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A grid of numbers, treated as one object

FRAME

An attendance register, a mark-sheet across five subjects for a class of thirty, a seating plan for an exam hall — each is a rectangular grid of numbers. This chapter treats that grid as ONE object, not a scattered collection of entries. A MATRIX packages the whole grid so it can be added, scaled, and multiplied as a single unit, the way a single number is.

Some of that arithmetic feels exactly like plain number arithmetic. Addition works entry by entry and commutes — order does not matter. Scalar multiplication spreads over addition, same as with plain numbers. One rule breaks: matrix multiplication does not commute in general. $A B$ and $B A$ can be totally different matrices. One can even be defined while the other is not.

This chapter builds that arithmetic from scratch. It covers what a matrix actually is, and how to add, scale, and multiply two matrices correctly. It also covers which special shapes come up again and again — identity, zero, symmetric, skew-symmetric — and what it means for a square matrix to have an inverse.

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Order, entries, and named matrix shapes

KEY-TERM

A matrix is a rectangular array of numbers, laid out in rows and columns. A matrix with $m$ rows and $n$ columns has ORDER $m \times n$ — rows first, then columns. The entry in row $i$, column $j$ is written $a_{i j}$. Note the space between $i$ and $j$ inside the brackets: they are two separate labels, not one merged symbol.

Two matrices are EQUAL only when both conditions hold at once: they share the same order, AND every matching entry agrees. $a_{i j} = b_{i j}$ for every $i$, $j$. Matching entries this way is what lets an equation between two matrices be solved for unknown entries. Match order first, then match entry by entry.

KEY-TERM

Certain matrix shapes turn up so often in this chapter that they earn their own names. A ROW matrix has exactly one row, order $1 \times n$; a COLUMN matrix has exactly one column, order $m \times 1$. A SQUARE matrix has equal rows and columns, order $n \times n$.

Within a square matrix, three more names nest inside each other. A DIAGONAL matrix has every off-diagonal entry zero: $a_{i j} = 0$ whenever $i \neq j$. A SCALAR matrix is a diagonal matrix whose diagonal entries are all the same number. The IDENTITY matrix $I$ is the scalar matrix whose diagonal entries are all $1$ — for example, $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ at order $2 \times 2$.

*A ZERO matrix, every entry $0$, is the one named shape that is not restricted to square* — it can have any order at all.

None of these names is a fact about one entry; each is a fact about the whole grid, which is why a single nonzero entry off the main diagonal rules out three of the four rings at once. Invertibility appears nowhere on this picture and that is worth noticing, because it cuts across every ring: a diagonal matrix can be singular, and a matrix with no name at all can have an inverse.

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Transpose, symmetric, and skew-symmetric matrices

KEY-TERM

The TRANSPOSE of a matrix $A$, written $A^T$, swaps its rows and columns: row $i$ of $A$ becomes column $i$ of $A^T$. If $A$ has order $m \times n$, then $A^T$ has order $n \times m$ — the shape itself flips.

A SQUARE matrix $A$ is SYMMETRIC when it equals its own transpose, $A^T = A$. In other words, $a_{i j} = a_{j i}$ for every $i$, $j$ — the matrix mirrors across its main diagonal. $A$ is SKEW-SYMMETRIC when $A^T = -A$ instead, which forces $a_{i j} = -a_{j i}$.

*Setting $i = j$ in that skew-symmetric condition gives $a_{i i} = -a_{i i}$.* That is only possible when $a_{i i} = 0$. Every diagonal entry of a skew-symmetric matrix is forced to be zero, with no exceptions.

Transposing twice puts every cell back where it began, so the operation undoes itself and a chain of transposes collapses to at most one. Watch the index order when you write entries down by hand: swapping the row and column labels is the quiet way to transpose a matrix you did not mean to transpose, and nothing in the arithmetic afterwards will show that it happened.

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Adding matrices and scaling by a number

CONCEPT

Two matrices add ENTRYWISE, and only when they share the same order. Entry $(i,j)$ of $A + B$ equals $a_{i j} + b_{i j}$. *There is no way to add a $2 \times 3$ matrix to a $3 \times 2$ matrix.* The orders must match exactly, position for position, before addition even makes sense.

SCALAR multiplication scales every entry by the same number: for a scalar $k$, entry $(i,j)$ of $k A$ equals $k \cdot a_{i j}$. $k A$ keeps the same order as $A$.

Both behave exactly like plain-number arithmetic. Addition is commutative, $A + B = B + A$, and associative — grouping does not matter. Scalar multiplication spreads over matrix addition. That changes once multiplication of two matrices enters the picture — this chapter turns to it next.

Scaling a matrix by a number leaves the footprint untouched, so an expression built only from sums and scalings needs one order check at the start and none afterwards. That check is worth doing first and once: a mark lost here is almost never lost in the arithmetic, it is lost by starting the arithmetic on two matrices that were never allowed to be added at all.
Worked example

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

  1. $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, $B = \begin{pmatrix} 5 & -1 \\ 0 & 2 \end{pmatrix}$
    Given matrices; compute $A+B$ first, then $2A - B$.
  2. $A + B = \begin{pmatrix} 1+5 & 2+(-1) \\ 3+0 & 4+2 \end{pmatrix} = \begin{pmatrix} 6 & 1 \\ 3 & 6 \end{pmatrix}$
    Add entry by entry — no row mixes with another row.
  3. $2A = \begin{pmatrix} 2 & 4 \\ 6 & 8 \end{pmatrix}$
    Scale every entry of $A$ by $2$.
  4. $2A - B = \begin{pmatrix} 2-5 & 4-(-1) \\ 6-0 & 8-2 \end{pmatrix} = \begin{pmatrix} -3 & 5 \\ 6 & 6 \end{pmatrix}$
    *Subtract $B$ from $2A$ entry by entry, the same way addition worked.*

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Row times column

CONCEPT

Matrix multiplication is NOT entrywise — this is the one place matrix arithmetic stops looking like plain numbers. The product $A B$ is built row-times-column. Entry $(i,j)$ of $A B$ is the dot product of row $i$ of $A$ with column $j$ of $B$: multiply matching entries and add them.

That row-times-column rule only works when the number of columns of $A$ equals the number of rows of $B$. That shared count is exactly what gets summed over, entry by entry. If $A$ is $m \times n$ and $B$ is $n \times p$, the product $A B$ exists and has order $m \times p$.

*If the inner numbers do not match, $A B$ is simply undefined — no matter how $B A$ behaves.* Checking that the sizes match comes first, before a single entry is multiplied.

The shared edge is not only a permission slip. Its height is how many products get added together to make each entry of the answer, so the picture says how long the arithmetic will be as well as whether it may start. Butt a third block on the right and the same rule applies at the new join, which is why a chain can be grouped either way: the joins do not move when the bracketing does.
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 $3 \times 2$)
    Given matrices; check the compatibility condition before multiplying.
  2. $3 = 3$ — the column count of $A$ matches the row count of $B$
    $A B$ exists, and will have order $2 \times 2$.
  3. entry $(1,1) = 1 \cdot 7 + 2 \cdot 9 + 3 \cdot 11 = 7+18+33 = 58$
    Row 1 of $A$ dot column 1 of $B$.
  4. entry $(1,2) = 1 \cdot 8 + 2 \cdot 10 + 3 \cdot 12 = 8+20+36 = 64$
    Row 1 of $A$ dot column 2 of $B$.
  5. entry $(2,1) = 4 \cdot 7 + 5 \cdot 9 + 6 \cdot 11 = 28+45+66 = 139$
    Row 2 of $A$ dot column 1 of $B$.
  6. entry $(2,2) = 4 \cdot 8 + 5 \cdot 10 + 6 \cdot 12 = 32+50+72 = 154$
    Row 2 of $A$ dot column 2 of $B$.
  7. *$A B = \begin{pmatrix} 58 & 64 \\ 139 & 154 \end{pmatrix}$*
    All four entries assembled; $B A$, if it were even formed, would come out $3 \times 3$ — a completely different shape from $A B$.

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What survives, what doesn’t

CONCEPT

Two familiar rules survive the move to matrix multiplication. Associativity: $(A B)C = A(B C)$, so a chain of products can be grouped either way without changing the answer. Distributivity: $A(B+C) = A B + A C$, and $(A+B)C = A C + B C$ — multiplication spreads over addition exactly as expected.

The rule that does NOT survive is commutativity: in general $A B \neq B A$. The two products can be completely different, or one can be defined while the other is not. The order of a matrix product is never optional, and it can never be swapped out of habit the way a plain-number multiplication can.

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A B is not B A

MISCONCEPTION

Plain numbers multiply in either order without changing the answer. That habit is hard to shake. It invites treating $A B$ and $B A$ as the same matrix too, just written two ways.

Take $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$. Computing both products entry by entry gives $A B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ but $B A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$ — two different matrices, not the same product read two ways.

Multiplication order is never optional. $A B$ and $B A$ can disagree outright, or one can even be undefined while the other is not. The matching condition from the last section is checked on its own for each order — never assumed to hold both ways at once.

Some pairs do commute, and knowing which ones saves time rather than costing it: the identity commutes with everything, a matrix commutes with its own inverse, and any two scalar matrices commute with each other. Everywhere else the order is part of the expression, so an answer that reordered a product to make the algebra tidier went wrong before the arithmetic started.
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. $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$
    Given matrices from the trap’s counterexample; compute both products entry by entry.
  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 $A$ dot 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 $A$ dot 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 $(1,2) = 1 \cdot 1 + 0 \cdot 1 = 1$
    Row 1 of $B$ dot 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 $B$ dot each column of $A$; so $B A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$.
  6. *$A B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \neq \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} = B A$*
    Confirmed entry by entry, even though both products are defined and both come out $2 \times 2$.

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Invertible matrices

CONCEPT

A square matrix $A$ is INVERTIBLE when some matrix $B$ of the same order satisfies $A B = B A = I$. That $B$ is called the INVERSE of $A$, written $A^{-1}$. Not every square matrix has one — a matrix with no inverse is called SINGULAR.

When an inverse exists, it is UNIQUE. Suppose both $B$ and $C$ satisfy $A B = B A = I$ and $A C = C A = I$. Then $B = B I = B(A C) = (B A)C = I C = C$. That chain of equal steps forces $B$ and $C$ to be the same matrix — a matrix can never have two different inverses.

This chapter states only the definition and the uniqueness fact. The METHOD for actually computing $A^{-1}$, via the adjoint, belongs to the next chapter.

Not every square matrix has a partner like this. One that flattens the plane onto a line has thrown information away and nothing can put it back, which is what singular means and what the zero-divisor figure later in the chapter draws. Where a partner does exist there is only ever one of it, and the chapter proves that in three lines using nothing but the grouping rule.

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Symmetric plus skew-symmetric

CONCEPT

Any square matrix $A$ can be written as a sum of a symmetric matrix and a skew-symmetric matrix, in exactly one way. Define $P = (1/2)(A + A^T)$ and $Q = (1/2)(A - A^T)$; adding them back gives $P + Q = A$ by construction.

$P$ is symmetric: $P^T = (1/2)(A^T + A) = P$, since transpose distributes over addition and $(A^T)^T = A$. $Q$ is skew-symmetric by the same kind of check: $Q^T = (1/2)(A^T - A) = -Q$.

Every square matrix, however messy, splits into one symmetric piece and one skew-symmetric piece that add back to it. This is not a special property of tidy matrices. The formulas for $P$ and $Q$ work on any square matrix at all.

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 $A^T = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}$
    Given matrix; compute the symmetric part $P$ and skew-symmetric part $Q$.
  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 $A^T$ entrywise, then halve.
  3. Check $P^T = \begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix} = P$
    *$P$ is confirmed 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 $A$ entrywise, 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 confirmed skew-symmetric.*
  6. $P + Q = \begin{pmatrix} 2 & 2 \\ 2 & 4 \end{pmatrix} + \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} = A$
    Adding the two parts back recovers $A$ exactly, confirming the split is exact.

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A B = 0 does not force A = 0 or B = 0

MISCONCEPTION

With plain numbers, a product equal to zero forces one of the factors to be zero. That habit invites assuming $A B = 0$ must force $A = 0$ or $B = 0$ too.

Take $A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$, neither one the zero matrix. Their product comes out $A B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ anyway — matrices can act as ZERO DIVISORS, a behaviour plain numbers never show.

*$A B = 0$ only says the product is zero.* It says nothing about the factors on their own. A matrix product can vanish even when both factors are packed with nonzero entries.

The habit this breaks is cancelling. From a plain-number equation you may divide both sides by a nonzero factor; from a matrix equation you may not, because a product can be the zero matrix without either factor being zero, and by the same token two different matrices can give the same product against a third. Cancel only once you know the factor you are cancelling has an inverse.
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
    Given matrices; compute $A B$ entry by entry.
  2. entry $(1,1) = 1 \cdot 1 + 1 \cdot (-1) = 1 - 1 = 0$
    Row 1 of $A$ dot column 1 of $B$.
  3. entry $(1,2) = 1 \cdot 1 + 1 \cdot (-1) = 0$
    Row 1 of $A$ dot column 2 of $B$.
  4. entry $(2,1) = 1 \cdot 1 + 1 \cdot (-1) = 0$
    Row 2 of $A$ dot column 1 of $B$.
  5. entry $(2,2) = 1 \cdot 1 + 1 \cdot (-1) = 0$
    Row 2 of $A$ dot column 2 of $B$.
  6. *$A B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$*
    Every entry of $A B$ is $0$, even though $A$ and $B$ both have every entry nonzero.

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One object, its own arithmetic

RECAP

A matrix packages a whole grid of numbers into one object with its own arithmetic. Addition and scalar multiplication behave exactly like plain numbers, entrywise. Nothing about this chapter’s first two operations should surprise you.

Multiplication does not. It only works when the first matrix’s column count matches the second matrix’s row count. Even when both $A B$ and $B A$ are defined, they can be different matrices. *Never assume $A B$ equals $B A$, and never assume $A B = 0$ forces $A = 0$ or $B = 0$.* Both habits carry over from plain numbers, and both fail here.

Named shapes — identity, zero, symmetric, skew-symmetric — come up again and again because they simplify this arithmetic. An invertible matrix’s inverse, when it exists, is unique. And every square matrix, however untidy, splits cleanly into a symmetric part and a skew-symmetric part that add back to it exactly.

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Practice set

Exercise 3.1 practises reading off a matrix’s order and entries, and building a matrix from a given rule. It covers naming the types — row, column, square, diagonal, scalar, identity, zero — and uses matrix equality to solve for 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 $B = \begin{pmatrix} 4 & -2 & 7 \\ 0 & 5 & -3 \\ 9 & 1 & 6 \end{pmatrix}$, 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 $B = [a_{i j}]$ 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. $81$
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. $B = \begin{pmatrix} 0 & -1 \\ 3 & 2 \\ 8 & 7 \end{pmatrix}$
  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 — $81$.

Exercise 3.2 practises matrix addition, scalar multiplication, and matrix multiplication on concrete matrices. It checks that the matrix sizes match before multiplying, and uses the associative and distributive properties. *It also tests whether $A B = B A$ each time, since it usually does not.*

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 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. $A+B = \begin{pmatrix} 4 & 0 \\ 4 & 9 \end{pmatrix}$; $3A-C = \begin{pmatrix} 10 & 6 \\ 1 & 9 \end{pmatrix}$; $A B = \begin{pmatrix} 9 & 4 \\ 13 & 18 \end{pmatrix}$; $B A = \begin{pmatrix} 1 & -6 \\ 14 & 26 \end{pmatrix}$ — $A B \neq B A$.
  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. $A B = \begin{pmatrix} 4 & 4 \\ 7 & 2 \end{pmatrix}$, $B A = \begin{pmatrix} 2 & 4 \\ 7 & 4 \end{pmatrix}$ — $A B \neq B A$.
  21. C — any positive integer.
  22. C — $2 \times 3$.

Exercise 3.3 practises finding a matrix’s transpose, and testing whether a given square matrix is symmetric or skew-symmetric. It also has students decompose a square matrix into its symmetric and skew-symmetric parts and verify the two pieces add back to the original.

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. $x = 5$, $y = 5$

Exercise 3.4 practises checking whether a given pair of square matrices satisfies $A B = B A = I$, which is what invertibility actually means. It also has students use the uniqueness fact to reason about candidate inverses, without yet computing an inverse from scratch — that method 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 $A$ (it is $\text{adj} A$, missing the division by $|A| = -2$).
  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.

The Miscellaneous Exercise mixes every idea in the chapter into single multi-step problems, rather than drilling one skill at a time. A typical problem might ask for an order check before multiplying, a transpose, and a symmetric/skew-symmetric split. It might also ask whether a stated product could possibly equal the zero matrix without either factor being zero.

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 $\begin{pmatrix} x & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -9 \end{pmatrix} \begin{pmatrix} x \\ 1 \end{pmatrix} = 0$.
  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. a diagonal matrix
    2. the zero matrix
    3. a square 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 — $4I$.

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