not force A = O or B = O

✕MISCONCEPTION
(x − 2)(x − 3) = 0
→
one factor is 0
→
solve
→
x = 2 or x = 3
The Class 10 step: a product is 0, so one factor is 0, so x is 2 or 3.

In Class 10 you solved quadratic equations by factorising. When a product of two brackets was zero, you said one of the brackets must be zero. That step works because two numbers that are not zero never multiply to zero. You have used it so often that it feels like a law of arithmetic. For matrices it is false.

The step went like this. A product of two brackets was zero:

$(x - 2)(x - 3) = 0$.

So one bracket is zero:

$x - 2 = 0$ or $x - 3 = 0$.

Solve each equation:

$x = 2$ or $x = 3$.

Neither A nor B has a zero entry, yet every entry of A B is 0. In the top left, 1 + (−1) = 0.
B flattens the unit square onto a line through (1, minus 1). A then takes that whole line to the point (0, 0).

Now take these two matrices:

$A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$.

Neither one is the zero matrix. Every entry of each is $1$ or $-1$. Yet row times column gives

$A B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O$.

Each entry is $1 - 1$. Two matrices that are not zero can multiply to the zero matrix.

A B and A O both come out O, so A B = A O. Yet B is not O: you cannot cancel A here.

*So from $A B = O$* you cannot say that $A = O$ or $B = O$.

A second habit fails with it. From $A B = A C$ you cannot cancel $A$ to get $B = C$. Here is a case:

$A B = O$ and $A O = O$.

*Both products are $O$,* so

$A B = A O$, and still $B \neq O$.

*For a square $A$,* you may cancel $A$ only when $A$ has an inverse. Here is the cancelling, one line at a time:

Why it works

Cancel $A$ when $A^{-1}$ exists

  1. $A B = A C$
    Start from two products with the same left factor $A$.
  2. $A^{-1} (A B) = A^{-1} (A C)$
    Multiply both sides on the left by $A^{-1}$, the same side on both.
  3. $(A^{-1} A) B = (A^{-1} A) C$
    Regroup: grouping holds for matrices.
  4. $I B = I C$
    Replace $A^{-1} A$ by $I$, because $A^{-1}$ is the inverse of $A$.
  5. *$B = C$*
    Multiplying by $I$ changes nothing.

Reading a zero product

When a product of matrices is zero, look at the factors themselves. The product alone does not tell you that either factor is zero.

Your turn: with the same $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, work out $A C$ for $C = \begin{pmatrix} 2 & -3 \\ -2 & 3 \end{pmatrix}$. What does your answer say about cancelling $A$?

Check yourself
  1. TRAP: ‘numbers that multiply to zero force one factor to be zero, so $A B = O$ should force $A = O$ or $B = O$ too.’ Is this trap correct for matrices?
    1. yes: the same rule that holds for numbers holds for matrices
    2. yes, but only when $A$ and $B$ are both square
    3. no: matrices can be zero divisors
    4. no, because $A B = O$ is actually impossible for nonzero matrices
    Check your answer
    1. yes: the same rule that holds for numbers holds for matrices — The number-arithmetic rule does NOT carry over, matrices can be zero divisors, exactly what the trap denies.
    2. yes, but only when $A$ and $B$ are both square — The counterexample’s matrices are both $2 \times 2$, both square, squareness does not rescue the trap.
    3. ✓ no: matrices can be zero divisors — (C) Matrices can be ZERO DIVISORS, a behaviour ordinary numbers never show. $A B = O$ says nothing about the factors individually.
    4. no, because $A B = O$ is actually impossible for nonzero matrices — $A B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ for this nonzero $A$, $B$, the ‘impossible’ claim is directly contradicted.
  2. For $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$, both nonzero, it is given that $A B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$. What does this pair confirm?
    1. that $A$ and $B$ must both be symmetric
    2. that $A B = B A$ for this pair
    3. that $A$ and $B$ must be inverses of each other
    4. a zero product does not force a zero factor
    Check your answer
    1. that $A$ and $B$ must both be symmetric — Symmetry is never checked in this example; the point is entirely about the zero product, not about either matrix’s own shape.
    2. that $A B = B A$ for this pair — $B A = \begin{pmatrix} 2 & 2 \\ -2 & -2 \end{pmatrix}$, which is not $O$. So $A B \neq B A$ for this pair.
    3. that $A$ and $B$ must be inverses of each other — $A B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$, not the identity, this pair is not a pair of inverses.
    4. ✓ a zero product does not force a zero factor — (D) Two matrices packed with nonzero entries can still multiply out to zero, the zero product tells you nothing about the factors themselves.
  3. Given $A B = O$, which statement is SAFE?
    1. $A = O$ or $B = O$ need not hold
    2. neither $A$ nor $B$ can be invertible
    3. at least one of $A$, $B$ must equal the zero matrix
    4. $A$ and $B$ must have the same order as each other
    Check your answer
    1. ✓ $A = O$ or $B = O$ need not hold — (A) A zero product does not force a zero factor. The pair above has no zero entry at all, and still $A B = O$.
    2. neither $A$ nor $B$ can be invertible — $A = I$ and $B = O$ give $A B = O$, and $I$ is invertible. So a zero product does not stop a factor from being invertible.
    3. at least one of $A$, $B$ must equal the zero matrix — Forcing a zero factor is exactly the plain-number habit the counterexample refutes, it is not a safe conclusion for matrices.
    4. $A$ and $B$ must have the same order as each other — $A B$ needs only the inner numbers to match. A $2 \times 3$ matrix times a $3 \times 2$ matrix can give $O$, and those two orders differ.
  4. A student sees $A B = O$ and concludes $A$ must be singular. Using the zero-divisor fact above, is this reasoning safe in general?
    1. yes, but only when $B$ is also singular
    2. no, because $A B = O$ is impossible unless $A$ is singular
    3. no: this alone does not decide whether $A$ is singular
    4. yes: a zero product always means the left factor is singular
    Check your answer
    1. yes, but only when $B$ is also singular — Nothing about $B$’s own singularity is required for $A B = O$ with nonzero $A$, $B$, the counterexample above needs no such condition.
    2. no, because $A B = O$ is impossible unless $A$ is singular — $A B = O$ with neither $A$ nor $B$ singular is exactly the zero-divisor behaviour this chapter demonstrates , it is not impossible.
    3. ✓ no: this alone does not decide whether $A$ is singular — (C) The zero-divisor fact only says the PRODUCT can be zero with nonzero factors, it does not, on its own, pin the blame on $A$ over $B$, or force either to be singular.
    4. yes: a zero product always means the left factor is singular — A zero product never “always” points to $A$ specifically, the blame, if any, cannot be assigned without checking $A$ directly.
Worked example

Compute $A B$ for $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$

  1. $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$: neither one the zero matrix
    Neither matrix is $O$. Work out $A B$ entry by entry.
  2. Entry $(1,1) = 1 \cdot 1 + 1 \cdot (-1) = 1 - 1 = 0$
    Row 1 of $A$ times column 1 of $B$.
  3. Entry $(1,2) = 1 \cdot 1 + 1 \cdot (-1) = 0$
    Row 1 of $A$ times column 2 of $B$.
  4. Entry $(2,1) = 1 \cdot 1 + 1 \cdot (-1) = 0$
    Row 2 of $A$ times column 1 of $B$.
  5. Entry $(2,2) = 1 \cdot 1 + 1 \cdot (-1) = 0$
    Row 2 of $A$ times column 2 of $B$.
  6. *$A B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$*
    Every entry is $0$, so $A B = O$. Yet no entry of $A$ or $B$ is $0$.
  7. $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 - 1 \\ 1 - 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
    Check with one column. Both columns of $B$ are $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$, so this one product checks all four entries.
Check yourself
  1. The worked example computes entry $(1,1)$ of $A B$, for $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$, as $1 \cdot 1 + 1 \cdot (-1) = 0$. Which row and column produced it?
    1. the full matrix $A$ summed against the full matrix $B$
    2. row 1 of $A$ and column 1 of $B$
    3. only the first entries of $A$ and $B$, ignoring the rest of the row and column
    4. row 1 of $B$ and column 1 of $A$
    Check your answer
    1. the full matrix $A$ summed against the full matrix $B$ — Only ONE row of $A$ and ONE column of $B$ are needed for a single entry, the rest of both matrices plays no part in it.
    2. ✓ row 1 of $A$ and column 1 of $B$ — (B) Entry $(1,1)$ follows the usual rule, row 1 of $A$ dotted with column 1 of $B$, giving $1 \cdot 1 + 1 \cdot (-1) = 0$.
    3. only the first entries of $A$ and $B$, ignoring the rest of the row and column — The full row and full column both matter, using only the first entries of each drops most of the dot product.
    4. row 1 of $B$ and column 1 of $A$ — Row of $B$, column of $A$ describes $B A$’s rule for this position, not $A B$’s.
  2. The worked example finds every entry of $A B$ equal to $0$, for $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$. What does this confirm about $A$ and $B$ themselves?
    1. both have every entry nonzero, yet multiply to zero
    2. that $A$ and $B$ are secretly the same matrix
    3. that one of $A$, $B$ must actually be the zero matrix in disguise
    4. that $A$ and $B$ cannot both be $2 \times 2$
    Check your answer
    1. ✓ both have every entry nonzero, yet multiply to zero — (A) Every entry of $A$ and every entry of $B$ is nonzero, and the product is still the zero matrix, the zero-divisor behaviour in full.
    2. that $A$ and $B$ are secretly the same matrix — $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$ differ in their second row, they are not the same matrix.
    3. that one of $A$, $B$ must actually be the zero matrix in disguise — Every entry of $A$ and $B$ is $1$ or $-1$, never $0$, neither is secretly the zero matrix.
    4. that $A$ and $B$ cannot both be $2 \times 2$ — Both matrices are explicitly given at order $2 \times 2$, nothing about this result changes that.
  3. Take $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$ as before, but now consider $B A$ instead of $A B$. Without computing it fully, what can you say?
    1. $B A$ must equal $A B$, since both are zero
    2. $B A$ is undefined, since $A B = O$ already
    3. $B A$ must be checked on its own
    4. $B A$ must also be nonzero, to balance $A B = O$
    Check your answer
    1. $B A$ must equal $A B$, since both are zero — Nothing established that $B A$ must equal $A B$, that is exactly the assumption the commutativity warnings above tell you never to make.
    2. $B A$ is undefined, since $A B = O$ already — $A$ and $B$ are both $2 \times 2$, so $B A$ is defined regardless of what $A B$ came out to.
    3. ✓ $B A$ must be checked on its own — (C) $A B = O$ is a fact about that one order alone, $B A$ is a different computation entirely and has to be checked on its own.
    4. $B A$ must also be nonzero, to balance $A B = O$ — There is no “balancing” requirement in matrix arithmetic, $B A$’s value depends only on $B$ and $A$’s entries, not on what $A B$ happened to equal.

↑ Back to top

One object, its own arithmetic

■RECAP
Matrices Order, entries, and named matrix shapes + Adding matrices and scaling by a number Row times column AB ≠ BA What survives, what doesn’t Transpose, symmetric, and skew- symmetric matrices A = P + Q Symmetric plus skew- symmetric A A−1 = I Invertible matrices
Eight cards, read left to right. Order, entries and named shapes come first, and three arrows meet at invertible matrices.

The question at the start was this. A mark-sheet, a price list and a bill are grids of numbers. If you treat a whole grid as one thing, which of your arithmetic still works? And what changes?

works as with numbers
A + B = B + Ak(A + B) = kA + kB(A B) C = A (B C)A (B + C) = A B + A C
breaks
A B = B AA B = O gives A = O or B = O
What carried over from numbers, and what did not. Adding, scaling and grouping work as before; swapping (the commutative law) and the zero-product step fail.

Here is the answer. Adding and scaling work box by box, as with numbers, and every habit survives. Multiplying is row times column, and grouping and spreading still hold. Two habits from numbers break. The order of the factors in a product matters. And a zero product does not need a zero factor. Keep those two in mind, and the rest is the arithmetic you already know.

Check yourself
  1. This recap line says multiplying is ‘row times column, when the inner numbers match.’ Which earlier fact does this restate?
    1. the zero-divisor behaviour of matrix multiplication
    2. the compatibility condition gating $A B$
    3. the symmetric-plus-skew-symmetric decomposition
    4. the uniqueness of a matrix inverse
    Check your answer
    1. the zero-divisor behaviour of matrix multiplication — Zero divisors concern what a defined product can equal, compatibility concerns whether the product is defined AT ALL, a prior question.
    2. ✓ the compatibility condition gating $A B$ — (B) This is exactly the compatibility rule from the multiplication section, restated here in different words.
    3. the symmetric-plus-skew-symmetric decomposition — The decomposition splits a matrix into symmetric and skew-symmetric parts, unrelated to whether a product is even defined.
    4. the uniqueness of a matrix inverse — Inverse uniqueness is a different fact, about how many matrices can satisfy $A B=B A=I$, not about column and row counts matching.
  2. This recap line says ‘a zero product does not need a zero factor.’ Which worked example backs this up with a concrete pair?
    1. the worked example decomposing $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$
    2. the worked example using $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$
    3. the worked example computing $A B$ for $2 \times 3$ and $3 \times 2$ matrices
    4. the worked example using $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, $B = \begin{pmatrix} 5 & -1 \\ 0 & 2 \end{pmatrix}$
    Check your answer
    1. the worked example decomposing $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ — That worked example splits a matrix into symmetric and skew-symmetric parts, unrelated to a zero product.
    2. ✓ the worked example using $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$ — (B) That pair, both nonzero, multiplies out to the zero matrix, the exact concrete evidence behind that warning.
    3. the worked example computing $A B$ for $2 \times 3$ and $3 \times 2$ matrices — That worked example multiplies a $2 \times 3$ by a $3 \times 2$ matrix to a nonzero result, it does not touch the zero-divisor case.
    4. the worked example using $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, $B = \begin{pmatrix} 5 & -1 \\ 0 & 2 \end{pmatrix}$ — That worked example computes $A+B$ and $2A-B$, addition and scaling, with no product involved at all.
  3. Identity, zero, symmetric and skew-symmetric matrices each have a shape you can see. Which fact uses a shape to save a check?
    1. a nonzero diagonal entry rules out skew-symmetric at a glance
    2. the compatibility condition for $A B$ to exist
    3. the grouping rule $(A B) C = A (B C)$ for products
    4. the definition of matrix equality requiring matching order and entries
    Check your answer
    1. ✓ a nonzero diagonal entry rules out skew-symmetric at a glance — (A) A skew-symmetric matrix must have a zero diagonal. So one nonzero diagonal entry rules it out at a glance, with no entry-by-entry check.
    2. the compatibility condition for $A B$ to exist — Compatibility applies to any two matrices, named shape or not, it is a general rule, not a shape-specific shortcut.
    3. the grouping rule $(A B) C = A (B C)$ for products — Grouping a product either way gives the same answer for any matrices that can be multiplied. It is a general rule and uses no shape.
    4. the definition of matrix equality requiring matching order and entries — Matrix equality is defined the same way for every matrix, it is not a shortcut that comes from a named shape.
Check yourself: the whole chapter
  1. A matrix packages many numbers into one object. That object gets its own arithmetic. Which statement about it is true?
    1. Adding, scaling and multiplying all commute. A matrix acts just like a number.
    2. None of the three commute. A matrix acts nothing like a number.
    3. Adding does not commute. Scaling and multiplying do.
    4. Adding and scaling commute. Multiplying does not, in general.
    Check your answer
    1. Adding, scaling and multiplying all commute. A matrix acts just like a number. — Adding and scaling really do behave like ordinary arithmetic. Multiplying is the exception , order changes the answer.
    2. None of the three commute. A matrix acts nothing like a number. — Adding and scaling are entrywise. They behave exactly like ordinary numbers. Only multiplying breaks the pattern.
    3. Adding does not commute. Scaling and multiplying do. — Adding is entrywise. Order never matters there: $A + B = B + A$. It is multiplying that fails to commute.
    4. ✓ Adding and scaling commute. Multiplying does not, in general. — (D) Adding and scaling carry over unchanged from ordinary arithmetic. Multiplying is the one operation that does not commute in general.
  2. Matrix $A$ and matrix $B$ both have order $2 \times 2$. Every number in $A$ also appears in $B$, and vice versa, but arranged in different positions. Are $A$ and $B$ equal?
    1. Yes: matching order and having the same numbers overall is enough, whatever their positions.
    2. No: equal matrices need the same order AND every corresponding entry to match.
    3. Yes: matching order is enough by itself; once the orders agree, the entries do not need checking.
    4. No: equal matrices must also both be square.
    Check your answer
    1. Yes: matching order and having the same numbers overall is enough, whatever their positions. — Position matters. $a_{i j}$ must equal $b_{i j}$ at every single spot, not just appear somewhere in the grid.
    2. ✓ No: equal matrices need the same order AND every corresponding entry to match. — (B) Equality needs both tests: the same order, and every entry matching in its own position, $a_{i j} = b_{i j}$ for every $i$, $j$.
    3. Yes: matching order is enough by itself; once the orders agree, the entries do not need checking. — Matching order only confirms the two matrices are comparable. Equality still needs every entry to match, position by position.
    4. No: equal matrices must also both be square. — Squareness plays no part in the definition. Even a $2 \times 3$ matrix can equal another $2 \times 3$ matrix, equality never asks for a square shape.
  3. A $3 \times 3$ matrix has every off-diagonal entry $0$, and its diagonal entries are $5, 5, 5$. Which named types does it belong to?
    1. Diagonal only: since the diagonal entries are not $1$, it cannot also be scalar.
    2. Scalar and identity both, since every diagonal entry is the same number.
    3. Diagonal and scalar, but not identity.
    4. None of the three: a matrix cannot be diagonal and scalar at the same time.
    Check your answer
    1. Diagonal only: since the diagonal entries are not $1$, it cannot also be scalar. — Scalar only needs the diagonal entries to equal EACH OTHER, at any value. Identity is the stricter case where that shared value is $1$.
    2. Scalar and identity both, since every diagonal entry is the same number. — Equal diagonal entries make it scalar, not automatically identity. Identity is the one special case where that shared value is exactly $1$.
    3. ✓ Diagonal and scalar, but not identity. — (C) Off-diagonal entries are all $0$, so it is diagonal. Diagonal entries are all equal ($5$), so it is also scalar. They are not $1$, so it is not identity.
    4. None of the three: a matrix cannot be diagonal and scalar at the same time. — The three types nest inside each other: every identity is scalar, and every scalar is diagonal. A matrix can carry all three labels at once.
  4. A skew-symmetric matrix $A$ has order $3 \times 3$. What must its diagonal entries be?
    1. All zero: skew-symmetry forces each one to equal its own negative.
    2. All equal to each other, though not necessarily zero.
    3. All equal to $1$: skew-symmetric matrices are a variant of the identity matrix.
    4. Unconstrained: the skew-symmetric condition only limits entries off the diagonal.
    Check your answer
    1. ✓ All zero: skew-symmetry forces each one to equal its own negative. — (A) Setting $i = j$ inside $a_{i j} = -a_{j i}$ gives $a_{i i} = -a_{i i}$. The only number equal to its own negative is $0$, so every diagonal entry must vanish.
    2. All equal to each other, though not necessarily zero. — The diagonal entries are not just equal to each other, each one equals its own negative, $a_{i i} = -a_{i i}$, which only $0$ satisfies.
    3. All equal to $1$: skew-symmetric matrices are a variant of the identity matrix. — Skew-symmetric and identity are unrelated named types. Skew-symmetric is defined by $A^T = -A$; identity by every diagonal entry being $1$ and every off-diagonal entry $0$.
    4. Unconstrained: the skew-symmetric condition only limits entries off the diagonal. — The condition $a_{i j} = -a_{j i}$ holds for every $i$ and $j$, diagonal positions included. Setting $i = j$ gives $a_{i i} = -a_{i i}$, forcing that entry to $0$.
  5. Matrix $A$ has order $2 \times 3$ and matrix $B$ has order $3 \times 2$. Can $A + B$ be computed?
    1. Yes: as long as both matrices have the same number of entries in total.
    2. Yes: pair up entries in the order they are written, regardless of row and column position.
    3. Yes, by first taking the transpose of one matrix so the orders match.
    4. No: addition needs both matrices to already share the same order.
    Check your answer
    1. Yes: as long as both matrices have the same number of entries in total. — Order means rows AND columns, not just total entry count. $2 \times 3$ and $3 \times 2$ both have six entries, but different shapes, so they still cannot be added.
    2. Yes: pair up entries in the order they are written, regardless of row and column position. — Addition is entrywise BY POSITION, row $i$, column $j$ of one matrix with row $i$, column $j$ of the other. There is no row $3$ in a $2 \times 3$ matrix to pair up.
    3. Yes, by first taking the transpose of one matrix so the orders match. — Taking a transpose changes which matrix you are adding. $A + B^T$ might be defined, but that is a different sum from $A + B$, which stays undefined.
    4. ✓ No: addition needs both matrices to already share the same order. — (D) Addition is defined only when the two matrices already share the same order, rows and columns both. $2 \times 3$ and $3 \times 2$ do not match, so $A + B$ is undefined.

↑ Back to top

Where you will meet this

You have already read a grid of numbers like this, long before you met the word matrix.

Your turn: A tuck shop sells samosas at $8$ rupees and tea at $6$ rupees. On one day it sells $15$ samosas and $20$ teas. Write this as a quantities row times a prices column, and find the day’s total sales. Answer: $15 \times 8 + 20 \times 6 = 120 + 120 = 240$ rupees.

LATER

Where the grid goes next

First-year university and engineering linear algebra generalises a matrix into a LINEAR TRANSFORMATION, a rule that moves points the way a matrix moves the unit square. It is the basis of computer graphics (moving pixels), MARKOV CHAINS (how a population shifts between states over time) and INPUT-OUTPUT ECONOMIC MODELS (how one industry’s output feeds another’s), not just the single production table evaluated earlier in this chapter, “The same idea, on a firm’s production table.”

↑ Back to top

Practice set: Exercise 3.1

Exercise 3.1 is about reading a matrix. You find its order and its entries, and you build a matrix from a rule for $a_{i j}$. You also name the types, and you use equality to find unknown entries.

Exercise 3.1: Order, type and equality of matrices
  1. practice For the matrix $A = \begin{pmatrix} 2 & -5 & 7 & 0 \\ -1 & 3 & 4 & 6 \\ 8 & 2 & -3 & 9 \end{pmatrix}$, state its order and the total number of elements it has.
  2. practice For the matrix $A = \begin{pmatrix} 4 & -2 & 7 \\ 0 & 5 & -3 \\ 9 & 1 & 6 \end{pmatrix}$, write the entries $a_{1 2}$, $a_{3 1}$ and $a_{2 3}$.
  3. practice If a matrix has $20$ elements, what are the possible orders it can have? What if it has $7$ elements instead?
  4. practice Construct the $2 \times 2$ matrix $A = [a_{i j}]$ whose entries are given by the rule $a_{i j} = 2 i - j$.
  5. practice Construct the $3 \times 2$ matrix $A = [a_{i j}]$ whose entries are given by the rule $a_{i j} = i^2 - j$.
  6. practice Name the type of each matrix: (i) $\begin{pmatrix} 3 & -1 & 2 \end{pmatrix}$, (ii) $\begin{pmatrix} 4 \\ 0 \\ -2 \end{pmatrix}$, (iii) $\begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}$, (iv) $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$.
  7. practice Find $x$ and $y$ if $\begin{pmatrix} x+2 & 3 \\ 1 & y-1 \end{pmatrix} = \begin{pmatrix} 5 & 3 \\ 1 & 4 \end{pmatrix}$.
  8. practice Find $x$, $y$ and $z$ if $\begin{pmatrix} x+y & x-y \\ 2 & z \end{pmatrix} = \begin{pmatrix} 8 & 2 \\ 2 & 5 \end{pmatrix}$.
  9. practice A matrix $A = [a_{i j}]_{m \times n}$ is a square matrix precisely when
    1. $m < n$
    2. $m > n$
    3. $m = n$
    4. none of these
  10. practice Which values of $x$ and $y$ make $\begin{pmatrix} 2x+1 & 4 \\ 0 & y-3 \end{pmatrix} = \begin{pmatrix} 7 & 4 \\ 0 & 2 \end{pmatrix}$ true?
    1. $x = 3$, $y = 5$
    2. $x = 4$, $y = 5$
    3. $x = 3$, $y = 2$
    4. not possible
  11. practice How many $2 \times 2$ matrices are possible if every entry must come from $\{1, 2, 3\}$?
    1. $12$
    2. $16$
    3. $64$
    4. $81$
  12. practice Table 3.1 (NCERT Introductory Microeconomics, Class 12) gives a firm’s output for each labour-capital pair. Its four entries for labour $L=3,4$ and capital $K=3,4$ are these. $L=3,K=3$ gives $30$; $L=3,K=4$ gives $40$; $L=4,K=3$ gives $40$; $L=4,K=4$ gives $50$. Write the $2 \times 2$ matrix of outputs, with labour down the rows and capital across, and state its order.
  13. practice Build the $2 \times 2$ output matrix $N = [n_{i j}]$ where $n_{i j}$ is Table 3.1’s output for labour $L=i+4$ and capital $K=j$, for $i, j \in \{1, 2\}$. The entries you need are these. $L=5,K=1$ gives $12$; $L=5,K=2$ gives $29$; $L=6,K=1$ gives $13$; $L=6,K=2$ gives $33$. Write $n_{1 2}$ and $n_{2 1}$.
Answers
  1. $12$ elements: order $3 \times 4$.
  2. $a_{1 2} = -2$, $a_{3 1} = 9$, $a_{2 3} = -3$.
  3. $20$ elements: $1 \times 20$, $2 \times 10$, $4 \times 5$, $5 \times 4$, $10 \times 2$, $20 \times 1$. $7$ elements: only $1 \times 7$ and $7 \times 1$, since $7$ is prime.
  4. $A = \begin{pmatrix} 1 & 0 \\ 3 & 2 \end{pmatrix}$
  5. $A = \begin{pmatrix} 0 & -1 \\ 3 & 2 \\ 8 & 7 \end{pmatrix}$
  6. (i) a row matrix, order $1 \times 3$. (ii) a column matrix, order $3 \times 1$. (iii) a scalar matrix (every diagonal entry $5$, off-diagonal $0$). (iv) the identity matrix $I$ of order $3 \times 3$.
  7. $x = 3$, $y = 5$
  8. $x = 5$, $y = 3$, $z = 5$
  9. (C) $m = n$.
  10. (A) $x = 3$, $y = 5$.
  11. (D) $81$.
  12. $\begin{pmatrix} 30 & 40 \\ 40 & 50 \end{pmatrix}$, order $2 \times 2$.
  13. $n_{1 2} = 29$, $n_{2 1} = 13$, and $N = \begin{pmatrix} 12 & 29 \\ 13 & 33 \end{pmatrix}$.

↑ Back to top

Practice set: Exercise 3.2

Exercise 3.2 practises the three operations. You add, scale and multiply, and you check the orders before each product. Some questions ask for both $A B$ and $B A$, so work out both.

Exercise 3.2: Addition, scalar multiplication and matrix multiplication
  1. practice Let $A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$, $B = \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}$, $C = \begin{pmatrix} -1 & 0 \\ 2 & 3 \end{pmatrix}$. Find (i) $A + B$, (ii) $3A - C$, (iii) $A B$, (iv) $B A$.
  2. practice Simplify $\begin{pmatrix} \cos^2 \theta & \sin^2 \theta \\ \sin^2 \theta & \cos^2 \theta \end{pmatrix} + \begin{pmatrix} \sin^2 \theta & \cos^2 \theta \\ \cos^2 \theta & \sin^2 \theta \end{pmatrix}$.
  3. practice Compute $\begin{pmatrix} 2 & -1 & 3 \end{pmatrix} \begin{pmatrix} 4 \\ 1 \\ -2 \end{pmatrix}$.
  4. practice Let $P = \begin{pmatrix} 1 & 0 & -2 \\ 3 & 1 & 4 \end{pmatrix}$ (order $2 \times 3$) and $Q = \begin{pmatrix} 2 & 1 \\ -1 & 3 \\ 0 & 2 \end{pmatrix}$ (order $3 \times 2$). Find $P Q$.
  5. practice For $A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$, $B = \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}$ and $D = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}$, verify that $(A B) D = A (B D)$.
  6. practice For $A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$, $B = \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}$ and $C = \begin{pmatrix} -1 & 0 \\ 2 & 3 \end{pmatrix}$, verify that $A (B + C) = A B + A C$.
  7. practice For $A = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & -2 \\ 3 & 5 \end{pmatrix}$, find $4A - 2B$.
  8. practice Find $X$ and $Y$ if $X + Y = \begin{pmatrix} 9 & -1 \\ 3 & 7 \end{pmatrix}$ and $X - Y = \begin{pmatrix} 5 & 3 \\ 1 & 1 \end{pmatrix}$.
  9. practice Find $X$ and $Y$ if $2X + 3Y = \begin{pmatrix} 7 & 4 \\ 2 & 10 \end{pmatrix}$ and $3X + 2Y = \begin{pmatrix} 8 & 1 \\ 3 & 5 \end{pmatrix}$.
  10. practice Find $X$ if $Y = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$ and $2X + Y = \begin{pmatrix} -1 & 4 \\ -3 & 0 \end{pmatrix}$.
  11. practice Find $x$ and $y$ if $\begin{pmatrix} 2 & 1 \\ 0 & x \end{pmatrix} + \begin{pmatrix} y & 3 \\ 1 & -2 \end{pmatrix} = \begin{pmatrix} 6 & 4 \\ 1 & 5 \end{pmatrix}$.
  12. practice Solve for $x$, $y$, $z$ and $t$ if $2 \begin{pmatrix} x & z \\ y & t \end{pmatrix} + \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 7 & 7 \\ 6 & 10 \end{pmatrix}$.
  13. practice Find $x$ and $y$ if $x \begin{pmatrix} 2 \\ 1 \end{pmatrix} + y \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 8 \\ 1 \end{pmatrix}$.
  14. practice If $F(x) = \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}$, prove that $F(x) F(y) = F(x+y)$.
  15. practice For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, find $A^2 - 5A + 6I$.
  16. practice For $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, show that $A^2 - 4A + 3I = O$.
  17. practice For $A = \begin{pmatrix} 2 & -1 \\ 2 & 0 \end{pmatrix}$, find $k$ such that $A^2 = k A - 2I$.
  18. practice A trust fund must invest ₹50,000 in two bonds paying $6\%$ and $9\%$ interest per year. Using matrix multiplication, find how much goes into each bond if the total annual interest must be (a) ₹3,900, (b) ₹4,200.
  19. practice A school shop has $8$ dozen notebooks, $6$ dozen pens and $5$ dozen erasers, selling at ₹50, ₹30 and ₹20 each respectively. Using matrix multiplication, find the total amount the shop receives from selling all of them.
  20. practice For $A = \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix}$, compute $A B$ and $B A$, and confirm they are not equal.
  21. practice Suppose $X$ has order $2 \times n$ and $Y$ has order $n \times 3$. For $X Y + Z$ to be defined, where $Z$ has order $2 \times 3$, what values can $n$ take?
    1. $n = 2$ only
    2. $n = 3$ only
    3. any positive integer
    4. $n$ must equal the number of columns of $Z$
  22. practice Let $X$ have order $2 \times p$ and $Y$ have order $p \times 3$. What is the order of $X Y$?
    1. $2 \times p$
    2. $p \times 3$
    3. $2 \times 3$
    4. $p \times p$
Answers
  1. $A+B = \begin{pmatrix} 4 & 0 \\ 4 & 9 \end{pmatrix}$; $3A-C = \begin{pmatrix} 10 & 6 \\ 1 & 9 \end{pmatrix}$; $A B = \begin{pmatrix} 9 & 4 \\ 13 & 18 \end{pmatrix}$; $B A = \begin{pmatrix} 1 & -6 \\ 14 & 26 \end{pmatrix}$, so $A B \neq B A$.
  2. $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$
  3. $1$
  4. $P Q = \begin{pmatrix} 2 & -3 \\ 5 & 14 \end{pmatrix}$
  5. Both sides equal $\begin{pmatrix} 18 & 21 \\ 26 & 67 \end{pmatrix}$: the associative law holds for this triple, as it does for any matrix product.
  6. Both sides equal $\begin{pmatrix} 10 & 10 \\ 20 & 30 \end{pmatrix}$: the distributive law holds.
  7. $\begin{pmatrix} 10 & 12 \\ -2 & 6 \end{pmatrix}$
  8. $X = \begin{pmatrix} 7 & 1 \\ 2 & 4 \end{pmatrix}$, $Y = \begin{pmatrix} 2 & -2 \\ 1 & 3 \end{pmatrix}$
  9. $X = \begin{pmatrix} 2 & -1 \\ 1 & -1 \end{pmatrix}$, $Y = \begin{pmatrix} 1 & 2 \\ 0 & 4 \end{pmatrix}$
  10. $X = \begin{pmatrix} -2 & 1 \\ -2 & -2 \end{pmatrix}$
  11. $x = 7$, $y = 4$
  12. $x = 2$, $y = 3$, $z = 3$, $t = 4$
  13. $x = 3$, $y = 2$
  14. Proved: $F(x) F(y) = \begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & y \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & x+y \\ 0 & 1 \end{pmatrix} = F(x+y)$.
  15. $\begin{pmatrix} 8 & 0 \\ 0 & 8 \end{pmatrix} = 8I$
  16. Shown: $A^2 = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$, so $A^2 - 4A + 3I = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix} - \begin{pmatrix} 8 & 4 \\ 4 & 8 \end{pmatrix} + \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O$.
  17. $k = 2$
  18. (a) ₹20,000 at $6\%$ and ₹30,000 at $9\%$. (b) ₹10,000 at $6\%$ and ₹40,000 at $9\%$.
  19. ₹8,160
  20. $A B = \begin{pmatrix} 4 & 4 \\ 7 & 2 \end{pmatrix}$, $B A = \begin{pmatrix} 2 & 4 \\ 7 & 4 \end{pmatrix}$, so $A B \neq B A$.
  21. (C) any positive integer.
  22. (C) $2 \times 3$.

↑ Back to top

Practice set: Exercise 3.3

Exercise 3.3 is about the transpose. You find transposes, check the transpose rules on numbers and test matrices for symmetry. You also split a square matrix into its symmetric and skew-symmetric parts.

Exercise 3.3: Transpose, symmetric and skew-symmetric matrices
  1. practice Find the transpose of each matrix: (i) $\begin{pmatrix} 5 & -2 & 7 \end{pmatrix}$, (ii) $\begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix}$, (iii) $\begin{pmatrix} -1 & 5 \\ 6 & 0 \\ 2 & 3 \end{pmatrix}$.
  2. practice For $A = \begin{pmatrix} 2 & -3 & 1 \\ 0 & 4 & -5 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 2 \end{pmatrix}$, find $A + 2B$.
  3. practice For $A = \begin{pmatrix} 1 & -2 \\ 0 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 4 & 1 \\ -1 & 2 \end{pmatrix}$, verify that $(A+B)^T = A^T + B^T$.
  4. practice For $A = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & -2 & 4 \end{pmatrix}$, verify that $(A B)^T = B^T A^T$.
  5. practice If $A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$, verify that $A A^T = I$.
  6. practice Show that $A = \begin{pmatrix} 4 & -2 & 3 \\ -2 & 5 & 1 \\ 3 & 1 & 6 \end{pmatrix}$ is a symmetric matrix.
  7. practice Show that $A = \begin{pmatrix} 0 & 3 & -5 \\ -3 & 0 & 2 \\ 5 & -2 & 0 \end{pmatrix}$ is a skew-symmetric matrix.
  8. practice For $A = \begin{pmatrix} 4 & 7 \\ 1 & 6 \end{pmatrix}$, find $(1/2)(A + A^T)$ and $(1/2)(A - A^T)$.
  9. practice Express $A = \begin{pmatrix} 3 & 5 \\ -1 & 2 \end{pmatrix}$ as the sum of a symmetric matrix and a skew-symmetric matrix.
  10. practice Express $A = \begin{pmatrix} 2 & 6 & -4 \\ 8 & 1 & 10 \\ 0 & -2 & 3 \end{pmatrix}$ as the sum of a symmetric matrix and a skew-symmetric matrix.
  11. practice If $A$ and $B$ are symmetric matrices of the same order, $A B - B A$ is
    1. a symmetric matrix
    2. a skew-symmetric matrix
    3. the zero matrix
    4. the identity matrix
  12. practice If $A = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ and $A + A^T = I$, then $\theta$ equals
    1. $\pi/6$
    2. $\pi/3$
    3. $\pi$
    4. $3 \pi/2$
  13. practice Find $x$ and $y$ so that $A = \begin{pmatrix} 3 & 2x-1 & 5 \\ x+4 & 7 & 3y-1 \\ 5 & y+9 & -2 \end{pmatrix}$ is symmetric.
Answers
  1. (i) $\begin{pmatrix} 5 \\ -2 \\ 7 \end{pmatrix}$. (ii) $\begin{pmatrix} 1 & 2 \\ -1 & 3 \end{pmatrix}$. (iii) $\begin{pmatrix} -1 & 6 & 2 \\ 5 & 0 & 3 \end{pmatrix}$.
  2. $\begin{pmatrix} 4 & -3 & 5 \\ -2 & 10 & -1 \end{pmatrix}$
  3. Both sides equal $\begin{pmatrix} 5 & -1 \\ -1 & 5 \end{pmatrix}$.
  4. Both sides equal $\begin{pmatrix} 6 & 3 & -3 \\ -4 & -2 & 2 \\ 8 & 4 & -4 \end{pmatrix}$.
  5. Verified: every entry of $A A^T$ collapses to $\cos^2 \theta + \sin^2 \theta = 1$ on the diagonal and $0$ off it, so $A A^T = I$ for every $\theta$.
  6. Shown: $a_{1 2} = a_{2 1} = -2$, $a_{1 3} = a_{3 1} = 3$, $a_{2 3} = a_{3 2} = 1$, so $A^T = A$.
  7. Shown: every diagonal entry is $0$, and $a_{1 2} = -a_{2 1}$, $a_{1 3} = -a_{3 1}$, $a_{2 3} = -a_{3 2}$, so $A^T = -A$.
  8. $(1/2)(A+A^T) = \begin{pmatrix} 4 & 4 \\ 4 & 6 \end{pmatrix}$ (symmetric), $(1/2)(A-A^T) = \begin{pmatrix} 0 & 3 \\ -3 & 0 \end{pmatrix}$ (skew-symmetric).
  9. $A = \begin{pmatrix} 3 & 2 \\ 2 & 2 \end{pmatrix} + \begin{pmatrix} 0 & 3 \\ -3 & 0 \end{pmatrix}$
  10. $A = \begin{pmatrix} 2 & 7 & -2 \\ 7 & 1 & 4 \\ -2 & 4 & 3 \end{pmatrix} + \begin{pmatrix} 0 & -1 & -2 \\ 1 & 0 & 6 \\ 2 & -6 & 0 \end{pmatrix}$
  11. (B) a skew-symmetric matrix.
  12. (B) $\pi/3$.
  13. $x = 5$, $y = 5$

↑ Back to top

Practice set: Exercise 3.4

Exercise 3.4 asks whether one matrix undoes another. You check both products, $A B$ and $B A$, against $I$. Working out an inverse from scratch belongs to the next chapter.

Exercise 3.4: Invertible matrices
  1. practice For $A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$, verify that $A B = B A = I$, and say what this tells you about $A$.
  2. practice For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, check whether $B = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$ satisfies $A B = I$.
  3. practice Find $k > 0$ so that $A = (1/k) \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}$ satisfies $A \cdot \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix} = I$.
  4. practice Matrices $B$ and $C$ both satisfy $A B = B A = I$ and $A C = C A = I$ for the same square matrix $A$. Without computing anything new, what must be true of $B$ and $C$, and why?
Answers
  1. $A B = B A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$, so $A$ is invertible with $A^{-1} = B$.
  2. No: $A B = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} = -2I \neq I$, so $B$ is not the inverse of $A$. The inverse is $-B/2$, since $A(-B/2) = (-B/2)A = I$, which you can check by multiplying.
  3. $k = 10$
  4. $B = C$: the inverse of a matrix, when it exists, is unique, so two matrices that both satisfy the defining equation for $A^{-1}$ must be the same matrix.

↑ Back to top

Practice set: miscellaneous

The Miscellaneous Exercise mixes the whole chapter. One problem may need an order check, a product and a transpose. Some ask you to prove a rule for every matrix of a given kind.

Miscellaneous Exercise: Matrices
  1. practice If $A$ and $B$ are symmetric matrices of the same order, prove that $A B - B A$ is skew-symmetric.
  2. practice Find $x > 0$ so that $A = (1/x) \begin{pmatrix} 3 & 4 \\ -4 & 3 \end{pmatrix}$ satisfies $A A^T = I$.
  3. practice Find the values of $x$ so that $\begin{pmatrix} x & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -9 \end{pmatrix} \begin{pmatrix} x \\ 1 \end{pmatrix} = O$.
  4. practice For $A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}$, show that $A^2 - 5A + 7I = O$.
  5. practice A trader sells three products, $x$, $y$ and $z$, in two cities. City I sells $8000$, $3000$ and $15000$ units respectively; City II sells $5000$, $12000$ and $9000$ units. The selling prices per unit are ₹3, ₹2 and ₹1, and the costs per unit are ₹2.50, ₹1.50 and ₹0.50. Using matrix multiplication, find the gross profit in each city.
  6. practice Find the matrix $X$ so that $X \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \end{pmatrix} = \begin{pmatrix} 2 & 3 & 2 \\ 1 & 5 & 15 \end{pmatrix}$.
  7. practice If a matrix $A$ is both symmetric and skew-symmetric, then $A$ must be
    1. the identity matrix
    2. the zero matrix
    3. a nonzero scalar matrix
    4. none of these
  8. practice If $A^2 = A$, then $(2I - A)^2 + 3A$ equals
    1. $I$
    2. $2I$
    3. $4I$
    4. $A$
Answers
  1. Proved: $(A B - B A)^T = (A B)^T - (B A)^T = B^T A^T - A^T B^T = B A - A B = -(A B - B A)$, using $A^T = A$ and $B^T = B$.
  2. $x = 5$
  3. $x = 3$ or $x = -3$
  4. Shown: $A^2 = \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix}$, so $A^2 - 5A + 7I = \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} - \begin{pmatrix} 15 & 5 \\ -5 & 10 \end{pmatrix} + \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O$.
  5. ₹13,000 profit in City I and ₹13,000 profit in City II.
  6. $X = \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}$. Checked against the third column too: $3a+4b = 6-4 = 2$ and $3c+4d = 3+12 = 15$, both matching.
  7. (B) the zero matrix.
  8. (C) $4I$.
Board paper practice
  1. practice A matrix $A = [a_{i j}]_{2 \times 2}$ has entries given by $a_{i j} = (i+j)^2/2$. Which of these is $A$? (1 mark)
    1. $\begin{pmatrix} 2 & 4.5 \\ 4.5 & 8 \end{pmatrix}$
    2. $\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$
    3. $\begin{pmatrix} 2 & 4 \\ 4 & 8 \end{pmatrix}$
    4. $\begin{pmatrix} 4.5 & 2 \\ 2 & 8 \end{pmatrix}$
  2. practice Assertion (A). There are $16$ matrices of order $2 \times 2$ whose every entry is $0$ or $1$. Reason (R). Each of the $4$ entries can independently be chosen in $2$ ways, giving $2^4$ such matrices. (1 mark)
    1. Both A and R are true and R is the correct explanation of A.
    2. Both A and R are true but R is not the correct explanation of A.
    3. A is true but R is false.
    4. A is false but R is true.
  3. practice If $\begin{pmatrix} x+y & 2 \\ 5 & x y \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 5 & 8 \end{pmatrix}$, find $x$ and $y$. (2 marks)
  4. practice If $A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}$, show that $A^2 - 5A + 7I = O$. Hence find $A^{-1}$. (3 marks)
  5. practice If $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix}$, verify that $(A B)^T = B^T A^T$. (5 marks)
  6. practice Let $A = \begin{pmatrix} 3 & 5 \\ 1 & -1 \end{pmatrix}$. (i) Find $A^T$. (1 mark) (ii) Write $A = P + Q$ where $P = (A+A^T)/2$ and $Q = (A-A^T)/2$. (2 marks) (iii) Verify that $P$ is symmetric and $Q$ is skew-symmetric. (1 mark)
  7. practice Give an example of two matrices $A$ and $B$, neither the zero matrix, such that $A B = O$. Verify your answer. (2 marks)
Answers
  1. (A) $\begin{pmatrix} 2 & 4.5 \\ 4.5 & 8 \end{pmatrix}$.
  2. (A) Both A and R are true and R is the correct explanation of A.
  3. $x=2,y=4$ or $x=4,y=2$.
  4. Shown: $A^2-5A+7I = O$. Both $A \times [(5I-A)/7]$ and $[(5I-A)/7] \times A$ equal $I$, so $A^{-1} = 1/7 \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix}$.
  5. Verified: $(A B)^T = B^T A^T = \begin{pmatrix} 4 & 10 \\ 5 & 11 \end{pmatrix}$.
  6. (i) $A^T = \begin{pmatrix} 3 & 1 \\ 5 & -1 \end{pmatrix}$. (ii) $P = \begin{pmatrix} 3 & 3 \\ 3 & -1 \end{pmatrix}$, $Q = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}$. (iii) $P^T=P$, $Q^T=-Q$, both verified.
  7. $A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$, $B = \begin{pmatrix} 2 & -1 \\ -2 & 1 \end{pmatrix}$; $A B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.

↑ Back to top