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One number, one big question
Class 10’s pairs of lines: two cross once, two run parallel and never meet, two lie on top of each other.
In Class 10 you solved pairs of linear equations in two unknowns. Each equation is a straight line on a graph. Two lines can do only three things. They can cross at one point. They can run side by side and never meet. Or they can lie on top of each other. So a pair of equations has one answer, no answer, or endless answers. You could tell which before you solved anything.
Chapter 3’s A is invertible: times this matrix it gives I. The all-ones matrix is not; its answers always have two equal rows.
Chapter 3 met the same kind of question with matrices. A square matrix that has an inverse is called INVERTIBLE: multiply it by its inverse $A^{-1}$ and you get back to where you started, $A A^{-1} = I$. Other matrices are not invertible at all. The zero matrix $O$ is one of them.
The entries of both add up to 12. One is invertible and one is not. A single number, worked out from the entries, tells which.
Here is the question for the whole chapter. Is there a quick test that tells you which kind of matrix you have? You want one number, worked out from the entries, before you try to find $A^{-1}$.
Naming that single number
There is such a number, called the DETERMINANT. It exists only for a square matrix. By the last page you will know what it says about the inverse. You will also see what it says about a pair of lines, and about the area of a triangle.
A SCHOLAR INDIA REMEMBERS
Brahmagupta worked at Bhillamala, now Bhinmal in Rajasthan, around 628 CE. In his Brahmasphutasiddhanta he gave the first full rules for zero, and for working with negative numbers, as numbers in their own right. Ask his question of the number you are about to meet. When a determinant comes out $0$, has nothing happened? Or has it told you something? Watch for the answer. Here a $0$ says more than any other value: no inverse, and not exactly one answer.
Brahmagupta · the brahminy kite
A matrix as a picture
*A $2 \times 2$ matrix* can be drawn on graph paper. Take $A = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}$. Its two rows are the pairs $(3, 1)$ and $(1, 2)$, and each pair is a point. Draw an arrow from $(0, 0)$ to each point. The two arrows make a parallelogram, and its area is 5. That area is the determinant of $A$: $3 \cdot 2 - 1 \cdot 1 = 5$. If the two arrows lie along one line, the parallelogram is flat, and the determinant is $0$.
Chapter 3 drew a matrix another way, as a move of the plane. Multiply $A$ by the column $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, row times column:
*So $A$ moves* the point $(1, 0)$ to $(3, 1)$ and the point $(0, 1)$ to $(1, 2)$. It moves the whole unit square, area 1, onto the parallelogram on its two columns. That is the picture behind Chapter 3’s leaning squares.
*For this $A$* the rows and the columns are the same two pairs, so both pictures give the same parallelogram, the same unit square sheared into a parallelogram that Chapter 3 drew. Its area is the absolute value of the determinant, with the sign showing which way the arrows turn. The figure below shows it.
Your turn: find the area of the parallelogram on the rows $(2, 1)$ and $(1, 1)$.
The unit square, area 1. Beside it, the rows of A as arrows from (0, 0) make a parallelogram of area 5, the determinant of A.
Check yourself
What single question does the determinant of a square matrix answer first?
whether the matrix is symmetric
whether it is invertible
how many rows the matrix has
how many entries the matrix has
Check your answer
whether the matrix is symmetric — Symmetry compares $a_{i j}$ to $a_{j i}$ directly. A symmetric matrix can have any determinant, including zero, the two facts do not track each other.
✓ whether it is invertible — (B) The determinant is one number computed from a square matrix; it is nonzero exactly when the matrix is invertible.
how many rows the matrix has — Row count is fixed the moment the matrix is written down, no determinant is needed to find it. The determinant answers a different question: does an inverse exist.
how many entries the matrix has — Entry count is $n^2$, visible without computing anything. The determinant is a single computed number, not a count.
Geometrically, what does the determinant of a $2 \times 2$ matrix measure?
the area of the parallelogram on its two rows, with a sign
the largest entry in the matrix
the angle the transformation rotates by
the number of solutions a linear system has, since a bigger determinant should allow more solutions
Check your answer
✓ the area of the parallelogram on its two rows, with a sign — (A) For a $2 \times 2$ matrix the determinant is the area of the parallelogram on its two rows, and its sign says which way round the rows go.
the largest entry in the matrix — No single entry decides the determinant’s geometric meaning. The area comes from every entry acting together, not from the largest one.
the angle the transformation rotates by — A rotation matrix has determinant $1$ regardless of the angle, the determinant carries no rotation information at all.
the number of solutions a linear system has, since a bigger determinant should allow more solutions — Solution count is something this chapter reaches later, by way of the invertibility test. The direct geometric reading of $|A|$ is a signed area.
A matrix has a zero determinant. What does this force?
the matrix cannot be multiplied by another matrix
the matrix must be the zero matrix
no inverse exists: the matrix is singular
the matrix has fewer rows than columns
Check your answer
the matrix cannot be multiplied by another matrix — Multiplying two matrices only needs matching inner dimensions. A singular matrix multiplies just fine, it simply cannot be inverted.
the matrix must be the zero matrix — A zero determinant only requires the rows to be dependent, two identical rows already force it, not every entry to be zero.
✓ no inverse exists: the matrix is singular — (C) A zero determinant is exactly the invertibility test failing, the matrix is singular and no inverse exists.
the matrix has fewer rows than columns — A determinant is only ever defined for a square matrix, so row and column counts are already equal before this question can be asked.
You have already made a check like this, long before you met the word determinant. Try each one below.
Area of a triangle from its vertices (Class 11, Mathematics (NCERT Class 11), page 152). Here: this area formula reappears here as a 3×3 determinant instead of the coordinate sum, same vertices, same number. Check: the triangle with vertices $(4,4)$ $(3,-2)$ $(-3,16)$ has area $(1/2)|4(-2-16)+3(16-4)+(-3)(4+2)| = 27$.
Zero area means three points are collinear (Class 11, Mathematics (NCERT Class 11), page 152). Here: that same zero-area test decides collinearity here with a 3×3 determinant instead of the coordinate formula. Check: the points $(1,1)$ $(2,2)$ $(3,3)$ give area $(1/2)|1(2-3)+2(3-1)+3(1-2)| = 0$. They are collinear.
Solving a pair of linear equations (Class 10, Mathematics (NCERT Class 10), page 30). Here: once that pair is written as one matrix equation, $X = A^{-1} B$ reaches the same numbers this elimination or substitution already finds. Check: $x+y=6$ and $x-y=2$ give $x=4$ and $y=2$.
One, none or infinitely many solutions (Class 10, Mathematics (NCERT Class 10), page 25). Here: this same solution question is settled here by one number, the coefficient determinant, not by drawing two lines. Check: $x+y=2$ and $2x+2y=4$ describe one line, so $(0,2)$ and $(1,1)$ both solve the pair.
Multiplication distributes over addition (Class 11, Mathematics (NCERT Class 11), page 78). Here: expanding a $3×3$ determinant along a row multiplies each entry by its own $2×2$ cofactor; inside one term, $a_{1 1}(a_{2 2}a_{3 3} - a_{2 3}a_{3 2}) = a_{1 1}a_{2 2}a_{3 3} - a_{1 1}a_{2 3}a_{3 2}$ is this same distributive law, applied to a bracket instead of a sum. Check: $2(3 + 4) = 2 \times 3 + 2 \times 4$, both giving $14$.
If any of these felt new, read the page named before going on.
Class 10 test: if the x-coefficient and y-coefficient ratios differ, the lines cross once. Here 2 and 1 differ; 1/2 and 1/2 do not.
Go back to the Class 10 pair of equations. Before you solved them, you could tell what kind of answer to expect. You compared the numbers in front of x with each other. Then you did the same for the numbers in front of y. If the two ratios were different, the lines crossed once. Then there was exactly one answer.
Cross-multiplied, the test gives 2 times 1 minus 1 times 1 = 1 for the crossing pair and 0 for the parallel pair.
The test written with letters
Write the pair with letters, as $a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$. Then turn the Class 10 test into one number, line by line:
Why it works
Turn the Class 10 test into one number
$a_1 / a_2 \neq b_1 / b_2$ The Class 10 test: the lines cross once when these two ratios differ.
$a_1 b_2 \neq a_2 b_1$ Cross-multiply: multiply both sides by $a_2 b_2$, which needs $a_2 b_2 \neq 0$.
*$a_1 b_2 - a_2 b_1 \neq 0$* Take $a_2 b_1$ from both sides: the whole test is one number, not 0. This form still works when $a_2$ or $b_2$ is $0$, where the ratios cannot be written: for $x = 1$, $y = 2$ it gives $1 \cdot 1 - 0 \cdot 0 = 1$, not $0$.
Check: $2 \cdot 1 - 1 \cdot 1 = 1 \neq 0$ for $2 x + y = 8$, $x + y = 5$ The crossing pair of the figure: the number is not 0, and the lines cross once.
Cross-multiplied, a1/a2 not equal b1/b2 becomes a1 b2 minus a2 b1 not equal 0. That number is the determinant.
Now put the numbers in front of x and y in a grid, $\begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}$. The test number $a_1 b_2 - a_2 b_1$ comes straight from that grid. You multiply down the main diagonal. Then you subtract the product along the other diagonal. That number has a name.
3 times 2 minus 4 times 1 = 2. A 1 by 1 determinant is its one entry: minus 3 means minus 3, not 3.
What a determinant is
The determinant of a square matrix is one number, worked out from its entries by a fixed rule. For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the rule is $a d - b c$. Multiply the two entries on the main diagonal, $a$ and $d$. Multiply the two on the other diagonal, $b$ and $c$. Take the second product away from the first. That one number tells you whether $A$ has an inverse. It has one when the determinant is not $0$. It has none when the determinant is $0$. The rest of the chapter shows why.
It is written $|A|$, or $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$, or $\det A$. All three mean the same number. The bars do not mean absolute value here, and a determinant can be negative. For a $1 \times 1$ matrix $\begin{pmatrix} a \end{pmatrix}$, the determinant is just the entry $a$.
Let us check one. Take $A = \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix}$. Then $|A| = 3 \cdot 2 - 4 \cdot 1 = 6 - 4 = 2$.
Rectangles of 6 and 4 leave 6 minus 4, or 2. The parallelogram on the rows (3, 4) and (1, 2) has area 2.
Here is the picture from the opening section again. Draw the two rows of $A$ as arrows from $(0, 0)$, here to $(3, 4)$ and $(1, 2)$. The two arrows make a thin parallelogram, and its area is 2, the same as $|A|$. *For every $2 \times 2$ matrix*, that area is the absolute value of $|A|$, with any minus sign dropped. Which arrow comes first decides the sign, and a later section comes back to that.
The determinant is 1: the lines cross once. It is 0: they do not. The determinant is the Class 10 test turned into one number.
In Class 10 terms, a determinant that is not 0 means the two lines cross once. That is the first job this number does.
Your turn: find $|A|$ for $A = \begin{pmatrix} 4 & 6 \\ 2 & 3 \end{pmatrix}$. What does it tell you about the lines $4 x + 6 y = 1$ and $2 x + 3 y = 5$?
Four matrices sorted by their determinants, 13, 7, 3 and 0: the first three have an inverse, and D, with determinant 0, has none.
Check yourself
For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, what does the determinant $|A|$ equal?
$a b - c d$
$a d + b c$
$a c - b d$
$a d - b c$
Check your answer
$a b - c d$ — The formula multiplies down each DIAGONAL ($a,d$ and $b,c$), never along a row ($a,b$ or $c,d$).
$a d + b c$ — The two diagonal products are subtracted, not added, $a d - b c$, never $a d + b c$.
$a c - b d$ — The formula pairs entries diagonally, not by column, $a$ pairs with $d$, and $b$ pairs with $c$.
✓ $a d - b c$ — (D) The $2 \times 2$ determinant is the main-diagonal product minus the other diagonal’s product.
For $A = \begin{pmatrix} 3 & 4 \\ 1 & 2 \end{pmatrix}$, what is $|A|$?
$10$
$2$
$-2$
$8$
Check your answer
$10$ — $3 \cdot 2 + 4 \cdot 1 = 10$ adds the diagonal products; the formula subtracts the second from the first, giving $6-4=2$.
$-2$ — Flipping the subtraction order gives $4-6=-2$, the main-diagonal product ($3 \cdot 2$) must come first.
$8$ — $8$ is not even $3 \cdot 2$, the formula needs BOTH diagonal products, $3 \cdot 2$ and $4 \cdot 1$, combined by subtraction, not one product reported alone.
For a $1 \times 1$ matrix $A = \begin{pmatrix} a \end{pmatrix}$, what does the determinant $|A|$ equal?
$1$
$0$
$a^2$
$a$
Check your answer
$1$ — The matrix being $1 \times 1$ describes its shape, not the value stored inside it, the determinant reports that stored value, $a$.
$0$ — Nothing forces a $1 \times 1$ determinant to be zero, it equals whatever the single entry $a$ is.
$a^2$ — No squaring is involved, a $1 \times 1$ determinant is the entry itself, unchanged.
✓ $a$ — (D) For a $1 \times 1$ matrix the determinant is just the single entry itself.
A $3 \times 3$ determinant is worked out from smaller $2 \times 2$ ones. Two words name the pieces.
Pick an entry $a_{i j}$ of a square matrix. Cover its row and its column. The determinant of what is left is the MINOR of that entry, written $M_{i j}$.
Cover row 1 and column 2. Only the entry 2 is left. That is the minor of position one, two.
*Take $A = \begin{pmatrix} 4 & 7 \\ 2 & 6 \end{pmatrix}$.* It has four entries, so it has four minors. Cover the row and the column of each entry in turn:
Worked example
Find the four minors of $A = \begin{pmatrix} 4 & 7 \\ 2 & 6 \end{pmatrix}$
$M_{1 2} = 2$ Cover row 1 and column 2: only the entry 2 is left, and that is the minor.
$M_{1 1} = 6$ Cover row 1 and column 1: 6 is left.
$M_{2 1} = 7$ Cover row 2 and column 1: 7 is left.
$M_{2 2} = 4$ Cover row 2 and column 2: 4 is left.
One minor per entry, covering its own row and column each time: 6, 2 on top, 7, 4 below.
Giving the minor a sign
The COFACTOR of the entry is its minor with a sign: $C_{i j} = (-1)^{i + j} M_{i j}$. NCERT writes the cofactor as $A_{i j}$. This book writes $C_{i j}$, so that it is not mixed up with the entry $a_{i j}$. When $i + j$ is even, the sign is $+$. When $i + j$ is odd, the sign is $-$.
The corner of a chessboard: light, dark, light along the top, and each row starts on the other colour.
Here are the signs for a $3 \times 3$ matrix: $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$. They look like the squares of a chessboard, with $+$ in the top-left corner. The sign changes at every step along a row or down a column.
The cofactor signs sit like those squares: plus on the light ones, minus on the dark. Position one, two is minus.
Apply the signs to the minors: where i + j is odd, the sign changes. The 2 at top right becomes minus 2, the 7 at bottom left becomes minus 7, and 6 and 4 stay.
So for the matrix above, put the sign on each minor. Only the two places with an odd $i + j$ change sign:
Worked example
Give each minor of $A = \begin{pmatrix} 4 & 7 \\ 2 & 6 \end{pmatrix}$ its sign
$C_{1 1} = (-1)^{1 + 1} M_{1 1} = 6$ $1 + 1 = 2$ is even, so the sign is $+$ and the minor stays 6.
$C_{1 2} = (-1)^{1 + 2} M_{1 2} = -2$ $1 + 2 = 3$ is odd, so the sign is $-$ and 2 becomes $-2$.
$C_{2 2} = (-1)^{2 + 2} M_{2 2} = 4$ $2 + 2 = 4$ is even, so 4 stays.
Your turn: in $\begin{pmatrix} 1 & 0 & 2 \\ 3 & 1 & 4 \\ 0 & 5 & 1 \end{pmatrix}$, find the minor and the cofactor of the entry 4.
Check yourself
What is the minor $M_{i j}$ of the entry $a_{i j}$?
the value of the entry $a_{i j}$ itself
the sum of every entry in row $i$
the determinant left after deleting row $i$ and column $j$
the determinant left after deleting row $j$ and column $i$
Check your answer
the value of the entry $a_{i j}$ itself — The minor is a whole smaller determinant, not the single entry $a_{i j}$, it is what remains after that entry’s row and column are struck out.
the sum of every entry in row $i$ — Summing a row has nothing to do with the minor, the minor is the determinant of the smaller matrix left after removing row $i$ and column $j$ entirely.
✓ the determinant left after deleting row $i$ and column $j$ — (C) The minor is exactly the smaller determinant left behind once the entry’s own row and column are removed.
the determinant left after deleting row $j$ and column $i$ — $M_{i j}$ deletes row $i$ and column $j$, the first index is always the row, the second the column, never swapped.
How is the cofactor $C_{i j}$ related to the minor $M_{i j}$?
$C_{i j} = (-1)^{i+j+1} M_{i j}$
$C_{i j} = (-1)^{i j} M_{i j}$
$C_{i j} = (-1)^{i+j} M_{i j}$
$C_{i j} = M_{i j}$
Check your answer
$C_{i j} = (-1)^{i+j+1} M_{i j}$ — Adding 1 to the exponent flips every sign in the pattern, position $(1,1)$ would come out $-$ instead of the correct $+$.
$C_{i j} = (-1)^{i j} M_{i j}$ — The exponent is the sum $i+j$, not the product $i j$. The product $i j$ and the sum $i+j$ give the same sign at only one of the nine places, $(2,2)$.
✓ $C_{i j} = (-1)^{i+j} M_{i j}$ — (C) The cofactor is the signed minor: multiply the minor by $(-1)^{i+j}$.
$C_{i j} = M_{i j}$ — Dropping the sign turns the cofactor back into a plain minor, the whole point of $(-1)^{i+j}$ is the checkerboard of $+$ and $-$ it introduces.
What sign does the cofactor carry at position $(2,3)$?
depends on the value of the entry
$+$
$-$, since $i+j=2+3=5$ is odd
$0$
Check your answer
depends on the value of the entry — The sign comes purely from the position $(i,j)$ through $(-1)^{i+j}$, it never depends on what number sits at that entry.
$+$ — The sign alternates by position, it is only $+$ when $i+j$ is even. At $(2,3)$, $i+j=5$ is odd, so the sign is $-$.
✓ $-$, since $i+j=2+3=5$ is odd — (C) $i+j = 2+3 = 5$ is odd, so $(-1)^5 = -1$.
$0$ — $(-1)^{i+j}$ is always $+1$ or $-1$, it is never $0$, at any position.
What does expanding a $3 \times 3$ determinant along different rows or columns give?
the same value every time, so pick the one with the most zeros
a different value each time, so the first row is always safest
the same value, but only when the matrix has no zero entries
a different value, and only the first COLUMN is guaranteed correct
Check your answer
✓ the same value every time, so pick the one with the most zeros — (A) Any row or column gives the identical determinant value; the choice is purely about which is easiest to compute.
a different value each time, so the first row is always safest — Every row or column gives the same value, so no row is ’safer’ than another for correctness, the first row is only convenient when it happens to have the most zeros.
the same value, but only when the matrix has no zero entries — The independence from choice of row or column holds for every matrix, zero entries do not break it. Zeros simply make one choice faster to compute.
a different value, and only the first COLUMN is guaranteed correct — There is nothing special about the first column, expanding along any row or any column gives the same determinant.
The determinant of a $3 \times 3$ matrix is built from one row. Multiply each entry of the first row by its own cofactor, and add the three products:
$|A| = a_{1 1} C_{1 1} + a_{1 2} C_{1 2} + a_{1 3} C_{1 3}$.
This is called EXPANDING the determinant along the first row. Each cofactor is a $2 \times 2$ determinant with its sign. *So one $3 \times 3$ determinant* becomes three $2 \times 2$ ones.
Along row 1, each entry times its cofactor: 2 times 12 + 0 + 1 times minus 3 = 21. The zero needs no working.
Row 2 has two zeros, so one term is left: 3 times 7 = 21. The same answer, for a third of the work.
The same sum works along any row or any column. You always get the same number. A zero entry in the row or column you expand along makes that term $0$, so you have less to work out.
Let us try one with zeros in it. Take $A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & 4 \end{pmatrix}$. Row 2 has two zeros, so expand along row 2.
Worked example
Find $|A|$ by expanding along row 2
$|A| = 0 \cdot C_{2 1} + 3 \cdot C_{2 2} + 0 \cdot C_{2 3}$ Expand along row 2: multiply each entry by its own cofactor, and add.
$|A| = 3 \cdot C_{2 2}$ The two zero entries make their terms $0$.
$C_{2 2} = M_{2 2}$ The sign at $(2, 2)$ is $+$, since $2 + 2$ is even.
$M_{2 2} = \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} = 2 \cdot 4 - 1 \cdot 1 = 7$ Cover row 2 and column 2, and work out the $2 \times 2$ determinant left.
*$|A| = 3 \cdot 7 = 21$* Put the cofactor back into line 2.
$2 \cdot (3 \cdot 4 - 0 \cdot 0) - 0 + 1 \cdot (0 \cdot 0 - 3 \cdot 1) = 24 - 3 = 21$ Check along row 1: the same 21, so the two rows agree.
Square matrices only
A $2 \times 3$ matrix has no determinant at all.
Your turn: expand the same $|A|$ down column 1. Do you get 21 again?
Row 1 and column 1 of this matrix both give 1. Column 1 has a zero in it, so it needs only two minors.
Check yourself
Expanding a $3 \times 3$ determinant along its first row turns the problem into what?
three $2 \times 2$ determinants, one per entry in that row
nine $1 \times 1$ determinants, one per entry in the matrix
one $2 \times 2$ determinant, taken from the middle of the matrix
three $3 \times 3$ determinants, one per row
Check your answer
✓ three $2 \times 2$ determinants, one per entry in that row — (A) Each entry in the first row contributes its cofactor, and each cofactor is itself built from a $2 \times 2$ minor.
nine $1 \times 1$ determinants, one per entry in the matrix — Expansion works along ONE row (or column), three entries, three $2 \times 2$ minors, not every entry in the full $3 \times 3$ grid.
one $2 \times 2$ determinant, taken from the middle of the matrix — Each of the three first-row entries gets its OWN $2 \times 2$ minor, three separate minors are combined, not one taken from the middle.
three $3 \times 3$ determinants, one per row — The whole point of expansion is to shrink the problem, each first-row entry yields a smaller $2 \times 2$ minor, not another full $3 \times 3$.
Expanding $A = \begin{pmatrix} a_{1 1} & a_{1 2} & a_{1 3} \\ a_{2 1} & a_{2 2} & a_{2 3} \\ a_{3 1} & a_{3 2} & a_{3 3} \end{pmatrix}$ along its first row, what does $|A|$ equal?
$a_{1 1} M_{1 1} + a_{1 2} M_{1 2} + a_{1 3} M_{1 3}$ — The formula uses COFACTORS, which already carry the alternating sign, using raw minors drops that sign and can give a wrong total.
$a_{2 1} C_{1 1} + a_{2 2} C_{1 2} + a_{2 3} C_{1 3}$ — The entries and their cofactors must come from the SAME row, mixing second-row entries with first-row cofactors breaks the formula.
✓ $a_{1 1} C_{1 1} + a_{1 2} C_{1 2} + a_{1 3} C_{1 3}$ — (C) Each first-row entry is multiplied by its own cofactor and the three products are summed.
$a_{1 1} C_{1 1} \cdot a_{1 2} C_{1 2} \cdot a_{1 3} C_{1 3}$ — The three entry-cofactor products are ADDED, never multiplied, each term contributes independently to the sum.
Why does picking a row or column with the most zeros make expansion faster?
a zero entry makes its own cofactor equal to zero too
zeros make the whole determinant equal to zero
zeros change the sign pattern of the cofactors
a zero entry kills its whole term with no extra step
Check your answer
a zero entry makes its own cofactor equal to zero too — The cofactor is built from the remaining entries after deleting a row and column, a zero entry says nothing about whether ITS cofactor is zero, only that the PRODUCT with that entry is zero.
zeros make the whole determinant equal to zero — One zero entry in a row does not make the determinant zero, it only removes ONE of the three terms in that row’s expansion; the other terms can still sum to a nonzero value.
zeros change the sign pattern of the cofactors — The sign pattern $(-1)^{i+j}$ depends only on position, a zero-valued entry does not change any cofactor’s sign.
✓ a zero entry kills its whole term with no extra step — (D) A zero entry multiplied by any cofactor is still zero, so that term drops out without any work.
Expanding a $3 \times 3$ determinant along one row produces how many terms to add?
6
3
2
9
Check your answer
6 — Six would come from combining a row and a column together, expansion uses only ONE row or ONE column, giving three terms, not six.
✓ 3 — (B) One row of a $3 \times 3$ matrix has three entries, each contributing one entry-cofactor term.
2 — Each of the THREE entries in the chosen row contributes its own entry-cofactor term, two terms would leave one entry out entirely.
9 — Nine is the entry count for the WHOLE matrix. Expansion uses only one row (or column) at a time, three entries, three terms.
Worked example
Expand a $3 \times 3$ determinant along the first row
$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$ You are given $A$. Expand along row 1, so you need the three cofactors of that row.
$M_{1 1} = \begin{vmatrix} 1 & 4 \\ 6 & 0 \end{vmatrix} = 1 \cdot 0 - 4 \cdot 6 = -24$ Cover row 1 and column 1. Work out the $2 \times 2$ determinant that is left.
$C_{1 1} = -24$, $C_{1 2} = -(-20) = 20$, $C_{1 3} = -5$ The signs along row 1 are $+$, $-$, $+$. So only $C_{1 2}$ changes sign.
$|A| = 1 \cdot (-24) + 2 \cdot 20 + 3 \cdot (-5) = -24 + 40 - 15$ Multiply each entry of row 1 by its own cofactor, and add.
*$|A| = 1$* The three terms add to 1.
$0 \cdot 18 + 1 \cdot (-15) + 4 \cdot 4 = 1$ Check along row 2. Its cofactors are $C_{2 1} = 18$, $C_{2 2} = -15$ and $C_{2 3} = 4$, and the sum is 1 again.
Check yourself
For $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$, expanding along the first row, what is the minor $M_{1 1}$?
$1$
$24$
$-24$
$-20$
Check your answer
$1$ — $1$ is the FINAL determinant $|A|$ after combining all three terms, $M_{1 1}$ is just one intermediate minor along the way.
$24$ — $1 \cdot 0 - 4 \cdot 6 = 0 - 24 = -24$, the sign is part of the value; dropping it changes the answer.
$-20$ — $-20$ is $M_{1 2}$ (delete column 2), not $M_{1 1}$ (delete column 1), the two minors use different rows/columns of the remaining matrix.
For $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$, what does the cofactor $C_{1 2}$ equal?
$-20$
$-24$
$20$
$40$
Check your answer
$-20$ — $M_{1 2} = -20$ is the raw minor; position $(1,2)$ has $i+j=3$, odd, so the sign flips it to $C_{1 2} = 20$.
$-24$ — $-24$ is $C_{1 1}$, a different position with its own minor and its own sign, not $C_{1 2}$.
✓ $20$ — (C) $M_{1 2} = -20$, and position $(1,2)$ has $i+j=3$, odd, so $C_{1 2} = -(-20) = 20$.
$40$ — $40 = 2 \cdot 20$ mixes in the entry $a_{1 2}=2$ too early, the cofactor $C_{1 2}$ itself is $20$; multiplying by the entry is the NEXT step, not part of the cofactor.
For $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$, what is $|A|$?
$-79$
$-24$
$79$
$1$
Check your answer
$-79$ — Using raw minors instead of signed cofactors gives $1(-24)+2(-20)+3(-5) = -79$ , only the middle term’s sign was wrong, but it changes the whole total.
$-24$ — $-24$ is only the FIRST of three terms, the full expansion adds all three entry-cofactor products, $-24+40-15$, before reaching $1$.
$79$ — $79$ comes from taking every product as positive, $24 + 40 + 15 = 79$, instead of carrying each term’s own sign through the sum.
Here is a working for the matrix of the worked example, $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$. Which step is wrong?
First try
I worked out the minors $M_{1 1}=-24$, $M_{1 2}=-20$ and $M_{1 3}=-5$ for the worked example’s matrix $A$, then multiplied each one straight into row 1 without checking any sign.
Second look
$C_{1 2}$ actually carries a minus sign, since $1+2=3$ is odd, so $C_{1 2} = -M_{1 2} = 20$, not $-20$. Using the raw minors instead of the signed cofactors swings the determinant from $1$ to $-79$, eighty apart.
A minor and its cofactor are the same number only when $i+j$ is even. Write the sign in before you multiply, never after.
The working step by step
Cover row 1 and each column in turn. The minors are $M_{1 1} = -24$, $M_{1 2} = -20$ and $M_{1 3} = -5$.
Multiply each entry of row 1 by its minor, and add: $1 \cdot (-24) + 2 \cdot (-20) + 3 \cdot (-5)$.
So $|A| = -24 - 40 - 15 = -79$.
First try
−79
Check
1
First try: use the minors as they are, and get minus 79. Check: position one, two carries a minus, so the answer is 1.
Only position one, two sits on a minus square. The two corner terms keep their sign either way, which is why the slip hides.
On one number line, signed cofactors give 1 and raw minors give minus 79. The two answers are 80 apart.
The three minors are right. The worked example found the same ones. But it found $|A| = 1$, not $-79$.
Where the sign actually sits
Look at the second step. It puts minors where the cofactors belong. The sign at $(1, 2)$ is $-$, because $1 + 2 = 3$ is odd. *So $C_{1 2} = 20$*, and the middle term is $+40$, not $-40$. The places $(1, 1)$ and $(1, 3)$ have an even $i + j$. Their signs are $+$, so there the minor and the cofactor are the same.
With signs: minus 24 + 40 minus 15 = 1. Without them: minus 79. One term differs; the answers are 80 apart.
That is why the mistake is hard to see. Two of the three terms come out right. One sign moves the answer by 80.
Write the sign immediately
Check the sign before you multiply, not after: by the time you have multiplied, a dropped sign already looks like part of the answer.
Check yourself
A student computes $|A|$ for $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$ using the raw minors $-24, -20, -5$ with no sign flip, getting $-79$. What went wrong?
the alternating sign $(-1)^{i+j}$ was skipped
the minors themselves were computed incorrectly
the wrong row was chosen for expansion
the entries $a_{1 1}, a_{1 2}, a_{1 3}$ were multiplied in the wrong order
Check your answer
✓ the alternating sign $(-1)^{i+j}$ was skipped — (A) The cofactor is the SIGNED minor; using the raw minor at position $(1,2)$ (odd, so it should flip) leaves the wrong sign in the sum.
the minors themselves were computed incorrectly — The minors $-24,-20,-5$ are correct as computed, the error is entirely in the next step, applying the alternating sign to turn them into cofactors.
the wrong row was chosen for expansion — The first row is a perfectly valid choice for expansion, the error is not which row was picked, it is that the sign step was skipped afterward.
the entries $a_{1 1}, a_{1 2}, a_{1 3}$ were multiplied in the wrong order — Multiplication order never changes a product’s value, the actual error is the missing sign flip, not the order the entries were multiplied in.
Comparing the correct expansion ($|A|=1$) against the sign-forgotten one ($-79$) for $A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$, which single term differs between the two?
the middle term, $(1,2)$
the last term, from position $(1,3)$
the first term, from position $(1,1)$
all three terms differ equally
Check your answer
✓ the middle term, $(1,2)$ — (A) Only $(1,2)$ has $i+j=3$, odd, the two corner positions $(1,1)$ and $(1,3)$ have even $i+j$ and keep their sign either way.
the last term, from position $(1,3)$ — $(1,3)$ has $i+j=4$, even, like $(1,1)$, its sign is unaffected by skipping the alternating rule.
the first term, from position $(1,1)$ — $(1,1)$ has $i+j=2$, even, its cofactor sign is $+$ regardless of whether the alternating rule is applied, so this term is unaffected.
all three terms differ equally — Only the odd-position term changes, the two even-position terms, $(1,1)$ and $(1,3)$, come out identical whether or not the sign rule is applied, which is exactly why the error is easy to miss.
Why can the sign-forgotten error be easy to miss on a small example?
rounding errors mask the mistake in small examples
small matrices never actually need cofactor signs
the student got lucky and the answer happened to still be correct, the way a guess sometimes lands right
the other two terms sit at even positions, so they come out right either way
Check your answer
rounding errors mask the mistake in small examples — There is no rounding anywhere in this computation, every value is an exact integer, so rounding cannot be why the error hides.
small matrices never actually need cofactor signs — Size is irrelevant: the alternating sign rule applies to every square matrix. The slip hides because two of the three terms sit at even positions, not because the matrix is small.
the student got lucky and the answer happened to still be correct, the way a guess sometimes lands right — In this example the sign-forgotten method does give a wrong answer, $-79$ for $1$, so luck did not save it. The slip is easy to miss because two of the three terms still come out right.
✓ the other two terms sit at even positions, so they come out right either way — (D) In row 1 only $(1,2)$ has an odd $i+j$. At $(1,1)$ and $(1,3)$ the sum is even, so the minor and the cofactor are the same number, and two of the three terms come out right either way.
Which habit correctly guards against the sign-forgotten error?
compute all the minors first, then decide on signs afterward for the whole set
write the sign beside each cofactor as it is worked out, not at the end
skip the sign whenever the minor is already a large number
only apply the sign to the first term in the expansion
Check your answer
compute all the minors first, then decide on signs afterward for the whole set — Deferring the sign to a separate later pass is exactly the workflow that lets it get forgotten, attaching the sign at the same moment as each minor closes that gap.
✓ write the sign beside each cofactor as it is worked out, not at the end — (B) Recording the sign at the moment each cofactor is computed catches a missed flip before it propagates into the final sum.
skip the sign whenever the minor is already a large number — The sign rule depends only on position $(i,j)$, never on how large the minor’s numeric value happens to be.
only apply the sign to the first term in the expansion — Every term gets its OWN sign from its OWN position, applying the sign only to the first term leaves the rest of the terms unsigned.
Four facts tell you how $|A|$ changes when you change a row. Here they are tools. You use them to make zeros before you expand, and to see the product rule at work. Each fact holds for columns too.
SWAP two rows, and $|A|$ changes sign. For a $2 \times 2$, $\begin{vmatrix} c & d \\ a & b \end{vmatrix} = c b - d a$, which is $-(a d - b c)$. For a $3 \times 3$, expand along the row that does not move. Each of its $2 \times 2$ minors has its two rows swapped, so each minor changes sign, and so does the total.
Two EQUAL rows give $|A| = 0$. Swapping them changes nothing, yet a swap must change the sign. So $|A| = -|A|$, and only $0$ does that.
SCALE one row by a number $k$, and $|A|$ is multiplied by $k$. Expand along that row, and every term carries one factor $k$.
ADD $k$ times one row to another row, and $|A|$ does not change. The extra terms add up to $k$ times a determinant with two equal rows, which is $0$.
Operation
Effect on $|A|$
Swap two rows
changes sign
Two equal rows
$|A| = 0$
Scale one row by $k$
multiplied by $k$
Add $k$ times one row to another
unchanged
Swap the two rows and the determinant changes sign: 5 becomes minus 5. Its absolute value stays the same.
The same parallelogram twice. Rows (3, 1) then (1, 2) give 5. Swapped, the turn runs clockwise and the determinant is minus 5.
Two equal rows give 0. Swapping them changes nothing, yet the swap must flip the sign, and only 0 equals its own negative.
Double row 1 and the determinant doubles: 5 becomes 10. Read backwards, a common factor of one row comes outside.
Add 2 times row 1 to row 2. The rows change, the determinant stays 5. This is the move that makes zeros.
Left: doubling the first row takes the area from 5 to 10. Right: adding twice the first row to the second keeps it at 5.
Subtract row 1 from rows 2 and 3. Column 1 becomes 1, 0, 0, and down it, one term is left: 1.
Let us use the last fact to make zeros. Take $\begin{vmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 1 & 4 & 8 \end{vmatrix}$.
Worked example
Make two zeros in column 1, then expand
$\begin{vmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 1 & 4 & 8 \end{vmatrix} = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 1 & 4 & 8 \end{vmatrix}$ Subtract row 1 from row 2: that is ADD with $k = -1$, so the value stays.
$= \begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 2 & 5 \end{vmatrix}$ Subtract row 1 from row 3 the same way.
$= 1 \cdot \begin{vmatrix} 1 & 2 \\ 2 & 5 \end{vmatrix}$ Expand down column 1: its two zeros make their terms $0$.
*$= 1 \cdot (1 \cdot 5 - 2 \cdot 2) = 1$* Work out the $2 \times 2$ determinant that is left.
$1 \cdot 4 - 2 \cdot 3 + 3 \cdot 1 = 1$ Check along row 1 of the first grid: three minors, 4, 3 and 1, and the same answer.
Scaling the whole matrix
SCALE also tells you what happens to $k A$. Multiplying every entry by $k$ multiplies every row by $k$. Take the factor $k$ out of one row at a time. For $3 A$, with three rows:
$|3 A| = 3 \cdot 3 \cdot 3 \cdot |A| = 27 |A|$.
*One $k$ per row.* For an $n \times n$ matrix, $|k A| = k^n |A|$.
Your turn: two rows of a $3 \times 3$ matrix are $(1, 2, 3)$ and $(2, 4, 6)$. What is $|A|$?
Check yourself
Swapping any two rows of a matrix does what to its determinant?
doubles its value
leaves it completely unchanged, the way the add rule does
flips its sign, not its absolute value
sets it to zero
Check your answer
doubles its value — Swapping rows changes only the SIGN of $|A|$, never its size, the magnitude stays exactly the same.
leaves it completely unchanged, the way the add rule does — Leaving the determinant unchanged is what the ADD rule does, adding a multiple of one row to another. Swapping two rows flips the sign instead.
✓ flips its sign, not its absolute value — (C) The SWAP rule: exchanging two rows (or two columns) reverses the sign of $|A|$, leaving its size unchanged.
sets it to zero — A determinant of zero comes from two IDENTICAL rows, not from swapping two different ones, swapping only flips the sign.
If two rows of a matrix are identical, what does $|A|$ equal?
$1$
undefined, since the matrix is not really square anymore
$0$, the only value that flips sign and stays the same
the value of one of the repeated rows
Check your answer
$1$ — $1$ is the identity for multiplication, not a consequence of two rows being identical, the swap-and-identical argument forces $0$, not $1$.
undefined, since the matrix is not really square anymore — Repeating a row does not change how many rows or columns the matrix has, it is still square, and its determinant is a well-defined number: $0$.
✓ $0$, the only value that flips sign and stays the same — (C) Swapping two identical rows must both flip the sign and leave the determinant unchanged, the only value that satisfies both is $0$.
the value of one of the repeated rows — The determinant is one number computed from the whole matrix, not a copy of any one row’s entries, identical rows force it specifically to $0$.
Multiplying every entry of one row by a scalar $k$ does what to $|A|$?
multiplies $|A|$ by $k$
adds $k$ to $|A|$
multiplies $|A|$ by $k^2$
leaves $|A|$ unchanged, since $k$ only affects one row
Check your answer
✓ multiplies $|A|$ by $k$ — (A) The SCALE rule: a common factor $k$ in one row (or column) can be pulled straight out of the determinant.
adds $k$ to $|A|$ — Scaling a row multiplies the determinant by $k$, adding $k$ describes a different kind of change entirely, not this rule.
multiplies $|A|$ by $k^2$ — Only ONE row is scaled by $k$, so only one factor of $k$ comes out, $k^2$ would be the result of scaling two rows, not one.
leaves $|A|$ unchanged, since $k$ only affects one row — The determinant depends on ALL rows together, so scaling even one row does change it, by the factor $k$, not by nothing.
Adding a scalar multiple of one row to another row does what to $|A|$?
flips its sign, the same as a swap
scales it by the multiple used
sets it to zero, the same as having identical rows
leaves it completely unchanged
Check your answer
flips its sign, the same as a swap — Only swapping two rows flips the sign. Adding a multiple of one row to another is a completely different operation, with no effect on $|A|$ at all.
scales it by the multiple used — Scaling changes a row’s OWN entries by a factor and scales $|A|$ by that factor. Adding a multiple of one row to a DIFFERENT row leaves $|A|$ untouched instead.
sets it to zero, the same as having identical rows — This rule does not create identical rows, so it does not force $0$ the way that case does, adding a multiple of one row to another leaves $|A|$ exactly as it was.
✓ leaves it completely unchanged — (D) The ADD rule is the one operation that costs nothing, it is exactly the move row-reduction uses to create zeros before expanding.
▲CONCEPT
One rule about products
Chapter 3 ended with a matrix and its inverse. Multiply a matrix by its inverse, and you get the identity, the matrix that changes nothing. The determinant has to fit with that fact. To see how, you need one rule about products.
Multiply the matrices first, or the determinants first: the determinant of M N is 6, and 2 times 3 is 6.
Parallelograms on the rows of M, N and their product M N have areas 2, 3 and 6, and 2 times 3 is 6.
Stating the product rule
For two square matrices $A$ and $B$ of the same order, $|A B| = |A| \cdot |B|$. The determinant of a product is the product of the determinants. This is the PRODUCT RULE.
Let us check it on numbers. Take $M = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}$ and $N = \begin{pmatrix} 1 & 0 \\ 1 & 3 \end{pmatrix}$.
Worked example
Check $|M N| = |M| \cdot |N|$ on numbers
$|M| = 2 \cdot 1 - 1 \cdot 0 = 2$ Work out $a d - b c$ for $M$.
$|N| = 1 \cdot 3 - 0 \cdot 1 = 3$ The same for $N$.
$|M N| = 3 \cdot 3 - 3 \cdot 1 = 6$ Take the determinant of the product.
*$|M N| = 6 = 2 \cdot 3 = |M| \cdot |N|$* The determinant of the product is the product of the two determinants.
The SWAP fact shows the rule at work. Let $E = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Row times column shows that $E A$ is $A$ with its two rows swapped.
Worked example
Check the rule on a swap
$|E A| = -|A|$ SWAP: swapping the two rows changes the sign.
$|E| = 0 \cdot 0 - 1 \cdot 1 = -1$ Work out $a d - b c$ for $E$.
*$|E A| = -1 \cdot |A| = |E| \cdot |A|$* So the product rule holds for $E$ and $A$.
The SCALE and ADD facts fit the rule in the same way.
A and A inverse multiply to I, so their determinants multiply to 1: here 5 times one fifth = 1.
1
→
× |A|
|A| = 5
→
× |A⁻¹|
|A⁻¹| = 1/5
→
1
Times the determinant of A, then of A inverse, and 1 comes back. The two determinants are reciprocals, like A and its inverse.
Now go back to the inverse. Take an invertible matrix $A$, and follow its determinant:
Why it works
An invertible matrix has $|A| \neq 0$
$A A^{-1} = I$ Start from what an inverse does.
$|A A^{-1}| = |I|$ Take the determinant of both sides: equal matrices have equal determinants.
$|A| \cdot |A^{-1}| = 1$ Use the product rule on the left, and $|I| = 1$ on the right.
*$|A| \neq 0$ and $|A^{-1}| = 1 / |A|$* Two numbers multiply to 1 only if neither is 0, so divide by $|A|$.
1
→
× |A|
|A| = 0
→
× ?
any number
If the determinant of A is 0, times 0 stays 0; no number times 0 gives 1. A matrix with determinant 0 has no inverse.
Put another way, if $|A| = 0$, no matrix $B$ can give $|A| \cdot |B| = 1$. So a matrix with $|A| = 0$ is not invertible.
Your turn: $|A| = 4$. What is $|A^{-1}|$?
Check yourself
For square matrices $A$ and $B$ of the same order, what does $|A B|$ equal?
$|A|$ alone, since multiplying by $B$ does not change the determinant
$|A| \cdot |B|$
$|A \cdot B|$ divided by $2$
$|A| + |B|$
Check your answer
$|A|$ alone, since multiplying by $B$ does not change the determinant — Multiplying by $B$ absolutely changes the determinant, by a factor of $|B|$, ignoring $B$’s contribution drops half the rule.
✓ $|A| \cdot |B|$ — (B) The determinant of a product is the product of the determinants , this holds for every pair of same-order square matrices.
$|A \cdot B|$ divided by $2$ — No division enters the product rule at all, $|A B|$ equals $|A| \cdot |B|$ exactly, with no extra factor.
$|A| + |B|$ — Addition describes a different (and false) claim about sums, covered elsewhere in this chapter, the product rule is genuinely multiplicative: $|A B| = |A| \cdot |B|$.
Given $|A| \neq 0$, what does the product rule say about $|A^{-1}|$?
$|A^{-1}| = |A|$
$|A^{-1}| = 1/|A|$
$|A^{-1}| = -|A|$
$|A^{-1}|$ cannot be found from $|A|$ alone
Check your answer
$|A^{-1}| = |A|$ — $|A| \cdot |A^{-1}|$ must equal $|I|=1$, that only holds if $|A^{-1}|$ is the RECIPROCAL of $|A|$, not equal to it (unless $|A|=1$).
✓ $|A^{-1}| = 1/|A|$ — (B) From $A \cdot A^{-1} = I$ and $|I|=1$, the product rule gives $|A| \cdot |A^{-1}| = 1$, so $|A^{-1}| = 1/|A|$.
$|A^{-1}| = -|A|$ — Negating $|A|$ does not satisfy $|A| \cdot |A^{-1}| = 1$ in general, the reciprocal is the value that makes the product equal $1$.
$|A^{-1}|$ cannot be found from $|A|$ alone — The product rule, applied to $A \cdot A^{-1} = I$, gives $|A^{-1}|$ directly as $1/|A|$, no extra information about $A^{-1}$ is needed.
How does the product rule confirm that a matrix is invertible exactly when $|A| \neq 0$?
it is a separate fact, unrelated to the product rule
the product rule only applies when $|A| \neq 0$ to begin with
if $A^{-1}$ existed with $|A|=0$, then $|A| \cdot |A^{-1}|$ would be $0$, never $1$
it only shows invertible matrices have SOME determinant, not that it must be nonzero
Check your answer
it is a separate fact, unrelated to the product rule — The link is direct: applying the product rule to $A \cdot A^{-1}=I$ is exactly the argument that rules out an inverse when $|A|=0$.
the product rule only applies when $|A| \neq 0$ to begin with — The product rule holds unconditionally for any two same-size square matrices, it is precisely BECAUSE it holds even when $|A|=0$ that the contradiction (needing $0=1$) proves no inverse can exist there.
✓ if $A^{-1}$ existed with $|A|=0$, then $|A| \cdot |A^{-1}|$ would be $0$, never $1$ — (C) $A \cdot A^{-1}=I$ forces $|A| \cdot |A^{-1}| = 1$ by the product rule, impossible if $|A|=0$, since anything times $0$ is $0$, not $1$.
it only shows invertible matrices have SOME determinant, not that it must be nonzero — The argument does more than show ’some’ determinant exists, it specifically shows $|A|=0$ is impossible for an invertible matrix, since it would force $0=1$.
✕MISCONCEPTION
(3 + 4)²
49
3² + 4²
25
√(9 + 16)
5
√9 + √16
7
3 plus 4 squared is 49, not 25. The root of 9 plus 16 is 5, not 7. The same wrong rule applies to determinants.
In Class 9 you learned that the square of a sum is not the sum of the squares. Take 3 and 4. Their sum is 7, and 7 squared is 49. But 3 squared plus 4 squared is only 25. Square roots behave the same way. The square root of 9 plus 16 is 5, not 3 plus 4.
So $(a + b)^2 \neq a^2 + b^2$, and $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$. Squaring works on the sum as a whole, not on its parts. So does taking a square root.
The same wrong rule for determinants
The determinant of a sum tempts the same wrong rule. The product rule says $|A B| = |A| \cdot |B|$. There is no such rule for addition. $|A + B|$ is not $|A| + |B|$.
Determinant of A plus determinant of B gives 4. Determinant of A plus B gives 13. Add the matrices first, then take one determinant.
Parallelograms for A, B and A plus B hold 2, 6 and 13 unit squares. The determinants, minus 2 and 6, add to just 4.
Let us test it on one pair. Take $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$. Predict $|A + B|$ using the rule above: work out $|A|$, $|B|$, and add them.
Worked example
Test $|A + B|$ against $|A| + |B|$
$|A| = 1 \cdot 4 - 2 \cdot 3 = -2$ Work out $a d - b c$ for $A$.
$|B| = 2 \cdot 3 - 0 \cdot 1 = 6$ The same for $B$.
$A + B = \begin{pmatrix} 3 & 2 \\ 4 & 7 \end{pmatrix}$ Add the matrices entry by entry, as in Chapter 3.
$|A + B| = 3 \cdot 7 - 2 \cdot 4 = 13$ Take one determinant of the sum.
*$|A| + |B| = -2 + 6 = 4$, not 13* The two numbers differ, so one pair is enough to show the rule is false.
No formula links the two
There is no formula for $|A + B|$ built from $|A|$ and $|B|$ alone. So add the matrices first. Then take one determinant.
Your turn: take $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Find $|A + B|$ and $|A| + |B|$.
A SCHOLAR INDIA REMEMBERS
You predicted that $|A + B|$ would be $|A| + |B| = -2 + 6 = 4$. Then you worked it out and got $13$. Which do you change, the prediction or the working? Check the working first: $21 - 8 = 13$. It holds. So the rule behind the prediction is what fails, and the rule goes. One case that disagrees is enough to drop it. No number of cases that agree would prove it.
Brahmagupta · the brahminy kite
Check yourself
Since $|A B| = |A| \cdot |B|$ is true, does $|A + B| = |A| + |B|$ also hold in general?
no, there is no general formula for $|A+B|$ from $|A|$ and $|B|$ alone
yes, the same multiplicative logic carries over to addition
yes, but only for $2 \times 2$ matrices
no, but only because $A$ and $B$ were the wrong kind of matrix in this example
Check your answer
✓ no, there is no general formula for $|A+B|$ from $|A|$ and $|B|$ alone — (A) The product rule is genuinely multiplicative; addition does not commute with the determinant the same way.
yes, the same multiplicative logic carries over to addition — The product rule being true does not carry over to addition, $|A B|=|A||B|$ and $|A+B|=|A|+|B|$ are separate claims, and only the first is true in general.
yes, but only for $2 \times 2$ matrices — Matrix size does not rescue the additive claim, $|A+B| \neq |A|+|B|$ in general, whatever the dimension.
no, but only because $A$ and $B$ were the wrong kind of matrix in this example — The additive claim fails in general, for essentially any pair of matrices, it is not a quirk of one particular example’s $A$ and $B$.
For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$, what is $|A+B|$?
$-2$
$6$
$4$
$13$
Check your answer
$-2$ — $-2$ is $|A|$ alone, correct as a determinant but it skips $B$ completely, there’s no way to reach $|A+B|$ without first forming the sum.
$6$ — Dropping $A$ entirely keeps only $|B|$. The task needs the combined matrix’s determinant, not either matrix’s own determinant in isolation.
$4$ — $4$ is $|A|+|B|$, the sum of the two SEPARATE determinants, the question asks for $|A+B|$, which requires adding the matrices first, then taking one determinant.
For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$, how does $|A+B|$ compare to $|A|+|B|$?
$13 = 4$ once rounding is accounted for
$13$ and $4$ are both wrong; the correct value is $|A||B|$
$13 \neq 4$: $|A+B|=13$ while $|A|+|B|=4$
they are equal only because $A$ and $B$ happen to be $2 \times 2$
Check your answer
$13 = 4$ once rounding is accounted for — Both $13$ and $4$ are exact whole numbers, there is no rounding involved anywhere in this computation, so no rounding can explain away the difference.
$13$ and $4$ are both wrong; the correct value is $|A||B|$ — $|A||B|=-12$ is the answer to a completely different question ($|A B|$, by the product rule), it has no bearing on $|A+B|$, which was computed directly as $13$.
✓ $13 \neq 4$: $|A+B|=13$ while $|A|+|B|=4$ — (C) $|A+B|=13$ while $|A|+|B| = -2+6=4$, a direct numerical counterexample to the additive claim.
they are equal only because $A$ and $B$ happen to be $2 \times 2$ — Matrix size is not the reason the claim fails, it is not true for $2 \times 2$ matrices, and it is not true for larger ones either.
Why is there no general formula for $|A+B|$ built from $|A|$ and $|B|$ alone?
the determinant is built from products, not entrywise addition
because $|A|$ and $|B|$ are not always both nonzero
because addition of matrices is not always defined
because the determinant is only defined up to a constant multiple
Check your answer
✓ the determinant is built from products, not entrywise addition — (A) The determinant is defined through products of entries across rows and columns; that structure interacts cleanly with matrix multiplication but not with entrywise addition.
because $|A|$ and $|B|$ are not always both nonzero — Whether $|A|$ or $|B|$ is zero or nonzero makes no difference, there is still no general formula for $|A+B|$ in terms of them.
because addition of matrices is not always defined — Matrix addition is perfectly well defined for same-size matrices, $A+B$ was computed with no trouble; the issue is that no formula in $|A|,|B|$ alone predicts $|A+B|$.
because the determinant is only defined up to a constant multiple — The determinant is one precise number for a given matrix, nothing about it is ’up to a constant.’ The real reason is structural: it does not commute with addition the way it does with multiplication.
Three points, one, one, two, two, three, three, lie on one line: no triangle. Move the third point to three, five: a thin triangle opens.
In Class 9 you plotted points on graph paper. Plot three points and join them, and you usually get a triangle. Now try the points (1, 1), (2, 2) and (3, 3). They lie on one straight line, so they make no triangle at all. The area is 0. Move the third point up to (3, 5), and a thin triangle opens.
The coordinates go in two columns, with a column of ones beside them. Delta is 2, and the area is half that: 1.
List the corners the other way round and delta becomes minus 2. The triangle has not changed, so the area uses the absolute value of delta.
Left: (1, 1), (2, 2), (3, 3) on one line, area 0. Right: the third point at (3, 5) gives a triangle, area 1.
One determinant gives the area from the three corners. For corners $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$, build
$\Delta = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$.
The coordinates fill two columns, and the third column is all 1s.
The area of the triangle is $(1/2) |\Delta|$. Here the bars mean the absolute value of the number $\Delta$. If you list the corners in the other order, the sign of $\Delta$ changes. An area is never negative, so you take the absolute value.
*Expand $\Delta$* down the column of 1s: $\Delta = x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)$, the same sum the Class 11 coordinate formula uses.
Checking both cases by hand
Let us check both cases. Expand each $\Delta$ along row 1.
*Area $= (1/2) \cdot 2 = 1$* The area is half the absolute value of $\Delta$.
Zero, one, two, two and four, three give delta equals 0. They lie on one straight line, so there is no triangle to measure.
*So $\Delta = 0$ exactly when the three points lie on one line. Points on one line are called COLLINEAR*. The same determinant answers two questions. How big is the triangle? And is there a triangle at all?
Your turn: are the points (1, 2), (2, 5) and (3, 8) collinear?
A SCHOLAR INDIA REMEMBERS
List the corners $(1, 1)$, $(4, 2)$, $(2, 4)$ and you get $\Delta = 8$. List them as $(1, 1)$, $(2, 4)$, $(4, 2)$ and you get $\Delta = -8$. Is a negative number allowed here? As an area, no. An area is never below $0$. But the minus is not a mistake either. It records which way round the corners you went. So keep the sign while you work, and drop it only for the area.
Brahmagupta · the brahminy kite
Check yourself
For vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$, with $\Delta = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$, what does the area of the triangle equal?
$|\Delta|$, without the factor of $1/2$
$(1/2) |\Delta|$, half the absolute value of the determinant
$\Delta$ itself, with no factor of $1/2$ and no absolute value
$(1/2) \Delta$, without the absolute value
Check your answer
$|\Delta|$, without the factor of $1/2$ — Without the $1/2$, the result is exactly TWICE the true area, the factor of $1/2$ is required to land on the actual area.
✓ $(1/2) |\Delta|$, half the absolute value of the determinant — (B) The area formula packages the three vertices into a $3 \times 3$ determinant, then takes half its absolute value.
$\Delta$ itself, with no factor of $1/2$ and no absolute value — $\Delta$ alone can be negative (depending on vertex order) and is exactly twice the actual area, both the $1/2$ and the absolute value are required.
$(1/2) \Delta$, without the absolute value — Without the absolute value, listing the vertices the other way round, clockwise instead of anticlockwise would make this ’area’ negative, area itself is never negative.
Why does the area formula take the absolute value of $\Delta$?
vertex order can flip $\Delta$’s sign, but area is never negative
to correct a rounding error introduced by the determinant computation
because $\Delta$ would otherwise be measured in different units than area
because $\Delta$ is always negative for a valid triangle
Check your answer
✓ vertex order can flip $\Delta$’s sign, but area is never negative — (A) $\Delta$ is a signed quantity by construction; the absolute value strips the sign so the result matches what area actually means.
to correct a rounding error introduced by the determinant computation — There is no rounding in computing $\Delta$, it is an exact value from exact arithmetic. The absolute value corrects a sign issue, not a rounding one.
because $\Delta$ would otherwise be measured in different units than area — There is no unit mismatch, $\Delta$ and $(1/2)|\Delta|$ are measured on the same scale; the absolute value only removes the sign, not a unit.
because $\Delta$ is always negative for a valid triangle — $\Delta$’s sign flips with the order the vertices are listed in, it is not always negative; the absolute value is needed precisely because the sign can go either way.
If $\Delta = 0$ for three given points, what does that mean?
the three points form a right triangle
the three points lie on a single straight line
the triangle has the smallest possible positive area
one of the three points must be the origin
Check your answer
the three points form a right triangle — $\Delta=0$ carries no information about any angle in a triangle, it specifically signals that the three points are collinear, not that a right angle exists.
✓ the three points lie on a single straight line — (B) A zero value for the same determinant used in the area formula means the three points sit on one line.
the triangle has the smallest possible positive area — $\Delta=0$ is exactly zero, not a small positive value, it means the three points do not form a triangle at all; they lie on one line.
one of the three points must be the origin — Three collinear points can sit anywhere in the plane, none of them needs to be the origin for $\Delta$ to come out zero.
How does the area-of-a-triangle computation relate to the rest of this chapter’s determinant work?
it needs an entirely separate computation method not covered elsewhere in the chapter
the same $3 \times 3$ mechanics, new geometry only
it only works for right triangles, unlike the general determinant methods
it replaces cofactor expansion with a shortcut that skips minors entirely
Check your answer
it needs an entirely separate computation method not covered elsewhere in the chapter — No new computation method is introduced, the area formula reuses the same $3 \times 3$ cofactor expansion and row/column properties already used for determinants.
✓ the same $3 \times 3$ mechanics, new geometry only — (B) The row/column properties and cofactor expansion already used for computing determinants apply unchanged here, only the interpretation (area, collinearity) is new.
it only works for right triangles, unlike the general determinant methods — The formula works for any triangle from any three points, not only right triangles, nothing in its derivation depends on any particular angle.
it replaces cofactor expansion with a shortcut that skips minors entirely — $\Delta$ is still a $3 \times 3$ determinant, expanded exactly the way earlier examples were, by minors and signed cofactors, with no shortcut that skips them.
The triangle on (1, 0), (2, 2) and (0, 3). Its determinant is 5, and half of 5 gives the area, 2.5 square units.
Worked example
Find the area of a triangle from its three vertices
Vertices $(1, 0)$, $(2, 2)$, $(0, 3)$ You are given three corners. Build $\Delta$ from them.
$\Delta = \begin{vmatrix} 1 & 0 & 1 \\ 2 & 2 & 1 \\ 0 & 3 & 1 \end{vmatrix}$ Put the coordinates in the first two columns and 1s in the third.
$\Delta = 1 \cdot \begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 2 \\ 0 & 3 \end{vmatrix}$ Expand along row 1. Its middle entry is $0$, so the middle term drops out.
$\Delta = 1(2 - 3) - 0 + 1(6 - 0) = -1 + 6 = 5$ Work out each $2 \times 2$ determinant and add.
Area $= (1/2) |\Delta| = (1/2)(5)$ The area is half the absolute value of $\Delta$.
*Area $= 2.5$ square units* $\Delta = 5$ is already positive, so its absolute value is 5.
$6 - 1.5 - 1 - 1 = 2.5$ Check on graph paper. The box from $(0, 0)$ to $(2, 3)$ has area 6. Cut away the three right triangles around the shape, of areas 1.5, 1 and 1. What is left is 2.5.
Check yourself
For vertices $(1,0)$, $(2,2)$, $(0,3)$, what value does expanding $\Delta = \begin{vmatrix} 1 & 0 & 1 \\ 2 & 2 & 1 \\ 0 & 3 & 1 \end{vmatrix}$ along the first row give?
$2.5$ — $2.5$ is the FINAL area, after halving, $\Delta$ itself, before the $1/2$ is applied, is $5$.
$-1$ — $-1$ is only the first term, $1(2-3)$, the expansion has a second nonzero term, $1(6-0)=6$, still to add.
$7$ — Adding the terms as if both were positive gives $1+6=7$, the first term is actually $-1$, so the correct sum is $-1+6=5$.
For vertices $(1,0)$, $(2,2)$, $(0,3)$, what is the triangle’s area?
$10$ square units
$5$ square units
$-2.5$ square units
$2.5$ square units
Check your answer
$10$ square units — Doubling $\Delta$ moves in the wrong direction, the formula HALVES $|\Delta|$ to get area, giving $2.5$, not $10$.
$5$ square units — $5$ is $\Delta$, before halving, the area is $(1/2)(5)=2.5$ square units, not $\Delta$ itself.
$-2.5$ square units — Area is never negative, the formula takes the absolute value of $\Delta$ before halving, so a negative result cannot be the final answer.
✓ $2.5$ square units — (D) Area $= (1/2)|\Delta| = (1/2)(5) = 2.5$ square units.
Using the same $\Delta$-from-three-points method, how would you check whether $(1,0)$, $(2,2)$, $(0,3)$ are collinear?
check whether any two of the three points share a coordinate
compute the distances between each pair of points and compare them
recompute $\Delta$ using a different set of vertices entirely
compute the same $\Delta$ and check whether it equals $0$
Check your answer
check whether any two of the three points share a coordinate — Two points sharing an x- or y-coordinate does not make three points collinear in general, the actual test is whether $\Delta=0$.
compute the distances between each pair of points and compare them — Comparing pairwise distances is a valid geometric idea but not the method this chapter builds, the determinant test uses the same $\Delta$ already computed for area.
recompute $\Delta$ using a different set of vertices entirely — The collinearity test uses the exact same three points already given, there is no need to substitute in a different set of vertices.
✓ compute the same $\Delta$ and check whether it equals $0$ — (D) Collinearity uses the identical determinant $\Delta$, a nonzero value (like $5$ here) rules collinearity out.
The ADJOINT of a square matrix $A$ is built from its cofactors in two steps. It is written $\text{adj} A$.
Replace every entry by its cofactor, then transpose. The cofactor 20 at position one, two ends up at position two, one of the adjoint.
First replace each entry $a_{i j}$ by its cofactor $C_{i j}$. This gives the COFACTOR MATRIX.
Then take its transpose, so that rows become columns, as in Chapter 3.
Skip the transpose and six of the nine entries are in the wrong place. Only the diagonal agrees, so the slip looks right.
So the entry in row $i$, column $j$ of $\text{adj} A$ is $C_{j i}$. The second step is the one that gets left out. Every cofactor can be right and the adjoint still wrong.
Working one example by hand
Let us build one. Take $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.
For a 2 by 2 matrix the two steps give a shortcut: swap the two diagonal entries and change the sign of the other two.
*For a $2 \times 2$ matrix* there is a quick way. Swap the two diagonal entries, and change the sign of the other two. For $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ that gives $\begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$ again. *A $3 \times 3$ matrix has* nine cofactors, and you need all nine.
Your turn: find $\text{adj} A$ for $A = \begin{pmatrix} 3 & -1 \\ 2 & 5 \end{pmatrix}$.
Check yourself
What is the adjoint $\text{adj} A$ of a square matrix $A$?
the transpose of $A$ itself, before any cofactors are taken
the matrix of minors, without applying any sign or transpose
the cofactor matrix, built first, then transposed
the cofactor matrix itself, with no transpose
Check your answer
the transpose of $A$ itself, before any cofactors are taken — The transpose is applied to the COFACTOR matrix, not to $A$ directly, transposing $A$ itself skips computing any cofactors at all.
the matrix of minors, without applying any sign or transpose — Raw minors are missing both required steps, the alternating sign (to become cofactors) and the transpose. Neither can be skipped.
✓ the cofactor matrix, built first, then transposed — (C) First build the cofactor matrix (replace each entry with its cofactor), then transpose it, swap rows and columns.
the cofactor matrix itself, with no transpose — The cofactor matrix is only the first step, the adjoint is its TRANSPOSE. Skipping the transpose swaps off-diagonal cofactors and gives the wrong matrix.
A student computes every cofactor of $A$ correctly but skips the transpose step. What happens to the result?
nothing: the result is still the correct adjoint
the diagonal entries become incorrect, but the off-diagonal ones stay fine
the off-diagonal cofactors sit in each other’s places, so the adjoint is wrong unless the cofactor matrix is symmetric
every cofactor’s sign flips, even though the values were computed correctly, since skipping the last step reverses them all
Check your answer
nothing: the result is still the correct adjoint — The transpose step is what defines the adjoint, skipping it leaves the cofactor matrix itself, which is a genuinely different matrix from $\text{adj} A$ whenever any off-diagonal cofactors differ.
the diagonal entries become incorrect, but the off-diagonal ones stay fine — A transpose always fixes the diagonal entries in place ($C_{i i}$ stays put) , it is the OFF-diagonal entries, at $(i,j)$ and $(j,i)$, that get swapped when the transpose is skipped.
✓ the off-diagonal cofactors sit in each other’s places, so the adjoint is wrong unless the cofactor matrix is symmetric — (C) The transpose exchanges the positions of off-diagonal entries; skipping it leaves them in the cofactor matrix’s original (untransposed) positions.
every cofactor’s sign flips, even though the values were computed correctly, since skipping the last step reverses them all — Signs come from the cofactor formula $(-1)^{i+j}$, computed before the transpose , skipping the transpose does not touch any sign, only the POSITIONS of the off-diagonal cofactors.
For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, what shortcut gives the adjoint?
swap the diagonal entries, negate the off-diagonal ones: $\text{adj} A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$
negate the diagonal entries, swap the off-diagonal ones: $\begin{pmatrix} -a & c \\ b & -d \end{pmatrix}$
swap every entry with its diagonal mirror, with no sign change at all: $\begin{pmatrix} d & c \\ b & a \end{pmatrix}$
leave every entry exactly where it is, only changing signs: $\begin{pmatrix} a & -b \\ -c & d \end{pmatrix}$
Check your answer
✓ swap the diagonal entries, negate the off-diagonal ones: $\text{adj} A = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ — (A) For $2 \times 2$, computing every cofactor and transposing reduces to this one memorable swap-and-negate pattern.
negate the diagonal entries, swap the off-diagonal ones: $\begin{pmatrix} -a & c \\ b & -d \end{pmatrix}$ — The shortcut swaps the DIAGONAL entries and negates the OFF-diagonal ones, this option does the opposite of each operation.
swap every entry with its diagonal mirror, with no sign change at all: $\begin{pmatrix} d & c \\ b & a \end{pmatrix}$ — The off-diagonal entries must be negated, not just repositioned, dropping the sign change leaves out half of what the shortcut does.
leave every entry exactly where it is, only changing signs: $\begin{pmatrix} a & -b \\ -c & d \end{pmatrix}$ — The diagonal entries must also be swapped ($a$ and $d$ trade places), negating without swapping leaves the diagonal in its original, wrong position.
Why is $\text{adj} A$ the matrix that builds $A^{-1}$?
because $A (\text{adj} A) = |A| I$
because $\text{adj} A$ already equals $A^{-1}$ for every invertible $A$
because $\text{adj} A$ is the transpose of $A$
because $\text{adj} A$ is the matrix of the minors of $A$
Check your answer
✓ because $A (\text{adj} A) = |A| I$ — (A) $A (\text{adj} A) = (\text{adj} A) A = |A| I$. So when $|A| \neq 0$, the matrix $(1/|A|) \text{adj} A$ times $A$, either way round, gives $I$, and that is $A^{-1}$.
because $\text{adj} A$ already equals $A^{-1}$ for every invertible $A$ — $\text{adj} A$ equals $A^{-1}$ only when $|A| = 1$. In general $A^{-1} = (1/|A|) \text{adj} A$, so the division by $|A|$ is still needed.
because $\text{adj} A$ is the transpose of $A$ — $\text{adj} A$ is the transpose of the cofactor matrix, not of $A$. For $A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$, $A^T = A$, but $\text{adj} A = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$.
because $\text{adj} A$ is the matrix of the minors of $A$ — The minors are only the first step. Each minor takes its sign to become a cofactor, and the cofactor matrix is then transposed.
▲CONCEPT
The inverse from the adjoint
Chapter 3 said what an inverse is. It is the matrix that, multiplied by the original matrix in either order, gives the identity. But Chapter 3 had no way to find it. The adjoint nearly does the job. It needs one division to finish it.
A times its adjoint gives minus 2 on the diagonal and 0 elsewhere: minus 2 I. The adjoint nearly inverts A, scaled by minus 2.
Here the determinant of the worked example is 1, so A times its adjoint is I itself: for this matrix, the adjoint is the inverse.
*Multiply $A$ by* its adjoint and see what comes out. Take $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $\text{adj} A = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$ from above.
$A (\text{adj} A) = \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} = -2 I$ Work out each entry: $-2$ on the diagonal, 0 elsewhere.
*$-2 = 1 \cdot 4 - 2 \cdot 3 = |A|$* The number on the diagonal is the determinant of $A$.
Why this always happens
The same happens for every square matrix: $A (\text{adj} A) = (\text{adj} A) A = |A| I$. Here is why. Each diagonal entry is a row of $A$ times its own cofactors. That is the expansion of $|A|$ along that row. Each entry off the diagonal is a row times the cofactors of a different row: for $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, one such entry is $1 \cdot (-2) + 2 \cdot 1 = 0$, the expansion of a determinant with rows $(1, 2)$ and $(1, 2)$, which repeats a row and so is always $0$.
Divide every entry of the adjoint by the determinant, minus 2. What is left inverts A: the inverse is minus one half times the adjoint.
When the determinant of S is 0, S times its adjoint is the zero matrix: nothing to divide by, so no inverse exists.
Rows (2, 4) and (1, 3) give area 2. Change the 3 to 2, and both rows lie on one line: area 0, no inverse.
Finishing the division
So the adjoint gives $|A|$ times the identity. Divide by $|A|$ to finish, line by line:
Why it works
Divide by $|A|$ to get $A^{-1}$
$A (\text{adj} A) = (\text{adj} A) A = |A| I$ Start from what the adjoint gives, in either order.
$A ((1/|A|) \text{adj} A) = ((1/|A|) \text{adj} A) A = I$ Divide both sides by $|A|$, which needs $|A| \neq 0$.
*$A^{-1} = (1/|A|) \text{adj} A$* A matrix that gives $I$ on both sides of $A$ is its inverse.
This works only when $|A| \neq 0$. If $|A| = 0$, there is nothing to divide by.
A matrix with $|A| = 0$ is called SINGULAR. A matrix with $|A| \neq 0$ is NON-SINGULAR. The product rule showed that an invertible matrix has $|A| \neq 0$. The adjoint shows that every matrix with $|A| \neq 0$ is invertible. So a square matrix is INVERTIBLE exactly when it is non-singular.
*For $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ above*, divide its adjoint by $|A| = -2$:
$A^{-1} = (-1/2) \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$ Divide $\text{adj} A$ by $|A| = -2$, which is not 0.
*$A^{-1} = \begin{pmatrix} -2 & 1 \\ 3/2 & -1/2 \end{pmatrix}$* Multiply each entry by $-1/2$.
$1 \cdot (-2) + 2 \cdot 3/2 = 1$ Check one entry of $A A^{-1}$: row 1 times column 1 gives 1, as it should.
One more rule follows. Take determinants on both sides of $A (\text{adj} A) = |A| I$:
Why it works
Find $|\text{adj} A|$ for an $n \times n$ matrix
$\det(A (\text{adj} A)) = \det(|A| I)$ Take the determinant of both sides.
$|A| \cdot |\text{adj} A| = |A|^n$ Use the product rule on the left, and $|k I| = k^n$ on the right.
*$|\text{adj} A| = |A|^{n - 1}$* Divide both sides by $|A|$, which needs $|A| \neq 0$.
One last rule is about products. Take invertible matrices $A$ and $B$ of the same order. Then $(A B)^{-1} = B^{-1} A^{-1}$. To see why, multiply $A B$ by $B^{-1} A^{-1}$:
Why it works
Why $(A B)^{-1} = B^{-1} A^{-1}$
$(A B)(B^{-1} A^{-1}) = A (B B^{-1}) A^{-1}$ Regroup the brackets: a product of matrices can be grouped either way, as in Chapter 3.
$= A I A^{-1}$ The middle $B B^{-1}$ is $I$.
*$= A A^{-1} = I$* $I$ changes nothing, and $A A^{-1}$ is $I$.
$(B^{-1} A^{-1})(A B) = B^{-1} (A^{-1} A) B = B^{-1} B = I$ The other order gives $I$ in the same way.
So the inverse of a product is the product of the two inverses, in the reverse order.
Your turn: is $\begin{pmatrix} 2 & 4 \\ 3 & 6 \end{pmatrix}$ invertible? Work out its determinant first.
The check: A times A inverse, row times column, gives I. In the top left, minus 2 plus 3 equals 1.
Check yourself
When $|A| \neq 0$, what does the inverse of $A$ equal?
$(1/|A|) A$
$\text{adj} A$ alone, with no scaling by $|A|$
$(1/|A|) \text{adj} A$
$|A| \cdot \text{adj} A$
Check your answer
$(1/|A|) A$ — The formula scales the ADJOINT, not $A$ itself, $(1/|A|) A$ is not generally equal to $A^{-1}$.
$\text{adj} A$ alone, with no scaling by $|A|$ — The adjoint alone is not the inverse, it must be scaled by $1/|A|$ to become $A^{-1}$.
✓ $(1/|A|) \text{adj} A$ — (C) The inverse formula scales the adjoint by the reciprocal of the determinant.
$|A| \cdot \text{adj} A$ — The formula uses the RECIPROCAL of $|A|$, $1/|A|$, not $|A|$ itself, multiplying instead of dividing gives a completely different matrix.
Why does the inverse formula break down when $|A|=0$?
because the transpose step cannot be performed when $|A|=0$
because the cofactors cannot be computed for a singular matrix
because $\text{adj} A$ itself becomes undefined when $|A|=0$
there is nothing to divide by: $1/|A|$ is undefined
Check your answer
because the transpose step cannot be performed when $|A|=0$ — Transposing a matrix is a purely mechanical rearrangement that works for any matrix, the value of $|A|$ has no bearing on whether it can be performed.
because the cofactors cannot be computed for a singular matrix — Cofactors are minors with a sign attached, computable for any square matrix, $|A|=0$ does not prevent any cofactor from being found; only the final division fails.
because $\text{adj} A$ itself becomes undefined when $|A|=0$ — $\text{adj} A$ is perfectly well defined even when $|A|=0$, it is the DIVISION by $|A|$ that breaks, not the adjoint itself.
✓ there is nothing to divide by: $1/|A|$ is undefined — (D) The formula requires dividing the adjoint by $|A|$; when $|A|=0$ that division is undefined, matching the fact that no inverse exists for a singular matrix.
After computing $A^{-1}$ by the adjoint formula, what is good practice to catch an arithmetic slip?
recompute $|A|$ a second time using a different row
check that $\text{adj} A$ has the same determinant as $A$
confirm that every entry of $A^{-1}$ is smaller than the corresponding entry of $A$
check that $A \cdot A^{-1} = I$: catches cofactor or transpose slips
Check your answer
recompute $|A|$ a second time using a different row — Recomputing $|A|$ only re-checks the determinant, not the whole inverse, multiplying $A$ by the computed $A^{-1}$ is what verifies the FINAL answer directly.
check that $\text{adj} A$ has the same determinant as $A$ — Nothing in the inverse formula claims $\text{adj} A$ and $A$ share a determinant, the actual check is multiplying $A$ by $A^{-1}$ and confirming the identity matrix comes out.
confirm that every entry of $A^{-1}$ is smaller than the corresponding entry of $A$ — There is no rule that an inverse’s entries must be smaller than the original’s, entries can be larger, smaller, or fractional. The real check is $A \cdot A^{-1} = I$.
✓ check that $A \cdot A^{-1} = I$: catches cofactor or transpose slips — (D) Multiplying back to the identity matrix is a one-line check that catches errors in the cofactors or the transpose.
How does ‘the inverse formula fails at $|A|=0$’ connect to the opening section, One number, one big question?
it is a completely new, unrelated test introduced only at this point in the chapter
it contradicts the opening section, which claimed every matrix has an inverse
the same invertibility test, now a formula
it only applies to $2 \times 2$ matrices, unlike the opening section’s general claim
Check your answer
it is a completely new, unrelated test introduced only at this point in the chapter — The opening section, One number, one big question, already named this test: a nonzero determinant means an inverse exists. This section supplies the formula that computes it.
it contradicts the opening section, which claimed every matrix has an inverse — The opening section says a zero determinant means no inverse exists. The formula’s breakdown at $|A|=0$ AGREES with it; it does not contradict it.
✓ the same invertibility test, now a formula — (C) The opening section says a nonzero determinant means an inverse exists; this section delivers the formula, and its breakdown at $|A|=0$ is that same test again.
it only applies to $2 \times 2$ matrices, unlike the opening section’s general claim — Neither the opening section’s claim nor the inverse formula is limited to $2 \times 2$ matrices. Both apply to a square matrix of any size.
Worked example
Build the adjoint of a $2 \times 2$ matrix, then invert it
$A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ First find $|A|$. The inverse exists only if it is not $0$.
$|A| = 2 \cdot 4 - 3 \cdot 1 = 8 - 3 = 5$ $|A| = 5$ is not $0$, so $A$ is invertible.
$\text{adj} A = \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$ The cofactors are $C_{1 1} = 4$, $C_{1 2} = -1$, $C_{2 1} = -3$ and $C_{2 2} = 2$. Transpose them to get $\text{adj} A$. The quick way gives the same: swap the diagonal, change the other two signs.
$A \cdot A^{-1} = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix} \cdot \begin{pmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ Multiply back to check. Each entry is a row of $A$ times a column of $A^{-1}$.
*$A \cdot A^{-1} = I$, confirming the inverse is correct* The product is $I$, so $A^{-1}$ is indeed the inverse of $A$.
$A^{-1} A = (1/5) \begin{pmatrix} 4 \cdot 2 - 3 \cdot 1 & 4 \cdot 3 - 3 \cdot 4 \\ -1 \cdot 2 + 2 \cdot 1 & -1 \cdot 3 + 2 \cdot 4 \end{pmatrix} = (1/5) \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = I$ Check the other order too. An inverse must give $I$ on both sides.
Check yourself
For $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$, what does the $2 \times 2$ shortcut give for $\text{adj} A$?
$\begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix}$
$\begin{pmatrix} 4 & 3 \\ 1 & 2 \end{pmatrix}$
$\begin{pmatrix} -2 & 3 \\ 1 & -4 \end{pmatrix}$
$\begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$
Check your answer
$\begin{pmatrix} 2 & -3 \\ -1 & 4 \end{pmatrix}$ — The diagonal entries must swap places too, $2$ and $4$ trade positions; this option leaves them where they started.
$\begin{pmatrix} 4 & 3 \\ 1 & 2 \end{pmatrix}$ — The off-diagonal entries ($3$ and $1$) must be negated, this option swaps the diagonal correctly but leaves the off-diagonal signs unchanged.
$\begin{pmatrix} -2 & 3 \\ 1 & -4 \end{pmatrix}$ — This applies negation and swapping to the wrong pair of entries, the shortcut swaps the diagonal and negates the OFF-diagonal, not the other way round.
✓ $\begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$ — (D) Swap the diagonal ($2$ and $4$), negate the off-diagonal ($3$ and $1$).
For $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$, with $|A|=5$ and $\text{adj} A = \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$, what is $A^{-1}$?
$\begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$ — This is $\text{adj} A$ before scaling, the inverse still needs every entry divided by $|A|=5$.
✓ $\begin{pmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{pmatrix}$ — (B) $A^{-1} = (1/5) \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$, every entry of the adjoint divided by $5$.
$\begin{pmatrix} 20 & -15 \\ -5 & 10 \end{pmatrix}$ — The adjoint should be DIVIDED by $|A|=5$, not multiplied by it, multiplying moves in the wrong direction entirely.
$\begin{pmatrix} 4/5 & 3/5 \\ -1/5 & -2/5 \end{pmatrix}$ — Every entry keeps its own sign from the adjoint while dividing by $5$, this option drops the negative sign on the $b$ position, which should stay $-3/5$.
For $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ and $A^{-1} = \begin{pmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{pmatrix}$, what should $A \cdot A^{-1}$ equal if the work is correct?
$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
$\begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}$
$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$
$\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$
Check your answer
✓ $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ — (A) Multiplying $A$ by the computed $A^{-1}$ should return the identity matrix, confirming the inverse is correct.
$\begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix}$ — The division by $|A|=5$ inside the inverse formula is exactly what cancels out to leave plain $1$s on the diagonal, the check target is the identity, not a $|A|$-scaled version of it.
$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ — The zero matrix would signal something is badly wrong (or that no inverse exists) , a CORRECT inverse multiplies back to the identity, not to zero.
$\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$ — $A \cdot A^{-1}$ returning $A$ itself is not the target, the check confirms the IDENTITY matrix comes out, not a copy of $A$.
Three pairs of lines: crossing once at (3, 2), parallel and never meeting, and one line drawn twice. The last two have determinant 0.
Think back to the Class 10 picture. Two lines on a graph can cross once, and then the pair of equations has one answer. They can be parallel, and then there is no answer. Or they can be the same line twice, and then every point on that line is an answer.
Naming the two outcomes
A system of equations with at least one answer is CONSISTENT. A system with no answer is INCONSISTENT. Write the system as $A X = B$, with $A$ square.
If $|A| \neq 0$, the system is consistent and has exactly one answer. That is the crossing case.
If $|A| = 0$, work out $(\text{adj} A) B$. If it is not the zero matrix $O$, the system is inconsistent.
If $|A| = 0$ and $(\text{adj} A) B = O$, the system may have endless answers or none. This test alone cannot tell you which.
Here is why the test works. Start from $A X = B$:
Why it works
Multiply $A X = B$ on the left by $\text{adj} A$
$(\text{adj} A) A X = (\text{adj} A) B$ Multiply both sides on the left by $\text{adj} A$: matrix order matters, so both sides take it on the left.
$|A| I X = (\text{adj} A) B$ The adjoint gives $(\text{adj} A) A = |A| I$.
*$|A| X = (\text{adj} A) B$* $I X = X$: the identity changes nothing.
$0 \cdot X = O$, so $(\text{adj} A) B = O$ When $|A| = 0$ the left side is $O$ for every $X$, so an answer needs the right side $O$ too.
The parallel lines: the determinant is 0 and the adjoint times B is not the zero matrix. The system has no solution.
Let us test the parallel pair $x + y = 5$ and $2 x + 2 y = 16$.
Worked example
Test $x + y = 5$ and $2 x + 2 y = 16$
$A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$, $B = \begin{pmatrix} 5 \\ 16 \end{pmatrix}$ $A$ holds the numbers in front of $x$ and $y$, and $B$ the right sides.
$|A| = 1 \cdot 2 - 1 \cdot 2 = 0$ Work out $|A|$ first: it is 0, so the test needs $(\text{adj} A) B$.
$\text{adj} A = \begin{pmatrix} 2 & -1 \\ -2 & 1 \end{pmatrix}$ Swap the diagonal entries and change the sign of the other two.
*$(\text{adj} A) B \neq O$: no answer* It is not the zero matrix, so the system is inconsistent and the lines never meet.
The same line twice: the determinant is 0 and the adjoint times B is the zero matrix. Every point of the line is a solution.
x plus y plus z cannot equal 1, 2 and 3 together: no solution exists. Yet adjoint times B is zero, so test cannot decide.
Now change the 16 to 10. Then $(\text{adj} A) B = \begin{pmatrix} 10 - 10 \\ -10 + 10 \end{pmatrix} = O$, and the test cannot decide alone. Here the two equations give one line, drawn twice. So every point on it is an answer.
Your turn: is the system $x + 3 y = 4$, $2 x + 6 y = 9$ consistent? Find $|A|$ first.
A SCHOLAR INDIA REMEMBERS
When $|A| = 0$, has nothing happened? Look again. That zero is a finding. It tells you the system does not have exactly one answer. It does not yet tell you whether it has none or endless ones. So take a second reading, the adjoint of $A$ times $B$. If that is not $O$, there is no answer. If it is $O$, go back to the equations themselves.
Brahmagupta · the brahminy kite
Check yourself
For a system written as $A X = B$, exactly when does a unique solution exist?
every entry of $A$ is nonzero
$|A| \neq 0$
$A$ has more columns than rows
$B$ is the zero vector
Check your answer
every entry of $A$ is nonzero — Individual entries being nonzero says nothing about $|A|$, a matrix with several zero entries can still have a nonzero determinant, and vice versa.
✓ $|A| \neq 0$ — (B) A nonzero coefficient determinant is exactly the condition for a unique solution, the same invertibility test used throughout.
$A$ has more columns than rows — $A$ must already be square for $|A|$ to be defined at all, row/column counts being unequal is not a meaningful condition here.
$B$ is the zero vector — Whether $B$ is the zero vector is a separate question from whether $A$ is invertible, the unique-solution condition is about $|A|$, not about $B$.
When $|A| = 0$, which check can prove that a system has no solution?
check whether $(\text{adj} A) B$ is the zero vector
recompute $|A|$ using a different row and see if the value changes
assume the system is always inconsistent whenever $|A|=0$
check whether $B$ itself is the zero vector
Check your answer
✓ check whether $(\text{adj} A) B$ is the zero vector — (A) If $(\text{adj} A) B$ is NOT zero, the system is inconsistent; if it IS zero, the singular case needs a further check the determinant test alone cannot settle.
recompute $|A|$ using a different row and see if the value changes — Any row or column expansion gives the same determinant value, recomputing $|A|$ a second way cannot produce a different answer, so it cannot resolve the singular case.
assume the system is always inconsistent whenever $|A|=0$ — $|A|=0$ alone does not decide consistency, the singular case may or may not be consistent, which is exactly why the $(\text{adj} A)B$ check is needed.
check whether $B$ itself is the zero vector — The real test is on $(\text{adj} A) B$, not on $B$ by itself, $B$ being nonzero does not by itself decide inconsistency once $|A|=0$.
The statement ’unique solution exists exactly when $|A| \neq 0$’ settles the question in which direction(s)?
only one way: $|A| \neq 0$ guarantees a solution, but $|A|=0$ says nothing
neither way: the determinant only ever hints at the answer, never settles it
both ways: $|A| \neq 0$ guarantees a unique solution, and $|A|=0$ rules one out
only the reverse direction: $|A|=0$ implies no unique solution, but $|A| \neq 0$ does not guarantee one
Check your answer
only one way: $|A| \neq 0$ guarantees a solution, but $|A|=0$ says nothing — ’Exactly when’ means both directions hold, $|A|=0$ does say something: it specifically rules out a UNIQUE solution (though the system might still be consistent with infinitely many, or inconsistent).
neither way: the determinant only ever hints at the answer, never settles it — ’Exactly when’ is a firm, precise condition, not a hint, $|A| \neq 0$ genuinely guarantees uniqueness, and $|A|=0$ genuinely rules it out.
✓ both ways: $|A| \neq 0$ guarantees a unique solution, and $|A|=0$ rules one out — (C) ’Exactly when’ is a two-way statement: it is both a sufficient and a necessary condition for a unique solution.
only the reverse direction: $|A|=0$ implies no unique solution, but $|A| \neq 0$ does not guarantee one — A nonzero $|A|$ directly guarantees a unique solution, $X=A^{-1}B$ can always be computed, so this option has the stronger direction backwards.
Why does the consistency test at $|A|=0$ reach for $\text{adj} A$ specifically, rather than some other quantity?
it is the piece of the blocked inverse formula that carries information
because computing $\text{adj} A$ is the only way to find $|A|$ in the first place
because $\text{adj} A$ is easier to compute than $A$ itself
because $\text{adj} A$ happens to equal $B$ whenever $|A|=0$
Check your answer
✓ it is the piece of the blocked inverse formula that carries information — (A) The inverse formula needs $|A| \neq 0$ to divide by; $\text{adj} A$ is the piece of that formula that still exists and carries information even when the division itself is blocked.
because computing $\text{adj} A$ is the only way to find $|A|$ in the first place — $|A|$ is computed directly by cofactor expansion, the adjoint is built FROM the cofactors afterward, not the other way round.
because $\text{adj} A$ is easier to compute than $A$ itself — The adjoint is not generally simpler to compute than $A$ itself, it is reused here because it is the piece of the inverse formula that survives when $|A|=0$ blocks the division.
because $\text{adj} A$ happens to equal $B$ whenever $|A|=0$ — $\text{adj} A$ and $B$ are unrelated quantities, nothing claims they become equal when $|A|=0$; the adjoint is simply reused as a diagnostic tool.
▲CONCEPT
Chapter 3’s bill read backwards: the totals are known, 140 and 90 rupees, but the prices of a pen and a notebook are not.
Remember the shop bill from Chapter 3. You bought 2 pens and 3 notebooks, and your friend bought 5 pens and 1 notebook. Quantities times prices gave the two bills, 140 rupees and 90 rupees. Now turn it round. Suppose you know the two bills, but not the prices. Can you find the prices?
In Chapter 3, quantities times prices gave the bills: 2 pens at 10 rupees and 3 notebooks at 40 come to 140 rupees.
Now the bills are known and the prices x and y are not. Written as matrices, that is A X = B.
Call the prices $x$ and $y$. It is the bill from Chapter 3, read backwards. Write the two bills as one matrix equation:
Worked example
Write the two bills as $A X = B$
$2 x + 3 y = 140$, $5 x + y = 90$ The bills: 2 pens and 3 notebooks cost 140 rupees, 5 pens and 1 notebook 90.
$\begin{pmatrix} 2 & 3 \\ 5 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 140 \\ 90 \end{pmatrix}$ Row times column on the left gives back both left sides.
*$A X = B$, with $A = \begin{pmatrix} 2 & 3 \\ 5 & 1 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \end{pmatrix}$, $B = \begin{pmatrix} 140 \\ 90 \end{pmatrix}$* Name the three matrices: quantities, prices and bills.
The general solving step
*If $|A| \neq 0$*, solve $A X = B$ for $X$:
Why it works
Solve $A X = B$ when $|A| \neq 0$
$A^{-1} A X = A^{-1} B$ Multiply both sides on the left by $A^{-1}$, which exists as $|A| \neq 0$.
$I X = A^{-1} B$ $A^{-1} A = I$: $A^{-1}$ is the inverse of $A$.
*$X = A^{-1} B$* $I X = X$, since $I$ changes nothing.
Every answer has to equal $A^{-1} B$. So when $|A| \neq 0$, there is exactly one answer.
The determinant is minus 13, not 0, so A inverse exists. A inverse times the bills gives the prices back: 10 and 40 rupees.
Let us solve it for the two bills.
Worked example
Find the prices from the bills
$|A| = 2 \cdot 1 - 3 \cdot 5 = -13$ Work out $|A|$ first: it is not 0, so $A^{-1}$ exists.
$\text{adj} A = \begin{pmatrix} 1 & -3 \\ -5 & 2 \end{pmatrix}$ Swap the diagonal entries and change the sign of the other two.
*$X = \begin{pmatrix} 10 \\ 40 \end{pmatrix}$* Divide each entry by $-13$: $x = 10$ and $y = 40$.
$2 \cdot 10 + 3 \cdot 40 = 140$ and $5 \cdot 10 + 40 = 90$ Put the prices back into both bills.
A pen costs 10 rupees and a notebook costs 40 rupees, as in Chapter 3.
Three unknowns work the same way: A is 3 by 3, X and B are 3 by 1. One row taller, nothing else new.
You need as many equations as unknowns, so that $A$ is square. You also need $|A| \neq 0$. With three unknowns, $A$ is $3 \times 3$ and $X$ is $3 \times 1$. Nothing else changes.
Shop 2’s bills, 129 and 95 rupees, with the same A inverse: pens at 12 rupees, notebooks at 35. One inverse serves every bill.
You work out $A^{-1}$ once. After that, each new set of bills $B$ needs just one more product.
Your turn: at a third shop, the same two baskets cost 90 rupees and 95 rupees. Which $A^{-1}$ do you need? What are that shop’s prices?
Check yourself
How is a linear system written as $A X = B$, with $|A| \neq 0$, solved?
$X = A^{-1} B$
$X = B A^{-1}$
$X = A B$
$X = B / A$
Check your answer
✓ $X = A^{-1} B$ — (A) Once $A$ is invertible, multiplying both sides by $A^{-1}$ isolates $X$ directly.
$X = B A^{-1}$ — Matrix multiplication is not commutative, $A^{-1}$ must multiply $B$ on the LEFT to match how it was isolated from $A X = B$, not on the right.
$X = A B$ — Solving for $X$ requires the INVERSE of $A$, not $A$ itself, $X=A B$ does not undo the multiplication by $A$ in the original equation.
$X = B / A$ — There is no such thing as dividing one matrix by another, the correct operation is multiplying by $A^{-1}$, which is how matrix ’division’ is actually expressed.
For a three-variable system, how do the shapes of $X$ and $A$ compare to the two-variable case?
the method changes entirely, requiring a different formula for three variables
both $X$ and $A$ stay the same order, and only $B$ grows
$X$ becomes $3 \times 1$, $A$ becomes $3 \times 3$, matching the extra variable
$X$ stays $2 \times 1$, but $A$ grows to $3 \times 3$
Check your answer
the method changes entirely, requiring a different formula for three variables — $X=A^{-1}B$ applies exactly the same way for two, three, or more variables, only the sizes of the matrices grow; no new formula is introduced.
both $X$ and $A$ stay the same order, and only $B$ grows — Adding a third unknown grows $X$ (one more entry) and $A$ (one more row and column) together, $B$ growing alone would leave the equation’s dimensions mismatched.
✓ $X$ becomes $3 \times 1$, $A$ becomes $3 \times 3$, matching the extra variable — (C) The method is identical; only the height of $X$ and the size of $A$ grow to match the number of variables.
$X$ stays $2 \times 1$, but $A$ grows to $3 \times 3$ — $X$ must list all three unknowns, so it grows to $3 \times 1$ along with $A$, it cannot stay at two rows once a third variable is added.
When is computing $A^{-1}$ once and reusing it especially worthwhile?
only when the system has exactly two variables
only when $B$ is the zero vector
only when $|A|$ is a whole number
solving several $B$’s against the same $A$
Check your answer
only when the system has exactly two variables — The advantage of reusing $A^{-1}$ has nothing to do with how many variables the system has, it comes from how many different $B$ vectors need solving, at any size.
only when $B$ is the zero vector — Whether $B$ happens to be zero is unrelated to the benefit of reusing $A^{-1}$ , the payoff comes from solving against several DIFFERENT right-hand sides, not from any one B’s value.
only when $|A|$ is a whole number — Whether $|A|$ comes out as a whole number or a fraction makes no difference to the benefit of reusing $A^{-1}$, the payoff is purely about how many different $B$’s are being solved.
✓ solving several $B$’s against the same $A$ — (D) $A^{-1}$ does not depend on $B$, so once it is found it can be multiplied against any number of different $B$ vectors.
How does the matrix-inverse method for $A X = B$ connect to the consistency test from earlier?
it needs $|A| \neq 0$, the same unique-solution condition
consistency is a stronger requirement than what the method needs, so the method applies more often
the method works whether or not the system is consistent, so consistency does not matter here
the two are unrelated: consistency is checked separately, after the method is already applied
Check your answer
✓ it needs $|A| \neq 0$, the same unique-solution condition — (A) $X=A^{-1}B$ presumes $A^{-1}$ exists, which is precisely the $|A| \neq 0$ condition the consistency section already established.
consistency is a stronger requirement than what the method needs, so the method applies more often — $|A| \neq 0$ is exactly the unique-solution condition from the consistency section , the method’s requirement and the consistency test are the same condition, not a stronger or weaker version of each other.
the method works whether or not the system is consistent, so consistency does not matter here — The method specifically requires $|A| \neq 0$, exactly the condition for a unique (consistent) solution. It does not work for an inconsistent or singular system.
the two are unrelated: consistency is checked separately, after the method is already applied — The method’s precondition ($|A| \neq 0$) is exactly the consistency test’s unique-solution condition, applying $X=A^{-1}B$ already presumes that test has passed.
For the example below, three copies of (2, 1) and two of (1, 1) reach B at (8, 5). Four copies of (2, 1) would reach only (8, 4).
Worked example: 2x + y = 8, x + y = 5. A inverse times (8, 5) gives x = 3, y = 2.
Worked example
Solve a linear system by the matrix-inverse method
$2x + y = 8$, $x + y = 5$ You are given two equations. Write them as $A X = B$.
$A = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \end{pmatrix}$, $B = \begin{pmatrix} 8 \\ 5 \end{pmatrix}$ $A$ holds the numbers in front of $x$ and $y$. $B$ holds the right-hand sides.
$|A| = 2 \cdot 1 - 1 \cdot 1 = 1$ $|A| = 1$ is not $0$, so $A$ is invertible. The system has exactly one answer.
$A^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$ Swap the diagonal entries and change the sign of the other two. Dividing by $|A| = 1$ changes nothing.
$X = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$, so $x = 3$, $y = 2$ Work out both entries.
*$2(3) + 2 = 8$ and $3 + 2 = 5$, both original equations hold* Put $x = 3$ and $y = 2$ back into both equations. Both hold.
The two-loop circuit worked below: a source and a resistor in each outer branch, one resistor shared between the loops, and the way each of the three currents flows.
Worked example
The same test, on a two-loop circuit
A two-loop circuit has three branch currents, $I_1$, $I_2$ and $I_3$ (amperes), all unknown Name the unknowns before writing any rule down.
At the shared junction, $I_1 + I_2 = I_3$ The junction rule from Physics 12, Chapter 3, Current Electricity, says the current flowing in equals the current flowing out.
Around loop 1 (resistor $R_1 = 1 \Omega$, the shared branch $R_3 = 1 \Omega$, a source of $7$ volts), the loop rule gives $I_1 + I_3 = 7$ The potential changes around this closed loop sum to zero, which is the loop rule applied to one path.
Around loop 2 (resistor $R_2 = 2 \Omega$, the same shared branch, a source of $6$ volts), the loop rule gives $2 I_2 + I_3 = 6$ A second closed path gives a second equation, the same rule applied again.
$A = \begin{pmatrix} 1 & 1 & -1 \\ 1 & 0 & 1 \\ 0 & 2 & 1 \end{pmatrix}$, $X = \begin{pmatrix} I_1 \\ I_2 \\ I_3 \end{pmatrix}$, $B = \begin{pmatrix} 0 \\ 7 \\ 6 \end{pmatrix}$ Collect the junction rule and both loop rules into one matrix equation $A X = B$, exactly as this chapter’s equation-solving section does.
$|A| = -5$, not $0$, so $A^{-1}$ exists The same determinant test applies here, to currents instead of prices.
*$I_1 = 3$, $I_2 = 1$, $I_3 = 4$ amperes, and $3 + 1 = 4$ checks the junction rule once more* Putting the solved currents back into the junction rule confirms the answer.
Check yourself
Solving $2x+y=8$, $x+y=5$ by the matrix-inverse method, with $A=\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ and $|A|=1$, what is $A^{-1}$?
$\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$
$\begin{pmatrix} 1 & 1 \\ -1 & -2 \end{pmatrix}$
$\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$
$\begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$
Check your answer
$\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$ — The off-diagonal entries must be negated as part of the shortcut, this option leaves them positive, unlike the correct $\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$.
$\begin{pmatrix} 1 & 1 \\ -1 & -2 \end{pmatrix}$ — The shortcut applies swap-the-diagonal and negate-the-off-diagonal consistently to both pairs, this option mixes the two operations rather than applying each correctly to its own pair.
✓ $\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$ — (C) Using the $2 \times 2$ shortcut with $|A|=1$: swap the diagonal, negate the off-diagonal, and dividing by $1$ changes nothing further.
$\begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$ — The diagonal entries ($2$ and $1$ from $A$) must swap places to become ($1$ and $2$), this option keeps them where $A$ had them.
Continuing the same system with $A^{-1}=\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$ and $B=\begin{pmatrix} 8 \\ 5 \end{pmatrix}$, what is $X=A^{-1}B$?
$\begin{pmatrix} 2 \\ 3 \end{pmatrix}$
$\begin{pmatrix} 13 \\ -6 \end{pmatrix}$
$\begin{pmatrix} 8 \\ 5 \end{pmatrix}$
$\begin{pmatrix} 3 \\ 2 \end{pmatrix}$
Check your answer
$\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ — The first row of the computation gives $x=3$ and the second gives $y=2$, this option reports them in the reversed order.
$\begin{pmatrix} 13 \\ -6 \end{pmatrix}$ — The first row is $1(8)+(-1)(5)=3$, not $8+5=13$, the negative sign on the second entry of $A^{-1}$’s first row must be kept, not dropped.
$\begin{pmatrix} 8 \\ 5 \end{pmatrix}$ — Multiplying by $A^{-1}$ transforms $B$ into $X$, reporting $B$ unchanged skips the matrix multiplication entirely.
How does the worked example verify that $x=3$, $y=2$ actually solves the original system?
by recomputing $A^{-1}$ a second time using a different method
substituting $x$ and $y$ back into both original equations
by checking that $|A|=1$ is a ’nice’ whole number
by confirming $X$ and $B$ have the same shape
Check your answer
by recomputing $A^{-1}$ a second time using a different method — Recomputing $A^{-1}$ only re-checks that intermediate matrix, the actual verification substitutes $x=3,y=2$ directly into $2x+y=8$ and $x+y=5$.
✓ substituting $x$ and $y$ back into both original equations — (B) $2(3)+2=8$ and $3+2=5$, both original equations check out with the computed values.
by checking that $|A|=1$ is a ’nice’ whole number — Whether $|A|$ happens to be a whole number says nothing about whether $x=3,y=2$ actually solves the system, only substitution into the original equations does that.
by confirming $X$ and $B$ have the same shape — $X$ and $B$ having matching shapes only confirms the multiplication was even possible, it says nothing about whether the specific values $x=3,y=2$ satisfy the two original equations.
Eight cards joined by arrows: 2 by 2 determinants, then cofactors, expansion, the adjoint, then products, area, the undecided case, and solving AX = B.
The question at the start was this. Is there one number, worked out from the entries, that tells you whether a matrix is invertible?
A $2 \times 2$ DETERMINANT is $a d - b c$. So $\begin{vmatrix} 3 & 4 \\ 1 & 2 \end{vmatrix} = 6 - 4 = 2$.
A $3 \times 3$ DETERMINANT is expanded along any row or column, with the checkerboard signs (plus, minus, plus along row 1). For $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 5 & 6 & 0 \end{pmatrix}$, $|A| = -24 + 40 - 15 = 1$.
PRODUCTS: $|A B| = |A| \cdot |B|$, but there is no such rule for addition. For $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $\begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix}$, $|A + B| = 13$, not $-2 + 6$.
THE INVERSE: $A^{-1} = (1/|A|) \text{adj} A$, only when $|A| \neq 0$. For $\begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$, $A^{-1} = (1/5) \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$.
EQUATIONS: if $|A| \neq 0$, then $A X = B$ has exactly one answer, $X = A^{-1} B$. For $2 x + y = 8$ and $x + y = 5$, $X = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$.
AREA: half the absolute value of $\Delta$. The corners $(1, 0)$, $(2, 2)$ and $(0, 3)$ give $\Delta = 5$, so the area is 2.5.
The opening question answered. Determinant not 0: one answer, X equals A inverse B. Determinant 0: no answer or endless answers. A nonzero adj A times B means no answer.
Answering the opening question
Here is the answer. The number is the determinant: zero exactly when the matrix is not invertible.
|A| ≠ 0: is invertible
1373
|A| = 0: is not invertible
0
Sorted by one number: determinant not 0 means invertible, determinant 0 means not. The entries alone never told you which.
Matrices on a number line: rows swapped at minus 1, a row repeated at 0, as given at 1, bottom row doubled at 2.
Check yourself
Across the whole chapter, what single computed number keeps deciding whether a matrix inverts, a triangle’s area, and a system’s solvability?
the value of the largest entry in the matrix
the number of rows in the matrix
the determinant, $|A|$
the adjoint, $\text{adj} A$
Check your answer
the value of the largest entry in the matrix — No single entry carries this information, it takes the whole matrix, combined into the determinant, to decide invertibility, area, or solvability.
the number of rows in the matrix — Row count is visible without any computation and has no bearing on invertibility, area, or solvability, those all hinge on the computed determinant, not on shape.
✓ the determinant, $|A|$ — (C) One number, computed from a square matrix, answers the invertibility question, feeds the area/collinearity formula, and settles a system’s unique solvability.
the adjoint, $\text{adj} A$ — The adjoint is a matrix, not a single number, and it is itself built FROM the determinant’s cofactors, the throughline is $|A|$, which the adjoint depends on.
Which pairing correctly describes how the determinant behaves under products versus addition, as this chapter’s recurring trap?
$|A B|=|A| \cdot |B|$ always holds; $|A+B| \neq |A|+|B|$ in general
$|A B|=|A| \cdot |B|$ and $|A+B|=|A|+|B|$ both always hold
neither $|A B|=|A| \cdot |B|$ nor $|A+B|=|A|+|B|$ holds in general
$|A+B|=|A|+|B|$ always holds; $|A B|=|A| \cdot |B|$ does not
Check your answer
✓ $|A B|=|A| \cdot |B|$ always holds; $|A+B| \neq |A|+|B|$ in general — (A) The one recurring trap here is treating the true multiplicative rule as if it also applied to addition, where it does not.
$|A B|=|A| \cdot |B|$ and $|A+B|=|A|+|B|$ both always hold — The additive claim is FALSE in general, the worked counterexample above ($|A+B|=13$, $|A|+|B|=4$) disproves it directly, even though the product rule is genuinely always true.
neither $|A B|=|A| \cdot |B|$ nor $|A+B|=|A|+|B|$ holds in general — The product rule $|A B|=|A||B|$ IS always true for same-size square matrices, only the additive version is false; the two are not equally unreliable.
$|A+B|=|A|+|B|$ always holds; $|A B|=|A| \cdot |B|$ does not — This has the pairing backwards, the PRODUCT rule is the one that always holds, and the ADDITIVE version is the one that fails.
What machinery does this chapter build to actually compute the determinant for any size of matrix?
estimating the determinant numerically without an exact formula
solving the characteristic equation for eigenvalues
cofactor expansion, using minors and the alternating sign
row reduction to echelon form, with no reference to minors or cofactors
Check your answer
estimating the determinant numerically without an exact formula — Every computation in this chapter is exact, cofactor expansion gives the precise determinant value, with no estimation or approximation involved.
solving the characteristic equation for eigenvalues — Eigenvalues and the characteristic equation do not appear anywhere in this chapter, the machinery built here is minors, cofactors, and expansion.
✓ cofactor expansion, using minors and the alternating sign — (C) Starting from the direct $1 \times 1$ and $2 \times 2$ cases, this generalizes to any size through minors, signed cofactors, and expansion along a row or column.
row reduction to echelon form, with no reference to minors or cofactors — Row reduction is a valid alternative method in general, but it is not the machinery built here, which develops minors, signed cofactors, and expansion along a row or column.
Check yourself: the whole chapter
For a square matrix $A$, which single fact about $|A|$ decides whether $A^{-1}$ exists?
Whether $|A|$ equals $0$ or not. Nonzero means invertible, zero means it does not.
Whether $|A|$ is positive or negative. Only a positive determinant gives an invertible matrix.
Whether $|A|$ is larger than the entries of $A$. A determinant “big enough” guarantees an inverse.
Whether $A$ is a symmetric matrix. Invertibility depends on the shape of $A$, not on $|A|$.
Check your answer
✓ Whether $|A|$ equals $0$ or not. Nonzero means invertible, zero means it does not. — (A) The determinant answers exactly one question: zero or nonzero. Sign and size play no part, $|A| \neq 0$ means $A^{-1}$ exists.
Whether $|A|$ is positive or negative. Only a positive determinant gives an invertible matrix. — Sign plays no part. $|A| = -7$ is exactly as invertible as $|A| = 7$, only zero fails the test.
Whether $|A|$ is larger than the entries of $A$. A determinant “big enough” guarantees an inverse. — There is no threshold to clear. Even $|A| = 0.001$ is invertible; only $|A| = 0$ is not.
Whether $A$ is a symmetric matrix. Invertibility depends on the shape of $A$, not on $|A|$. — Symmetry is a separate property altogether. Invertibility is decided by one thing only: whether $|A|$ is zero.
$1$ — A cofactor is a SIGNED minor. Deleting row $1$ and column $2$ from $A$ gives $M_{1 2} = 1$, and the sign at position $(1,2)$ is $-1$, so $C_{1 2} = -1$.
✓ $-1$ — (B) Deleting row $1$ and column $2$ leaves the $2 \times 2$ determinant $2 \cdot 3 - 5 \cdot 1 = 1$, which is $M_{1 2}$. The sign at $(1,2)$ is $(-1)^{1+2} = -1$, so $C_{1 2} = -1$.
$14$ — $M_{1 2}$ deletes row $1$ and column $2$, the first index is always the row. Deleting row $2$ and column $1$ finds $M_{2 1}$, a different number.
$-11$ — The $2 \times 2$ minor still uses the diagonal rule: one product minus the other, never plus.
Matrix $B$ is obtained from a $3 \times 3$ matrix $A$ by adding $5$ times row $2$ to row $1$. How does $|B|$ compare with $|A|$?
$|B| = 5 |A|$
$|B| = -|A|$
$|B| = |A|$
$|B| = |A| + 5$
Check your answer
$|B| = 5 |A|$ — Scaling a row is a different operation, and it is the one that multiplies $|A|$ by that constant. Adding a multiple of one row to another leaves $|A|$ untouched.
$|B| = -|A|$ — Swapping two rows is what flips the sign. Adding a multiple of one row to another is a separate operation, with a separate effect: none at all.
✓ $|B| = |A|$ — (C) Adding a multiple of one row to another row never changes the determinant. $|B| = |A|$.
$|B| = |A| + 5$ — The determinant is not adjusted by arithmetic on the side. Adding a multiple of one row to another leaves the value of $|A|$ exactly as it was.
To find the adjoint of a square matrix $A$, you first build the cofactor matrix: replace each entry with its cofactor. What is the next step?
Transpose the cofactor matrix: swap its rows and columns.
Nothing further: the cofactor matrix already is the adjoint.
Multiply the cofactor matrix by $|A|$.
Divide every entry of the cofactor matrix by the matching entry of $A$.
Check your answer
✓ Transpose the cofactor matrix: swap its rows and columns. — (A) The adjoint is the transpose of the cofactor matrix: build the cofactors first, then swap rows and columns.
Nothing further: the cofactor matrix already is the adjoint. — The cofactor matrix is only half the job. The adjoint is its TRANSPOSE, rows and columns swapped.
Multiply the cofactor matrix by $|A|$. — The determinant belongs to the next stage, $A^{-1} = (1/|A|)$ times the adjoint, not to building the adjoint itself.
Divide every entry of the cofactor matrix by the matching entry of $A$. — There is no entrywise division anywhere in this process. The adjoint comes from a transpose, a whole-matrix operation, not an entry-by-entry one.
The same one number turns up in everyday checks like these.
A park’s three corner posts. A city park is marked on a map by three corner posts. The park office sometimes needs to recheck them without re-measuring the ground. The maths: The area-of-triangle determinant from this book decides the park’s size. Its zero case would also catch three posts planted on one line by mistake. A park with corners $(0,0)$ $(6,0)$ $(0,8)$ has area $(1/2)|0(0-8)+6(8-0)+0(0-0)| = 24$ square units.
A stationery shop’s two bills. A stationery shop’s two bills list different combinations of notebooks and pens. Each bill totals a different amount. The maths: Writing both bills as one system and solving it the way $A X = B$ is solved here pins down each price uniquely. $2$ notebooks and $3$ pens for $160$ rupees, and $1$ notebook and $1$ pen for $60$ rupees, fix notebook $20$ rupees and pen $40$ rupees.
An auto-rickshaw’s fare slip. An auto-rickshaw’s fare slip lists one trip’s distance and fare. A second trip gives a different distance and a different fare. The maths: The same solving step used on the shop’s bills applies here too. It reads a fixed charge and a per-kilometre rate off the two trips. A trip of $8$ km costs $100$ rupees, and a trip of $12$ km costs $140$ rupees. These fix a fixed charge of $20$ rupees and a rate of $10$ rupees per km.
A festival fund’s recount. A festival fund’s collection sheet is counted once, then recounted on a second day and reported as exactly double. The maths: The determinant test for a system’s solution shows the second sheet only repeats the first, not a new fact. $x+y=500$ and $2x+2y=1000$ have a zero determinant and a matching right side, so infinitely many splits of $x$ and $y$ fit both sheets.
A classroom monitor’s row count. A classroom monitor counts two rows once. A second quick count of the same two rows gives a different total. The maths: That same determinant test for a system’s solution catches the contradiction directly. The zero determinant pairs with counts that do not actually agree. $x+y=10$ and $2x+2y=22$ have a zero determinant but $22$ is not twice $10$, so the two counts are inconsistent.
Your turn. A juice stall sells mango and orange juices. $2$ mango and $3$ orange juices cost $110$ rupees, and $1$ of each costs $50$ rupees. Use the determinant to check that the two prices are fixed, then find them. Answer: the determinant is $2 \times 1 - 3 \times 1 = -1$, nonzero, so the prices are fixed: mango $40$ rupees, orange $10$ rupees.
LATER
Where the determinant goes next
First-year university and engineering mathematics reuses this determinant as the Jacobian: one number that measures how a change of coordinates stretches area or volume. The same test applies. Where the Jacobian is zero, the change flattens area to nothing at that point, as a zero determinant flattens the parallelogram.
Exercise 4.1 asks you to work out $2 \times 2$ and $3 \times 3$ determinants. For a $3 \times 3$, expand along the row or column with the most zeros. A few questions ask for $|k A|$ as well.
Exercise 4.2 finds the area of a triangle from its three corners. Some questions use $\Delta = 0$ to test whether three points are collinear. One asks for the line through two points.
Exercise 4.2: Area of a triangle
practice Find the area of the triangle with vertices $(3,1)$, $(6,5)$ and $(9,2)$.
practice Find the area of the triangle with vertices $(-2,3)$, $(4,-5)$ and the origin $(0,0)$.
practice Find $\Delta$ for the points $(1,2)$, $(3,6)$ and $(5,10)$, and say what the value tells you about the three points.
practice Find $k$ if the area of the triangle with vertices $(k,0)$, $(6,0)$ and $(0,3)$ is $9$ square units.
practice Using determinants, find the equation of the line joining $(2,3)$ and $(5,9)$.
practice If the area of the triangle with vertices $(3,-4)$, $(5,2)$ and $(k,2)$ is $12$ square units, the possible values of $k$ are
$1, 9$
$-1, 9$
$1, -9$
$-1,-9$
Answers
$21/2$ square units.
$1$ square unit.
$\Delta = 0$, so the three points are collinear (they all lie on $y = 2x$).
$M_{1 1}=6, M_{1 2}=0, M_{1 3}=-3, M_{2 1}=0, M_{2 2}=3, M_{2 3}=0, M_{3 1}=-3, M_{3 2}=0, M_{3 3}=6$; the cofactors are the same numbers, since the sign flips only where a minor is already $0$.
Exercise 4.4 builds the adjoint and uses it to find the inverse. Check each inverse by multiplying it back into $A$. Some questions ask you to verify a rule, such as $A (\text{adj} A) = |A| I$.
Exercise 4.4: Adjoint and inverse
practice Find $\text{adj} A$ for $A = \begin{pmatrix} 3 & 1 \\ 2 & 4 \end{pmatrix}$.
Exercise 4.5 first asks whether a system is consistent. Work out $|A|$, and if it is $0$, work out $(\text{adj} A) B$. Then solve the systems with $|A| \neq 0$ by $X = A^{-1} B$.
Exercise 4.5: Consistency and solving systems by matrix method
practice Examine the consistency of the system $x + 2y = 3$, $3x + 4y = 7$.
practice Examine the consistency of the system $x + 2y = 4$, $2x + 4y = 5$.
practice Examine the consistency of the system $x + 3y = 5$, $2x + 6y = 10$.
practice Examine the consistency of the system $x + y + z = 6$, $x + 2y + 3z = 14$, $2x + y + 3z = 13$.
practice Examine the consistency of the system $x + y + z = 6$, $2x + 3y + 4z = 9$, $3x + 4y + 5z = 14$.
practice Examine the consistency of the system $2x - y + z = 3$, $x + 2y - z = 2$, $x + y + 2z = 6$.
practice Solve $3x + 2y = 7$, $5x + 3y = 11$ using the matrix method.
practice Solve $2x - y = 7$, $3x + 4y = 5$ using the matrix method.
practice Solve $4x + 3y = 11$, $2x + 5y = 9$ using the matrix method.
practice Solve $5x + 4y = 7$, $3x + 2y = 3$ using the matrix method.
practice Solve $x + y + z = 6$, $x + 2y + 3z = 14$, $2x + y + 3z = 13$ using the matrix method.
practice Solve $2x + y + z = 6$, $x + y + z = 4$, $x - y + 2z = 3$ using the matrix method.
practice Solve $x - y + 2z = 7$, $3x + 4y - 5z = -5$, $2x - y + 3z = 12$ using the matrix method.
practice Solve $2x + 3y + 3z = 5$, $x - 2y + z = -4$, $3x - y - 2z = 3$ using the matrix method.
practice The cost of $2$ pens, $3$ notebooks and $1$ eraser is ₹34. The cost of $3$ pens, $1$ notebook and $2$ erasers is ₹35. The cost of $1$ pen, $2$ notebooks and $3$ erasers is ₹27. Find the cost of each item using the matrix method.
Answers
$|A| = -2 \neq 0$, so the system is consistent with a unique solution.
$|A| = 0$ and $(\text{adj} A) B = \begin{pmatrix} 6 \\ -3 \end{pmatrix} \neq \begin{pmatrix} 0 \\ 0 \end{pmatrix}$, so the system is inconsistent.
$|A| = 0$ and $(\text{adj} A) B = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$, so the system is consistent (with infinitely many solutions, since the second equation is just twice the first).
$|A| = 3 \neq 0$, so the system is consistent with a unique solution.
$|A| = 0$ and $(\text{adj} A) B = \begin{pmatrix} -1 \\ 2 \\ -1 \end{pmatrix} \neq \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$, so the system is inconsistent.
$|A| = 12 \neq 0$, so the system is consistent with a unique solution.
$\begin{pmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{pmatrix}$
$(1/(x y z)) \begin{pmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{pmatrix}$
practice If $A$ is a $3 \times 3$ matrix with $|A| = 4$, then $|3A|$ equals
$12$
$36$
$108$
$4$
practice Find $x$ so that $A = \begin{pmatrix} 1 & 2 & x \\ 2 & 1 & 3 \\ 3 & 2 & 1 \end{pmatrix}$ is singular.
practice For $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$, use the determinant test to decide whether $A X = B$ is consistent.
Both sides of the first identity equal $\begin{pmatrix} 1/7 & 2/7 & 1/7 \\ 0 & 1/7 & 3/7 \\ 2/7 & 1/7 & 0 \end{pmatrix}$, and $(A^{-1})^{-1} = A$ by the definition of an inverse.
$-2(x+y)(x^2-x y+y^2)$
$x = 3$, $y = 4$, $z = 5$
Area of $A B C$ is $17/2$ square units. $\Delta_{A B D} = 0$, so $D$ does lie on line $A B$.
$|A|=0$ and $(\text{adj} A) B = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$, so the test cannot decide; the second equation is three times the first, so the system is consistent, with infinitely many solutions.
