Determinants: Class 12 Maths · Chapter 4

One number that decides areas, inverses and solutions

Mathematics Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

90 min
1 simulation
65 practice questions
24 MCQs

What you will learn in Determinants

Evaluate determinants of order 1, 2 and 3, expanding a 3 × 3 determinant along whichever row or column makes the work shortest
Use the rule that multiplying an n × n matrix by k multiplies its determinant by k to the power n, and solve equations whose unknown sits inside a determinant
Find the area of a triangle from its vertices with a determinant, test three points for collinearity and write the equation of a line through two points
Find the minors, cofactors and adjoint of a square matrix of order 2 or 3
Explain why a square matrix multiplied by its adjoint gives its determinant times the identity, and use this to decide whether the matrix is invertible and to find its inverse
Write a system of linear equations as AX = B, test it for consistency and solve it by the matrix method

Original companion text aligned to NCERT Mathematics Class 12, Chapter 4, pp. 76–103. Not affiliated with or endorsed by NCERT.

4.1 Introduction

Draw two straight lines on graph paper. They meet at exactly one point, or they never meet because they are parallel, or one lies exactly on top of the other. A pair of linear equations in $x$ and $y$ asks the same question in algebra: does it have one solution, none, or infinitely many?

For the pair $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ , a single number made from the coefficients answers this before you start solving. Multiply the first equation by $b_{2}$ and the second by $b_{1}$ , then subtract to remove $y$ : you get $(a_{1} b_{2} - a_{2} b_{1}) x = c_{1} b_{2} - c_{2} b_{1}$ . If $a_{1} b_{2} - a_{2} b_{1} \neq 0$ , you can divide, so $x$ is fixed and then $y$ is fixed: exactly one solution. If it is $0$ , the two lines have the same slope, so they are parallel or they coincide (Fig. 4.1).

1/3 $x - y = - 1$ meets $2 x + y = 7$ at $(2, 3)$ , and $a_{1} b_{2} - a_{2} b_{1} = (2) (- 1) - (1) (1) = - 3 \neq 0$ .

Fig. 4.1The number

a_{1} b_{2} - a_{2} b_{1}

is non-zero only when the lines cross.

A coordinate grid with the solid line 2x + y = 7. In the first step a dashed second line x − y = −1 crosses it at the point (2, 3), and the panel reads one solution with a1b2 − a2b1 = −3. In the second step the dashed line is 4x + 2y = 6, parallel to the first, and the panel reads no solution with the number equal to 0. In the third step the dashed line is 4x + 2y = 14, which lies exactly on the solid line, and the panel reads infinitely many solutions, again with the number 0.

Because $a_{1} b_{2} - a_{2} b_{1}$ depends only on the coefficient matrix $[\begin{array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \end{array}]$ , we attach it to that matrix and call it the determinant (सारणिक) of the matrix. We stay with real square matrices no bigger than $3 \times 3$ . First you learn to evaluate determinants quickly; then they earn their keep, giving the area of a triangle from its corners, a formula for the inverse of a matrix, and a test that tells you whether a system of equations can be solved before you try.

Quick check

1 of 2

For which pair of equations is

a_{1} b_{2} - a_{2} b_{1} = 0

x + 2 y = 5

and

2 x - y = 0

3 x + y = 1

and

x + 3 y = 1

x + 2 y = 5

and

2 x + 4 y = 7

x - y = 2

and

x + y = 2

Show answer

Answer (C) $x + 2 y = 5$ and $2 x + 4 y = 7$

Here

a_{1} b_{2} - a_{2} b_{1} = 1 \cdot 4 - 2 \cdot 2 = 0

: the second line has the same slope as the first. The others give

- 5

8

and

2

, so those pairs cross.

4.2 Determinant

Feed a square matrix in, and one number comes out: that output is its determinant. So the determinant is a function from square matrices to numbers. For the real matrices in this book the output is real; matrices with complex entries can have complex determinants. Three notations are common: $∣ A ∣$ , $\det A$ and $Δ$ .

Definition

Determinant

A rule that assigns to each square matrix $A$ a single number, written $∣ A ∣$ , $\det A$ or $Δ$ . In a determinant the entries sit between vertical bars, $∣ \begin{array}{cc} a & b \\ c & d \end{array} ∣$ ; in a matrix, between brackets.

Example $A = [\begin{array}{cc} 7 & 2 \\ 3 & 1 \end{array}]$ is a table of numbers, while $∣ A ∣ = ∣ \begin{array}{cc} 7 & 2 \\ 3 & 1 \end{array} ∣ = 1$ is one number.

4.2.1 Determinant of a matrix of order one

For a $1 \times 1$ matrix $A = [a]$ the determinant is the entry itself: $∣ A ∣ = a$ . So $\det [- 7] = - 7$ . This tiny case already shows that the bars do not mean absolute value.

4.2.2 Determinant of a matrix of order two

For a $2 \times 2$ matrix, multiply the entries on the main diagonal (top-left to bottom-right), multiply the entries on the other diagonal, and subtract the second product from the first (Fig. 4.2).

Order two

\begin{aligned} \det A & = ∣ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} ∣ \\ = a_{11} a_{22} - a_{21} a_{12} \end{aligned}

3/3Subtract: $∣ \begin{array}{cc} 5 & 3 \\ 2 & 4 \end{array} ∣ = 20 - 6 = 14$ .

Fig. 4.2Main diagonal product minus the other diagonal product.NCERT §4.2.2 · p. 77

The determinant with rows 5, 3 and 2, 4. In the first step the main diagonal, 5 and 4, is circled and an arrow runs along it; the panel shows 5 × 4 = 20. In the second step the other diagonal, 3 and 2, is circled with a dashed arrow and the panel adds 3 × 2 = 6. In the third step the panel shows the result 20 − 6 = 14.

Swapping the two rows swaps the two products, so the sign of the determinant flips. Which diagonal you subtract therefore matters.

Example 1NCERT Example 1 · p. 77

A numerical determinant

Evaluate $∣ \begin{array}{cc} 6 & - 2 \\ 5 & 3 \end{array} ∣$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 2NCERT Example 2 · p. 77

A determinant that ignores x

Show that $∣ \begin{array}{cc} x + 2 & x - 1 \\ x + 3 & x \end{array} ∣$ has the same value for every real number $x$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 2

Evaluate

∣ \begin{array}{cc} 4 & - 3 \\ 2 & 5 \end{array} ∣

Show answer

Answer 26

4 \times 5 - (- 3) \times 2 = 20 + 6 = 26

. The tempting

14

comes from forgetting that

- 3 \times 2

is negative, so subtracting it adds

6

4.2.3 Determinant of a matrix of order 3 × 3

A $3 \times 3$ determinant is built from $2 \times 2$ ones. Choose a row, say the first. Multiply each of its entries by the $2 \times 2$ determinant left over when that entry's row and column are deleted. Then combine the three products with the signs $+$ , $-$ , $+$ . This is called the expansion of the determinant along that row.

DerivationExpanding along the first row

Let $A = [a_{i j}]$ be $3 \times 3$ . The entry $a_{1 j}$ gets the sign $(- 1)^{1 + j}$ and the $2 \times 2$ determinant that remains after deleting row 1 and column $j$ :

\begin{aligned} ∣ A ∣ = & a_{11} ∣ \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} ∣ \\ - a_{12} ∣ \begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} ∣ \\ + a_{13} ∣ \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} ∣ \end{aligned}

Opening each small determinant gives six products:

Six-term form

\begin{aligned} ∣ A ∣ = & a_{11} a_{22} a_{33} - a_{11} a_{23} a_{32} \\ - a_{12} a_{21} a_{33} + a_{12} a_{23} a_{31} \\ + a_{13} a_{21} a_{32} - a_{13} a_{22} a_{31} \end{aligned}

Now try the second row, whose signs are $-$ , $+$ , $-$ : $- a_{21} (a_{12} a_{33} - a_{13} a_{32}) + a_{22} (a_{11} a_{33} - a_{13} a_{31}) - a_{23} (a_{11} a_{32} - a_{12} a_{31})$ . Multiply out and match terms: for example $- a_{23} a_{11} a_{32}$ is the second product in the six-term form, and $+ a_{23} a_{12} a_{31}$ is the fourth. All six agree, sign for sign. Expanding along the first column, $a_{11} (\dots) - a_{21} (\dots) + a_{31} (\dots)$ , gives the same six products again.

There is a reason. Each of the six products uses exactly one entry from every row and exactly one from every column. No row or column is special, so all six expansions (three rows, three columns) give the same value.

4/4 $Δ = - 4 - 8 + 63 = 51$ .

Fig. 4.3Each entry of the chosen row is multiplied by the determinant that is left when its row and column are removed.

The determinant with rows 2, 1, 3; 4, −1, 0; 1, 5, 2, with the signs plus, minus, plus above the columns. In the first step the entry 2 is circled, row 1 and column 1 are struck out, and the four remaining cells are shaded; the panel computes the minor (−1)(2) − (0)(5) = −2 and the term +2 × (−2) = −4. In the second step the same is done for the entry 1 in column 2: minor 8, term −1 × 8 = −8. In the third step for the entry 3 in column 3: minor 21, term +3 × 21 = 63. In the last step the panel adds −4 − 8 + 63 to give 51.

You do not need to compute $(- 1)^{i + j}$ each time. The signs form a checkerboard that starts with $+$ in the top-left corner, so each row and each column alternates (Fig. 4.4).

3/3Expanding along $C_{3}$ uses the signs $+, -, +$ .

Fig. 4.4The sign of the place in row

i

and column

j

(- 1)^{i + j}

A 3 by 3 grid of signs: plus, minus, plus in the first row; minus, plus, minus in the second; plus, minus, plus in the third, with the plus squares shaded. Rows are labelled R1 to R3 and columns C1 to C3. Next to it: the sign of place (i, j) is minus one to the power i + j, plus when i + j is even and minus when it is odd. In the second step row 2 is outlined and the panel reads: along R2, minus plus minus. In the third step column 3 is outlined: along C3, plus minus plus.

Example 3NCERT Example 3 · p. 80

Let the zeros do the work

Evaluate $Δ = ∣ \begin{array}{ccc} 2 & - 1 & 5 \\ 0 & 3 & 0 \\ 4 & 1 & - 2 \end{array} ∣$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 4NCERT Example 4 · p. 80

A determinant with trigonometric entries

Show that $Δ = ∣ \begin{array}{ccc} 1 & \sin θ & \cos θ \\ - \sin θ & 1 & 0 \\ - \cos θ & 0 & 1 \end{array} ∣$ does not depend on $θ$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 5NCERT Example 5 · p. 81

Solving for an unknown inside a determinant

Find all real $x$ for which $∣ \begin{array}{cc} 2 & x \\ x & 8 \end{array} ∣ = ∣ \begin{array}{cc} 3 & 1 \\ 5 & 4 \end{array} ∣$ .

Try it yourself first. Then reveal the solution one step at a time.

PropertyScaling a whole matrix

Multiply every entry of an $n \times n$ matrix $A$ by a number $k$ , and the determinant is multiplied by $k^{n}$ : $∣ k A ∣ = k^{n} ∣ A ∣$ .

A frequent slip is to write $∣ k A ∣ = k ∣ A ∣$ ; that holds only for $1 \times 1$ matrices. A picture shows why the power appears. For a $2 \times 2$ matrix, $∣ A ∣$ is the signed area of the parallelogram whose sides are the two columns of $A$ drawn as arrows. Doubling one column doubles the area; doubling both multiplies it by $2^{2}$ (Fig. 4.5). For $3 \times 3$ matrices the same idea works with volumes and $k^{3}$ .

3/3Doubling both columns multiplies them by $2^{2}$ : $∣ 2 A ∣ = 20 = 2^{2} ∣ A ∣$ .

Fig. 4.5Scaling every column by 2 scales the area, and the determinant, by 2 × 2 = 4.

A grid with two arrows from the origin, (3, 1) and (1, 2), forming a shaded parallelogram of area 5, which equals the determinant with rows 3, 1 and 1, 2. In the second step the first arrow is doubled to (6, 2); the parallelogram, drawn over a dashed outline of the original, has area 10. In the third step both arrows are doubled, to (6, 2) and (2, 4), and the area is 20, which is 4 × 5.

Practice · 8 questions

Practice 4.1: questions with answers

Short answerEasyNCERT Exercise 4.1, Q1
Evaluate $∣ \begin{array}{cc} - 3 & 7 \\ 2 & - 4 \end{array} ∣$ .
Show solution
$(- 3) (- 4) - 7 \times 2 = 12 - 14 = - 2$ .
2
Short answerMediumNCERT Exercise 4.1, Q2
Evaluate each determinant.
1. $∣ \begin{array}{cc} \sec θ & \tan θ \\ \tan θ & \sec θ \end{array} ∣$ (for $\cos θ \neq 0$ )
  Show solution
  Answer $1$
  $\sec^{2} θ - \tan^{2} θ = 1$ , from $1 + \tan^{2} θ = \sec^{2} θ$ .
2. $∣ \begin{array}{cc} x^{2} + 1 & x \\ x^{3} & x^{2} + 1 \end{array} ∣$
  Show solution
  Answer $2 x^{2} + 1$
  $(x^{2} + 1)^{2} - x \cdot x^{3} = x^{4} + 2 x^{2} + 1 - x^{4} = 2 x^{2} + 1$ .
Short answerEasyNCERT Exercise 4.1, Q3
Let $A = [\begin{array}{cc} 3 & - 1 \\ 2 & 5 \end{array}]$ . Show that $∣ 3 A ∣ = 9 ∣ A ∣$ .
Show solution
$∣ A ∣ = 15 - (- 2) = 17$ . $3 A = [\begin{array}{cc} 9 & - 3 \\ 6 & 15 \end{array}]$ , so $∣ 3 A ∣ = 135 - (- 18) = 153$ .
$153 = 9 \times 17$ , so $∣ 3 A ∣ = 9 ∣ A ∣ = 3^{2} ∣ A ∣$ , as the scaling rule predicts for order 2.
Short answerEasyNCERT Exercise 4.1, Q4
Let $A = [\begin{array}{ccc} 2 & 0 & 0 \\ 1 & - 1 & 0 \\ 4 & 3 & 5 \end{array}]$ . Show that $∣ 2 A ∣ = 8 ∣ A ∣$ .
Show solution
Along $R_{1}$ : $∣ A ∣ = 2 ∣ \begin{array}{cc} - 1 & 0 \\ 3 & 5 \end{array} ∣ = 2 (- 5) = - 10$ .
$2 A = [\begin{array}{ccc} 4 & 0 & 0 \\ 2 & - 2 & 0 \\ 8 & 6 & 10 \end{array}]$ , so $∣ 2 A ∣ = 4 ∣ \begin{array}{cc} - 2 & 0 \\ 6 & 10 \end{array} ∣ = 4 (- 20) = - 80$ .
$- 80 = 8 \times (- 10)$ , so $∣ 2 A ∣ = 2^{3} ∣ A ∣$ .
5
Short answerMediumNCERT Exercise 4.1, Q5
Evaluate each determinant, choosing a convenient row or column.
1. $∣ \begin{array}{ccc} 4 & 0 & - 1 \\ 2 & 0 & 3 \\ - 1 & 5 & 2 \end{array} ∣$
  Show solution
  Column 2 has two zeros. Its only non-zero entry, 5, sits at place $(3, 2)$ with sign $-$ : $- 5 ∣ \begin{array}{cc} 4 & - 1 \\ 2 & 3 \end{array} ∣ = - 5 (12 + 2) = - 70$ .
2. $∣ \begin{array}{ccc} 2 & - 3 & 1 \\ 4 & 1 & - 2 \\ 3 & 2 & 5 \end{array} ∣$
  Show solution
  Along $R_{1}$ : $2 (5 + 4) + 3 (20 + 6) + 1 (8 - 3) = 18 + 78 + 5 = 101$ .
3. $∣ \begin{array}{ccc} 0 & 2 & - 5 \\ - 2 & 1 & 4 \\ 5 & - 4 & 0 \end{array} ∣$
  Show solution
  Along $R_{1}$ (the first entry is 0): $- 2 ∣ \begin{array}{cc} - 2 & 4 \\ 5 & 0 \end{array} ∣ + (- 5) ∣ \begin{array}{cc} - 2 & 1 \\ 5 & - 4 \end{array} ∣ = - 2 (- 20) - 5 (8 - 5) = 40 - 15 = 25$ .
4. $∣ \begin{array}{ccc} 1 & - 2 & 3 \\ 0 & 4 & - 1 \\ 2 & - 3 & 0 \end{array} ∣$
  Show solution
  Along $C_{1}$ : $1 ∣ \begin{array}{cc} 4 & - 1 \\ - 3 & 0 \end{array} ∣ - 0 + 2 ∣ \begin{array}{cc} - 2 & 3 \\ 4 & - 1 \end{array} ∣ = 1 (0 - 3) + 2 (2 - 12) = - 3 - 20 = - 23$ .
Short answerMediumNCERT Exercise 4.1, Q6
Let $A = [\begin{array}{ccc} 2 & - 1 & 3 \\ 1 & 4 & - 2 \\ 4 & 7 & - 1 \end{array}]$ . Find $∣ A ∣$ , and compare row 3 with rows 1 and 2.
Show solution
Answer $∣ A ∣ = 0$ ; row 3 = row 1 + 2 × row 2.
Along $R_{1}$ : $2 (- 4 + 14) + 1 (- 1 + 8) + 3 (7 - 16) = 20 + 7 - 27 = 0$ .
Row 3 is $(2, - 1, 3) + 2 (1, 4, - 2) = (4, 7, - 1)$ , a combination of the other rows. Whenever one row is built from the others like this, the determinant is 0.
7
Short answerMediumNCERT Exercise 4.1, Q7
Find the values of $x$ .
1. $∣ \begin{array}{cc} x & 3 \\ 4 & x \end{array} ∣ = ∣ \begin{array}{cc} 3 & - 2 \\ 2 & 3 \end{array} ∣$
  Show solution
  Answer $x = \pm 5$
  $x^{2} - 12 = 9 + 4 = 13$ , so $x^{2} = 25$ and $x = 5$ or $x = - 5$ .
2. $∣ \begin{array}{cc} 3 x & 2 \\ x & 4 \end{array} ∣ = ∣ \begin{array}{cc} 9 & 3 \\ 2 & 4 \end{array} ∣$
  Show solution
  Answer $x = 3$
  $12 x - 2 x = 36 - 6$ , so $10 x = 30$ and $x = 3$ .
Quick answerEasyNCERT Exercise 4.1, Q8
If $∣ \begin{array}{cc} x & 2 \\ 4 & x \end{array} ∣ = ∣ \begin{array}{cc} 3 & 1 \\ 1 & 3 \end{array} ∣$ , then $x$ equals
$4$ only $\pm 2 \sqrt{2}$ $\pm 4$ $16$
Show answer
Answer (C) $\pm 4$
$x^{2} - 8 = 9 - 1 = 8$ , so $x^{2} = 16$ and $x = \pm 4$ . Both signs work; $\pm 2 \sqrt{2}$ forgets to move the $- 8$ across.

4.3 Area of a Triangle

A surveyor who knows the corners of a plot of land, or a phone app that reads them from GPS, finds the area by splitting the plot into triangles. The area of each triangle comes straight from the coordinates of its corners, and the formula is a determinant.

Here is the formula first; where it comes from follows. Let the corners be $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ .

Area of a triangle

\begin{array}{c} Δ = \frac{1}{2} ∣ \begin{array}{ccc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} ∣ \\ area of △ A B C = ∣ Δ ∣ \end{array}

Expanding along the first column turns the determinant into $x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})$ . That is the bracket in the coordinate-geometry area formula from Class 10, so the determinant is just a tidy way to store it.

DerivationWhy the determinant gives the area

Place the triangle above the $x$ -axis with $x_{1} < x_{2} < x_{3}$ and $B$ above the side $A C$ , and drop perpendiculars $A L$ , $B M$ , $C N$ to the axis (Fig. 4.6). The triangle is what remains of the two trapezia $A L M B$ and $B M N C$ after removing the trapezium $A L N C$ . A trapezium with parallel sides $p$ , $q$ and width $w$ has area $\frac{1}{2} (p + q) w$ , so the area of the triangle is

\begin{aligned} \frac{1}{2} (y_{1} + y_{2}) (x_{2} - x_{1}) \\ + & \frac{1}{2} (y_{2} + y_{3}) (x_{3} - x_{2}) \\ - & \frac{1}{2} (y_{1} + y_{3}) (x_{3} - x_{1}) \end{aligned}

Multiplying out, the terms $y_{1} x_{1}$ , $y_{2} x_{2}$ and $y_{3} x_{3}$ cancel, and what is left is $\frac{1}{2} [x_{1} (y_{3} - y_{2}) + x_{2} (y_{1} - y_{3}) + x_{3} (y_{2} - y_{1})] = - Δ$ . Going $A \to B \to C$ in this picture turns clockwise, and for clockwise order $Δ$ is negative, so the area is $- Δ = ∣ Δ ∣$ . Listing the vertices anticlockwise gives $+ Δ$ , and the other placements of a triangle work out the same way.

3/3Removing the trapezium under AC leaves exactly the triangle.

Fig. 4.6Area of ABC = ALMB + BMNC − ALNC.

Triangle ABC above the x-axis, with A on the left, B highest and C on the right; dashed perpendiculars drop from A, B and C to the points L, M and N on the axis. The first step shows the shaded triangle and the coordinates of the vertices. In the second step the two trapezia ALMB and BMNC are shaded in two colours and the panel reads ALMB + BMNC. In the third step the large trapezium ALNC is outlined with a dashed border, the triangle is shaded again, and the panel reads minus ALNC equals the area of ABC.

So the determinant has a sign as well as a size: it is positive when the vertices are listed anticlockwise and negative when they are listed clockwise (Fig. 4.7). Area cannot be negative, so we always take $∣ Δ ∣$ . When the area is given and a vertex is unknown, both signs are possible: solve $Δ = + (area)$ and $Δ = - (area)$ separately.

1/3Order $(1, 1) \to (5, 2) \to (2, 4)$ turns anticlockwise: the determinant is $+ 11$ , so $Δ = 5.5$ and the area is $5.5$ square units.

Fig. 4.7The sign of the determinant records the order of the vertices; its size gives the area.

A grid with the points (1, 1), (5, 2) and (2, 4). In the first step the triangle is shaded, the vertices are numbered 1st (1, 1), 2nd (5, 2), 3rd (2, 4), an arrow turns anticlockwise inside it, and the panel reads Δ = +11 and area 5.5. In the second step the order of the last two vertices is swapped, the outline is dashed, the arrow turns clockwise and the panel reads Δ = −11 with the same area. In the third step the third point moves to (3, 1.5) on the segment from (1, 1) to (5, 2); the triangle collapses to that segment and the panel reads Δ = 0, collinear.

The last picture gives a test. Three points are collinear exactly when $Δ = 0$ . It also gives the equation of a line: a point $P (x, y)$ lies on the line through $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ exactly when the three points are collinear, that is, when

Line through two points

∣ \begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array} ∣ = 0

Example 6NCERT Example 6 · p. 82

Area from three vertices

Find the area of the triangle with vertices $P (- 2, 1)$ , $Q (3, 4)$ and $R (5, - 2)$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 7NCERT Example 7 · p. 83

A line, then a point at a given area

(a) Use a determinant to find the equation of the line through $A (2, 1)$ and $B (4, 5)$ . (b) Find every point $D$ on the $y$ -axis for which triangle $A B D$ has area 6 square units.

Try it yourself first. Then reveal the solution one step at a time.

SimulationMove a vertex, predict the area

A coordinate grid from −6 to 8 across and −5 to 7 up, with three draggable points A(−2, −1), B(4, 2) and C(0, 4) joined into a triangle. Each side shows its length and each point shows its coordinates, which snap to whole numbers as you drag, so you can read off the vertices and compute the area determinant for any position.

Fig. 4.8Drag the vertices; the sides and coordinates update. Work out the determinant for each position you try.

Try this

Use the starting coordinates to show that the area is 12 square units. Then drag C to (2, 5). Predict the new area before you compute it, and explain the result using the line AB.
Drag C to (2, 1) or to (0, 0). What happens to the triangle, and what is the determinant now?
Drag C down to (0, −3), so it is on the other side of AB. Compute Δ with the vertices in the order A, B, C. What happened to its sign?

Practice · 5 questions

Practice 4.2: questions with answers

1
Short answerEasyNCERT Exercise 4.2, Q1
Find the area of the triangle with the given vertices.
1. $(0, 0)$ , $(5, 2)$ , $(1, 4)$
  square units
  Show solution
  $\frac{1}{2} ∣ \begin{array}{ccc} 0 & 0 & 1 \\ 5 & 2 & 1 \\ 1 & 4 & 1 \end{array} ∣ = \frac{1}{2} (20 - 2) = 9$ (expanding along $R_{1}$ , only the last entry survives). Area 9 square units.
2. $(- 3, 2)$ , $(4, 5)$ , $(1, - 2)$
  square units
  Show solution
  Determinant: $- 3 (5 + 2) - 2 (4 - 1) + 1 (- 8 - 5) = - 21 - 6 - 13 = - 40$ , so $Δ = - 20$ and the area is 20 square units.
3. $(1, - 2)$ , $(6, 3)$ , $(- 2, 4)$
  square units
  Show solution
  Determinant: $1 (3 - 4) + 2 (6 + 2) + 1 (24 + 6) = - 1 + 16 + 30 = 45$ , so the area is $22.5$ square units.
ProofMediumNCERT Exercise 4.2, Q2
Show that the points $P (2 p, 3 p - 1)$ , $Q (2 q, 3 q - 1)$ and $R (2 r, 3 r - 1)$ are collinear for all real $p$ , $q$ , $r$ .
Show solution
Expanding $∣ \begin{array}{ccc} 2 p & 3 p - 1 & 1 \\ 2 q & 3 q - 1 & 1 \\ 2 r & 3 r - 1 & 1 \end{array} ∣$ along $C_{3}$ (signs $+, -, +$ ):
$+ [2 q (3 r - 1) - 2 r (3 q - 1)] - [2 p (3 r - 1) - 2 r (3 p - 1)] + [2 p (3 q - 1) - 2 q (3 p - 1)]$ .
The brackets simplify to $2 r - 2 q$ , $2 r - 2 p$ and $2 q - 2 p$ , so the total is $(2 r - 2 q) - (2 r - 2 p) + (2 q - 2 p) = 0$ . Hence the points are collinear. (They all lie on $3 x - 2 y = 2$ .)
3
Short answerMediumNCERT Exercise 4.2, Q3
Find the possible values of k.
1. The triangle with vertices $(1, k)$ , $(4, - 2)$ , $(- 2, 4)$ has area 6 square units.
  Show solution
  Answer $k = - 1$ or $k = 3$
  $∣ \begin{array}{ccc} 1 & k & 1 \\ 4 & - 2 & 1 \\ - 2 & 4 & 1 \end{array} ∣ = 1 (- 2 - 4) - k (4 + 2) + 1 (16 - 4) = 6 - 6 k$ , so $Δ = 3 - 3 k$ .
  $3 - 3 k = 6$ gives $k = - 1$ ; $3 - 3 k = - 6$ gives $k = 3$ .
2. The triangle with vertices $(2, 1)$ , $(k, 3)$ , $(- 1, 5)$ has area 5 square units.
  Show solution
  Answer $k = 3$ or $k = - 2$
  $∣ \begin{array}{ccc} 2 & 1 & 1 \\ k & 3 & 1 \\ - 1 & 5 & 1 \end{array} ∣ = 2 (3 - 5) - 1 (k + 1) + 1 (5 k + 3) = 4 k - 2$ , so $Δ = 2 k - 1$ .
  $2 k - 1 = 5$ gives $k = 3$ ; $2 k - 1 = - 5$ gives $k = - 2$ .
4
Short answerEasyNCERT Exercise 4.2, Q4
Use determinants to find the equation of the line through the two points.
1. $(2, 3)$ and $(6, 11)$
  Show solution
  Answer $y = 2 x - 1$
  $∣ \begin{array}{ccc} x & y & 1 \\ 2 & 3 & 1 \\ 6 & 11 & 1 \end{array} ∣ = x (3 - 11) - y (2 - 6) + (22 - 18) = - 8 x + 4 y + 4 = 0$ , i.e. $y = 2 x - 1$ .
2. $(- 1, 4)$ and $(5, 1)$
  Show solution
  Answer $x + 2 y = 7$
  $x (4 - 1) - y (- 1 - 5) + (- 1 - 20) = 3 x + 6 y - 21 = 0$ , i.e. $x + 2 y = 7$ .
Quick answerMediumNCERT Exercise 4.2, Q5
The points $(2, 1)$ , $(5, 3)$ and $(k, k)$ form a triangle of area 4 square units. The possible values of $k$ are
$k = 7$ only $k = 3$ or $k = - 5$ $k = 7$ or $k = - 9$ $k = - 7$ or $k = 9$
Show answer
Answer (C) $k = 7$ or $k = - 9$
The determinant is $2 (3 - k) - 1 (5 - k) + (5 k - 3 k) = k + 1$ , so $Δ = \frac{k + 1}{2} = \pm 4$ , giving $k = 7$ or $k = - 9$ . Forgetting the $\frac{1}{2}$ gives $3$ or $- 5$ ; using only the plus sign loses $- 9$ .

4.4 Minors and Cofactors

The small determinants that appeared in every expansion keep coming back: in expansions, in the adjoint and in the inverse. They are worth naming, and naming them lets us write an expansion in one short line.

Definition

Minor

The minor $M_{i j}$ of the entry $a_{i j}$ is the determinant that remains after deleting row $i$ and column $j$ . So a minor is always one size smaller than the determinant it comes from: order $n - 1$ when the determinant has order $n \geq 2$ .

ExampleIn a $3 \times 3$ determinant every minor is a $2 \times 2$ determinant.

3/3 $3 + 2 = 5$ is odd, so $A_{32} = (- 1)^{5} M_{32} = - 22$ .

Fig. 4.9Minor: delete the entry's row and column. Cofactor: attach the sign

(- 1)^{i + j}

The determinant with rows 3, 0, −2; 5, 1, 4; −1, 6, 2. In the first step the entry 6 is circled and the panel says it sits in row 3 and column 2. In the second step row 3 and column 2 are struck out, the four remaining cells 3, −2, 5, 4 are shaded, and the panel computes (3)(4) − (−2)(5) = 22, the minor M32. In the third step the panel notes that 3 + 2 = 5 is odd, so the sign is minus and the cofactor A32 is −22.

Example 8NCERT Example 8 · p. 84

One minor

In $Δ = ∣ \begin{array}{ccc} 3 & 0 & - 2 \\ 5 & 1 & 4 \\ - 1 & 6 & 2 \end{array} ∣$ , find the minor of the entry 4.

Try it yourself first. Then reveal the solution one step at a time.

Definition

Cofactor

The cofactor of $a_{i j}$ is $A_{i j} = (- 1)^{i + j} M_{i j}$ : the minor with the checkerboard sign of its place attached.

ExampleIn Fig. 4.9, $M_{32} = 22$ and $A_{32} = - 22$ .

Cofactor

A_{i j} = (- 1)^{i + j} M_{i j}

Example 9NCERT Example 9 · p. 84

All minors and cofactors of a 2 × 2 determinant

Find the minors and cofactors of all the entries of $∣ \begin{array}{cc} 5 & - 3 \\ 2 & 7 \end{array} ∣$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 10NCERT Example 10 · p. 85

Minors and cofactors in letters

For $Δ = ∣ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} ∣$ , find the minors and cofactors of $a_{12}$ and $a_{32}$ .

Try it yourself first. Then reveal the solution one step at a time.

PropertyExpansion in cofactors

For a $3 \times 3$ matrix $A$ and any row $i$ , $∣ A ∣ = a_{i 1} A_{i 1} + a_{i 2} A_{i 2} + a_{i 3} A_{i 3}$ . For any column $j$ , $∣ A ∣ = a_{1 j} A_{1 j} + a_{2 j} A_{2 j} + a_{3 j} A_{3 j}$ .

What happens if you take the entries of one row but the cofactors of a different row? The answer is always zero, and it is the key to the inverse in the next section.

PropertyCofactors of the wrong line

If the entries of one row are multiplied by the cofactors of a different row and the products are added, the sum is $0$ ; the same holds for columns. For example, $a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13} = 0$ .

The cofactors $A_{11}$ , $A_{12}$ , $A_{13}$ are built from rows 2 and 3 only. So they are also the row-1 cofactors of the matrix $B$ made from $A$ by copying row 2 into row 1. Expanding $∣ B ∣$ along its first row gives exactly $a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13}$ .

Now expand the same $∣ B ∣$ along its third row instead. Each minor there comes from rows 1 and 2 of $B$ , which are equal, so each is like $∣ \begin{array}{cc} p & q \\ p & q \end{array} ∣ = p q - q p = 0$ . Hence $∣ B ∣ = 0$ , and the sum is $0$ .

3/3Row 2 entries times row-1 cofactors: $3 (- 10) + 0 (- 23) + 5 (6) = 0$ .

Fig. 4.10Entries of one row times the cofactors of another row: a determinant with two equal rows, which is 0.

The determinant with rows 1, 4, −2; 3, 0, 5; −1, 2, 6, with rows 2 and 3 shaded. The first step notes that the row-1 cofactors −10, −23 and 6 are built from rows 2 and 3 only. In the second step row 1 is replaced by a copy of row 2, 3, 0, 5, shaded in another colour; the cofactors are unchanged. In the third step the panel computes 3(−10) + 0(−23) + 5(6) = 0, because the new determinant has two equal rows.

Example 11NCERT Example 11 · p. 86

Every cofactor, and a sum that vanishes

Find all the minors and cofactors of $Δ = ∣ \begin{array}{ccc} 1 & 4 & - 2 \\ 3 & 0 & 5 \\ - 1 & 2 & 6 \end{array} ∣$ . Use them to find $Δ$ , and check that $a_{11} A_{12} + a_{21} A_{22} + a_{31} A_{32} = 0$ .

Try it yourself first. Then reveal the solution one step at a time.

The lab below does these calculations for any $3 \times 3$ matrix you type. Expand along each of the six lines in turn, then mix the entries of one line with the cofactors of another, and watch the total.

InteractiveCofactor lab

An interactive 3 by 3 matrix whose nine entries you can type, with presets for an example, a matrix with many zeros, a singular matrix and a random one. In the Expand tab you choose the row or column whose entries are used (R1 to C3) and whose cofactors are used. Three term cards show each entry, its minor worked out from the struck-out grid, its sign, its cofactor and the product, and tapping a card shades that minor in the matrix. The total is shown with a message: it equals the determinant when the line's own cofactors are used, and it is always 0 when another line's cofactors are used. The Adjoint tab shows the matrix of cofactors, its transpose adj A, the product A times adj A, and the inverse as fractions when the determinant is not zero.

Fig. 4.11Type a matrix, pick the row or column to expand along, and tap a term to see its minor. The second tab builds the adjoint and the inverse.

Quick check

1 of 4

∣ \begin{array}{ccc} 2 & 7 & 1 \\ 0 & 3 & - 4 \\ 5 & 1 & 6 \end{array} ∣

, find the cofactor

A_{21}

Show answer

Answer -41

Delete row 2 and column 1:

M_{21} = ∣ \begin{array}{cc} 7 & 1 \\ 1 & 6 \end{array} ∣ = 42 - 1 = 41

. Place

(2, 1)

has sign

-

, so

A_{21} = - 41

. The answer

41

is the minor, not the cofactor.

Practice · 5 questions

Practice 4.3: questions with answers

1
Short answerEasyNCERT Exercise 4.3, Q1
Write the minors and cofactors of every entry.
1. $∣ \begin{array}{cc} - 5 & 2 \\ 6 & 1 \end{array} ∣$
  Show solution
  Answer $M_{11} = 1, M_{12} = 6, M_{21} = 2, M_{22} = - 5$ ; $A_{11} = 1, A_{12} = - 6, A_{21} = - 2, A_{22} = - 5$
  Deleting row 1, column 1 leaves $1$ ; row 1, column 2 leaves $6$ ; row 2, column 1 leaves $2$ ; row 2, column 2 leaves $- 5$ . With signs: $A_{11} = 1$ , $A_{12} = - 6$ , $A_{21} = - 2$ , $A_{22} = - 5$ .
2. $∣ \begin{array}{cc} p & q \\ r & s \end{array} ∣$
  Show solution
  Answer $M_{11} = s, M_{12} = r, M_{21} = q, M_{22} = p$ ; $A_{11} = s, A_{12} = - r, A_{21} = - q, A_{22} = p$
  The same pattern with letters: minors $s, r, q, p$ and cofactors $s, - r, - q, p$ . Check: $p A_{11} + q A_{12} = p s - q r$ , the determinant.
2
Short answerMediumNCERT Exercise 4.3, Q2
Write the minors and cofactors of every entry.
1. $∣ \begin{array}{ccc} 2 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 3 \end{array} ∣$
  Show solution
  Answer $M_{11} = A_{11} = - 3$ , $M_{22} = A_{22} = 6$ , $M_{33} = A_{33} = - 2$ ; every other minor and cofactor is 0.
  $M_{11} = (- 1) (3) - 0 = - 3$ , $M_{22} = (2) (3) = 6$ , $M_{33} = (2) (- 1) = - 2$ . Each off-diagonal minor contains a whole row or column of zeros, so it is 0. Diagonal places have sign $+$ , so the cofactors equal the minors.
2. $∣ \begin{array}{ccc} 2 & - 1 & 0 \\ 1 & 3 & 4 \\ 0 & 5 & - 2 \end{array} ∣$
  Show solution
  AnswerMinors: $- 26, - 2, 5; 2, - 4, 10; - 4, 8, 7$ . Cofactors: $- 26, 2, 5; - 2, - 4, - 10; - 4, - 8, 7$ .
  Row 1: $M_{11} = - 6 - 20 = - 26$ , $M_{12} = - 2 - 0 = - 2$ , $M_{13} = 5 - 0 = 5$ .
  Row 2: $M_{21} = 2 - 0 = 2$ , $M_{22} = - 4 - 0 = - 4$ , $M_{23} = 10 - 0 = 10$ .
  Row 3: $M_{31} = - 4 - 0 = - 4$ , $M_{32} = 8 - 0 = 8$ , $M_{33} = 6 + 1 = 7$ .
  Cofactors flip the signs at places $(1, 2), (2, 1), (2, 3), (3, 2)$ . Check: $2 (- 26) + (- 1) (2) + 0 (5) = - 54 = Δ$ .
Short answerMediumNCERT Exercise 4.3, Q3
Using the cofactors of the entries of the third row, evaluate $Δ = ∣ \begin{array}{ccc} 4 & 1 & - 3 \\ 2 & 5 & 1 \\ 0 & - 2 & 6 \end{array} ∣$ .
Show solution
$A_{31} = + ∣ \begin{array}{cc} 1 & - 3 \\ 5 & 1 \end{array} ∣ = 16$ , $A_{32} = - ∣ \begin{array}{cc} 4 & - 3 \\ 2 & 1 \end{array} ∣ = - 10$ , $A_{33} = + ∣ \begin{array}{cc} 4 & 1 \\ 2 & 5 \end{array} ∣ = 18$ .
$Δ = 0 (16) + (- 2) (- 10) + 6 (18) = 20 + 108 = 128$ .
ProofHardNCERT Exercise 4.3, Q4
Using the cofactors of the entries of the first column, show that $∣ \begin{array}{ccc} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array} ∣ = (a - b) (b - c) (c - a)$ .
Show solution
$A_{11} = b c^{2} - b^{2} c$ , $A_{21} = - (a c^{2} - a^{2} c) = a^{2} c - a c^{2}$ , $A_{31} = a b^{2} - a^{2} b$ .
So $Δ = b c^{2} - b^{2} c + a^{2} c - a c^{2} + a b^{2} - a^{2} b$ .
Now expand the right side: $(a - b) (b - c) = a b - a c - b^{2} + b c$ , and multiplying by $(c - a)$ gives $- a^{2} b + a^{2} c + a b^{2} - a c^{2} - b^{2} c + b c^{2}$ , the same six terms.
Quick answerEasyNCERT Exercise 4.3, Q5
$Δ = ∣ a_{i j} ∣$ is a $3 \times 3$ determinant and $A_{i j}$ is the cofactor of $a_{i j}$ . Which expression equals $Δ$ ?
$a_{12} A_{11} + a_{22} A_{21} + a_{32} A_{31}$ $a_{21} A_{31} + a_{22} A_{32} + a_{23} A_{33}$ $a_{12} M_{12} + a_{22} M_{22} + a_{32} M_{32}$ $a_{12} A_{12} + a_{22} A_{22} + a_{32} A_{32}$
Show answer
Answer (D) $a_{12} A_{12} + a_{22} A_{22} + a_{32} A_{32}$
Option D is the expansion along column 2. A and B mix two different lines (each sums to 0), and C drops the signs of the cofactors.

4.5 Adjoint and Inverse of a Matrix

For a number $a \neq 0$ the inverse is simply $\frac{1}{a}$ . Is there a formula like that for a square matrix? There is, and it also picks out exactly the matrices that have no inverse at all. Its main ingredient is the cofactors, packed into a new matrix. (In Chapter 3 you found inverses by elementary operations instead.)

4.5.1 Adjoint of a matrix

Definition

Adjoint

For a square matrix $A$ , the adjoint $adj A$ is the transpose of the matrix whose entries are the cofactors $A_{i j}$ .

Adjoint

\begin{aligned} adj A & = {[\begin{array}{ccc} A_{11} & A_{12} & A_{13} \\ A_{21} & A_{22} & A_{23} \\ A_{31} & A_{32} & A_{33} \end{array}]}^{T} \\ = [\begin{array}{ccc} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end{array}] \end{aligned}

Read it as: the cofactors of row 1 of $A$ go down column 1 of $adj A$ , the cofactors of row 2 go down column 2, and so on (Fig. 4.12).

3/3Reflect in the main diagonal: rows become columns. The result is $adj A$ .

Fig. 4.12The adjoint is the transpose of the matrix of cofactors.

On the left the matrix A with rows 1, 0, 2; 3, 1, 0; 0, 2, 1, and an arrow to a matrix on the right. In the first step the entry 0 in row 1 is circled, the cells of its minor are shaded, and only its cofactor −3 appears on the right, with the note that it equals minus (3·1 − 0·0). In the second step the right matrix is filled with all nine cofactors: 1, −3, 6; 4, 1, −2; −2, 6, 1. In the third step a dashed line marks the main diagonal and the right matrix becomes its transpose, adj A, with rows 1, 4, −2; −3, 1, 6; 6, −2, 1.

Example 12NCERT Example 12 · p. 88

Adjoint of a 2 × 2 matrix

Find $adj A$ for $A = [\begin{array}{cc} 4 & - 1 \\ 3 & 2 \end{array}]$ .

Try it yourself first. Then reveal the solution one step at a time.

Theorem 1Multiplying by the adjoint

Multiplying any square matrix by its adjoint, on either side, gives its determinant times the identity matrix $I$ of the same size: $A (adj A) = ∣ A ∣ I = (adj A) A$ .

Take $n = 3$ . The entry in row $i$ , column $j$ of $A (adj A)$ is row $i$ of $A$ times column $j$ of $adj A$ . Column $j$ of $adj A$ holds the cofactors of row $j$ of $A$ , so this entry is $a_{i 1} A_{j 1} + a_{i 2} A_{j 2} + a_{i 3} A_{j 3}$ .

When $i = j$ it is the expansion of $∣ A ∣$ along row $i$ . When $i \neq j$ it pairs one row with the cofactors of another, which gives $0$ (Property). So the product has $∣ A ∣$ in every diagonal place and $0$ elsewhere: it is $∣ A ∣ I$ . Using columns instead of rows gives $(adj A) A = ∣ A ∣ I$ in the same way.

3/3 $A (adj A) = 13 I = ∣ A ∣ I$ .

Fig. 4.13Diagonal entries are expansions of

∣ A ∣

; off-diagonal entries mix two rows and vanish.

Three matrices in a row: A with rows 1, 0, 2; 3, 1, 0; 0, 2, 1, then adj A with rows 1, 4, −2; −3, 1, 6; 6, −2, 1, then an equals sign and the product. Row 1 of A is shaded throughout. In the first step column 1 of adj A is shaded and the product's top-left entry 13 appears, with the working 1(1) + 0(−3) + 2(6) = 13. In the second step column 2 is shaded and the next entry 0 appears, from 1(4) + 0(1) + 2(−2) = 0. In the third step the full product is shown, 13 on the diagonal and 0 elsewhere, labelled |A| I.

Singular matrices and the inverse

Definition

Singular matrix

A square matrix $A$ is singular if $∣ A ∣ = 0$ .

Example $[\begin{array}{cc} 3 & 6 \\ 1 & 2 \end{array}]$ is singular, since $3 \cdot 2 - 6 \cdot 1 = 0$ .

Definition

Non-singular matrix

A square matrix $A$ is non-singular if $∣ A ∣ \neq 0$ .

Example $[\begin{array}{cc} 2 & 5 \\ 1 & 4 \end{array}]$ is non-singular: its determinant is $3$ .

The next two facts are proved in higher mathematics; here we use them freely and check one on numbers.

Theorem 2Products of non-singular matrices

The product of two non-singular square matrices of equal size, taken in either order, is again non-singular.

Theorem 3Determinant of a product

For square matrices $A$ and $B$ of the same order, $∣ A B ∣ = ∣ A ∣ ∣ B ∣$ .

CorollaryDeterminant of the adjoint

For a non-singular $n \times n$ matrix $A$ , $∣ adj A ∣ = ∣ A ∣^{n - 1}$ ; for a $3 \times 3$ matrix this is $∣ A ∣^{2}$ .

Theorem 4When a matrix has an inverse

An inverse of a square matrix $A$ exists exactly when $∣ A ∣ \neq 0$ , and it is then given by $A^{- 1} = \frac{1}{∣ A ∣} adj A$ .

Inverse (when |A| ≠ 0)

A^{- 1} = \frac{1}{∣ A ∣} adj A

Two quick consequences. From $∣ A ∣ ∣ A^{- 1} ∣ = ∣ I ∣ = 1$ we get $∣ A^{- 1} ∣ = 1 / ∣ A ∣$ . And the inverse is unique, so whichever method you use, you must reach the same $A^{- 1}$ ; a singular matrix has no inverse at all, while every non-singular matrix has one.

Example 13NCERT Example 13 · p. 90

Verify Theorem 1, then invert

For $A = [\begin{array}{ccc} 1 & - 1 & 0 \\ 2 & 1 & 3 \\ 1 & 0 & 2 \end{array}]$ , check that $A (adj A) = ∣ A ∣ I$ and find $A^{- 1}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 14NCERT Example 14 · p. 91

The inverse of a product

For $A = [\begin{array}{cc} 3 & 1 \\ 5 & 2 \end{array}]$ and $B = [\begin{array}{cc} 2 & - 1 \\ 1 & 1 \end{array}]$ , verify that $(A B)^{- 1} = B^{- 1} A^{- 1}$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 15NCERT Example 15 · p. 92

An inverse from a matrix equation

Show that $A = [\begin{array}{cc} 5 & 2 \\ 3 & 1 \end{array}]$ satisfies $A^{2} - 6 A - I = O$ , and use this to find $A^{- 1}$ .

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

For

A = [\begin{array}{cc} 6 & - 2 \\ 4 & 1 \end{array}]

adj A

[\begin{array}{cc} 1 & 2 \\ - 4 & 6 \end{array}]

[\begin{array}{cc} 1 & - 4 \\ 2 & 6 \end{array}]

[\begin{array}{cc} 1 & - 2 \\ 4 & 6 \end{array}]

[\begin{array}{cc} - 1 & 2 \\ - 4 & - 6 \end{array}]

Show answer

Answer (A) $[\begin{array}{cc} 1 & 2 \\ - 4 & 6 \end{array}]$

Swap the diagonal entries (6 and 1) and change the signs of the others:

- (- 2) = 2

- (4) = - 4

. Option B is the matrix of cofactors before transposing.

Practice · 18 questions

Practice 4.4: questions with answers

Short answerEasyNCERT Exercise 4.4, Q1
Find the adjoint of $[\begin{array}{cc} 3 & - 4 \\ 5 & 2 \end{array}]$ .
Show solution
Answer $[\begin{array}{cc} 2 & 4 \\ - 5 & 3 \end{array}]$
Cofactors: $A_{11} = 2$ , $A_{12} = - 5$ , $A_{21} = 4$ , $A_{22} = 3$ . Transposing gives $[\begin{array}{cc} 2 & 4 \\ - 5 & 3 \end{array}]$ .
Short answerMediumNCERT Exercise 4.4, Q2
Find the adjoint of $[\begin{array}{ccc} 2 & 0 & - 1 \\ 1 & 3 & 2 \\ - 1 & 4 & 1 \end{array}]$ .
Show solution
Answer $[\begin{array}{ccc} - 5 & - 4 & 3 \\ - 3 & 1 & - 5 \\ 7 & - 8 & 6 \end{array}]$
Cofactors by rows: $A_{11} = 3 - 8 = - 5$ , $A_{12} = - (1 + 2) = - 3$ , $A_{13} = 4 + 3 = 7$ ; $A_{21} = - (0 + 4) = - 4$ , $A_{22} = 2 - 1 = 1$ , $A_{23} = - (8 - 0) = - 8$ ; $A_{31} = 0 + 3 = 3$ , $A_{32} = - (4 + 1) = - 5$ , $A_{33} = 6 - 0 = 6$ .
Transpose: $adj A = [\begin{array}{ccc} - 5 & - 4 & 3 \\ - 3 & 1 & - 5 \\ 7 & - 8 & 6 \end{array}]$ . (As a check, row 1 of $A$ times column 1 gives $- 10 + 0 - 7 = - 17 = ∣ A ∣$ .)
Short answerEasyNCERT Exercise 4.4, Q3
Check that $A (adj A)$ and $(adj A) A$ both equal $∣ A ∣ I$ when $A = [\begin{array}{cc} 3 & - 6 \\ - 2 & 4 \end{array}]$ .
Show solution
$∣ A ∣ = 12 - 12 = 0$ and $adj A = [\begin{array}{cc} 4 & 6 \\ 2 & 3 \end{array}]$ .
$A (adj A) = [\begin{array}{cc} 12 - 12 & 18 - 18 \\ - 8 + 8 & - 12 + 12 \end{array}] = O$ and $(adj A) A = [\begin{array}{cc} 12 - 12 & - 24 + 24 \\ 6 - 6 & - 12 + 12 \end{array}] = O$ . Both equal $0 \cdot I = ∣ A ∣ I$ .
Short answerMediumNCERT Exercise 4.4, Q4
Check that $A (adj A)$ and $(adj A) A$ both equal $∣ A ∣ I$ when $A = [\begin{array}{ccc} 2 & 1 & - 1 \\ 0 & 3 & 1 \\ 4 & - 2 & 1 \end{array}]$ .
Show solution
$∣ A ∣ = 2 (3 + 2) - 1 (0 - 4) + (- 1) (0 - 12) = 10 + 4 + 12 = 26$ .
$adj A = [\begin{array}{ccc} 5 & 1 & 4 \\ 4 & 6 & - 2 \\ - 12 & 8 & 6 \end{array}]$ .
Multiplying in either order gives $[\begin{array}{ccc} 26 & 0 & 0 \\ 0 & 26 & 0 \\ 0 & 0 & 26 \end{array}] = 26 I$ . For example, row 1 of $A$ times column 1 of $adj A$ is $10 + 4 + 12 = 26$ , and row 1 times column 2 is $2 + 6 - 8 = 0$ .
Short answerEasyNCERT Exercise 4.4, Q5
Find the inverse of $[\begin{array}{cc} 3 & 5 \\ 1 & 2 \end{array}]$ , if it exists.
Show solution
Answer $[\begin{array}{cc} 2 & - 5 \\ - 1 & 3 \end{array}]$
$∣ A ∣ = 6 - 5 = 1$ , so $A^{- 1} = adj A = [\begin{array}{cc} 2 & - 5 \\ - 1 & 3 \end{array}]$ .
Short answerEasyNCERT Exercise 4.4, Q6
Find the inverse of $[\begin{array}{cc} - 2 & 3 \\ 4 & - 5 \end{array}]$ , if it exists.
Show solution
Answer $[\begin{array}{cc} \frac{5}{2} & \frac{3}{2} \\ 2 & 1 \end{array}]$
$∣ A ∣ = 10 - 12 = - 2$ and $adj A = [\begin{array}{cc} - 5 & - 3 \\ - 4 & - 2 \end{array}]$ , so $A^{- 1} = - \frac{1}{2} adj A = [\begin{array}{cc} \frac{5}{2} & \frac{3}{2} \\ 2 & 1 \end{array}]$ .
Short answerMediumNCERT Exercise 4.4, Q7
Find the inverse of $[\begin{array}{ccc} 2 & 1 & 3 \\ 0 & 1 & - 2 \\ 0 & 0 & 4 \end{array}]$ , if it exists.
Show solution
Answer $\frac{1}{8} [\begin{array}{ccc} 4 & - 4 & - 5 \\ 0 & 8 & 4 \\ 0 & 0 & 2 \end{array}]$
$∣ A ∣ = 2 \times 1 \times 4 = 8$ (expand down column 1).
Cofactors: $A_{11} = 4$ , $A_{12} = 0$ , $A_{13} = 0$ , $A_{21} = - 4$ , $A_{22} = 8$ , $A_{23} = 0$ , $A_{31} = - 5$ , $A_{32} = 4$ , $A_{33} = 2$ .
$A^{- 1} = \frac{1}{8} [\begin{array}{ccc} 4 & - 4 & - 5 \\ 0 & 8 & 4 \\ 0 & 0 & 2 \end{array}] = [\begin{array}{ccc} \frac{1}{2} & - \frac{1}{2} & - \frac{5}{8} \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & \frac{1}{4} \end{array}]$ . The inverse is again upper triangular.
Short answerMediumNCERT Exercise 4.4, Q8
Find the inverse of $[\begin{array}{ccc} 1 & 0 & 0 \\ 2 & - 1 & 0 \\ 4 & 3 & 2 \end{array}]$ , if it exists.
Show solution
Answer $[\begin{array}{ccc} 1 & 0 & 0 \\ 2 & - 1 & 0 \\ - 5 & \frac{3}{2} & \frac{1}{2} \end{array}]$
$∣ A ∣ = 1 \times (- 1) \times 2 = - 2$ .
$adj A = [\begin{array}{ccc} - 2 & 0 & 0 \\ - 4 & 2 & 0 \\ 10 & - 3 & - 1 \end{array}]$ .
$A^{- 1} = - \frac{1}{2} adj A = [\begin{array}{ccc} 1 & 0 & 0 \\ 2 & - 1 & 0 \\ - 5 & \frac{3}{2} & \frac{1}{2} \end{array}]$ .
Short answerMediumNCERT Exercise 4.4, Q9
Find the inverse of $[\begin{array}{ccc} 1 & 2 & - 1 \\ 3 & 0 & 2 \\ - 2 & 1 & 1 \end{array}]$ , if it exists.
Show solution
Answer $\frac{1}{19} [\begin{array}{ccc} 2 & 3 & - 4 \\ 7 & 1 & 5 \\ - 3 & 5 & 6 \end{array}]$
Along $R_{2}$ : $∣ A ∣ = - 3 (2 + 1) + 0 - 2 (1 + 4) = - 9 - 10 = - 19$ .
$adj A = [\begin{array}{ccc} - 2 & - 3 & 4 \\ - 7 & - 1 & - 5 \\ 3 & - 5 & - 6 \end{array}]$ .
$A^{- 1} = - \frac{1}{19} adj A = \frac{1}{19} [\begin{array}{ccc} 2 & 3 & - 4 \\ 7 & 1 & 5 \\ - 3 & 5 & 6 \end{array}]$ .
Short answerMediumNCERT Exercise 4.4, Q10
Find the inverse of $[\begin{array}{ccc} 2 & - 1 & 1 \\ 1 & 3 & - 2 \\ 0 & 4 & 1 \end{array}]$ , if it exists.
Show solution
Answer $\frac{1}{27} [\begin{array}{ccc} 11 & 5 & - 1 \\ - 1 & 2 & 5 \\ 4 & - 8 & 7 \end{array}]$
Along $C_{1}$ : $∣ A ∣ = 2 (3 + 8) - 1 (- 1 - 4) + 0 = 22 + 5 = 27$ .
$adj A = [\begin{array}{ccc} 11 & 5 & - 1 \\ - 1 & 2 & 5 \\ 4 & - 8 & 7 \end{array}]$ , so $A^{- 1} = \frac{1}{27} adj A$ .
Short answerMediumNCERT Exercise 4.4, Q11
Find the inverse of $[\begin{array}{ccc} \cos α & 0 & \sin α \\ 0 & 1 & 0 \\ - \sin α & 0 & \cos α \end{array}]$ .
Show solution
Answer $[\begin{array}{ccc} \cos α & 0 & - \sin α \\ 0 & 1 & 0 \\ \sin α & 0 & \cos α \end{array}]$
Along $R_{2}$ : $∣ A ∣ = 1 (\cos^{2} α + \sin^{2} α) = 1$ .
The cofactors give $adj A = [\begin{array}{ccc} \cos α & 0 & - \sin α \\ 0 & 1 & 0 \\ \sin α & 0 & \cos α \end{array}]$ , and since $∣ A ∣ = 1$ this is $A^{- 1}$ . It is the transpose of $A$ : the inverse of a rotation turns back the other way.
Short answerMediumNCERT Exercise 4.4, Q12
Let $A = [\begin{array}{cc} 2 & 5 \\ 1 & 3 \end{array}]$ and $B = [\begin{array}{cc} 4 & 3 \\ 3 & 2 \end{array}]$ . Verify that $(A B)^{- 1} = B^{- 1} A^{- 1}$ .
Show solution
$A B = [\begin{array}{cc} 23 & 16 \\ 13 & 9 \end{array}]$ , $∣ A B ∣ = 207 - 208 = - 1$ , so $(A B)^{- 1} = [\begin{array}{cc} - 9 & 16 \\ 13 & - 23 \end{array}]$ .
$∣ A ∣ = 1$ , $A^{- 1} = [\begin{array}{cc} 3 & - 5 \\ - 1 & 2 \end{array}]$ ; $∣ B ∣ = - 1$ , $B^{- 1} = [\begin{array}{cc} - 2 & 3 \\ 3 & - 4 \end{array}]$ .
$B^{- 1} A^{- 1} = [\begin{array}{cc} - 6 - 3 & 10 + 6 \\ 9 + 4 & - 15 - 8 \end{array}] = [\begin{array}{cc} - 9 & 16 \\ 13 & - 23 \end{array}] = (A B)^{- 1}$ .
Short answerMediumNCERT Exercise 4.4, Q13
If $A = [\begin{array}{cc} 2 & - 1 \\ 3 & 1 \end{array}]$ , show that $A^{2} - 3 A + 5 I = O$ . Hence find $A^{- 1}$ .
Show solution
Answer $A^{- 1} = \frac{1}{5} [\begin{array}{cc} 1 & 1 \\ - 3 & 2 \end{array}]$
$A^{2} = [\begin{array}{cc} 1 & - 3 \\ 9 & - 2 \end{array}]$ , so $A^{2} - 3 A + 5 I = [\begin{array}{cc} 1 - 6 + 5 & - 3 + 3 \\ 9 - 9 & - 2 - 3 + 5 \end{array}] = O$ .
$∣ A ∣ = 5 \neq 0$ . Multiplying by $A^{- 1}$ : $A - 3 I + 5 A^{- 1} = O$ , so $A^{- 1} = \frac{1}{5} (3 I - A) = \frac{1}{5} [\begin{array}{cc} 1 & 1 \\ - 3 & 2 \end{array}]$ .
Short answerMediumNCERT Exercise 4.4, Q14
Determine $a$ and $b$ so that $A^{2} + a A + b I = O$ , where $A = [\begin{array}{cc} 4 & 3 \\ 1 & 2 \end{array}]$ .
Show solution
Answer $a = - 6$ , $b = 5$
$A^{2} = [\begin{array}{cc} 19 & 18 \\ 6 & 7 \end{array}]$ . The top-right entry of $A^{2} + a A + b I$ is $18 + 3 a = 0$ , so $a = - 6$ .
The top-left entry is $19 + 4 a + b = 19 - 24 + b = 0$ , so $b = 5$ . Check the bottom row: $6 - 6 = 0$ and $7 - 12 + 5 = 0$ .
Short answerHardNCERT Exercise 4.4, Q15
For $A = [\begin{array}{ccc} 2 & 1 & 0 \\ 1 & - 1 & 1 \\ 0 & 1 & 2 \end{array}]$ , show that $A^{3} - 3 A^{2} - 2 A + 8 I = O$ . Hence find $A^{- 1}$ .
Show solution
Answer $A^{- 1} = \frac{1}{8} [\begin{array}{ccc} 3 & 2 & - 1 \\ 2 & - 4 & 2 \\ - 1 & 2 & 3 \end{array}]$
$A^{2} = [\begin{array}{ccc} 5 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 5 \end{array}]$ and $A^{3} = [\begin{array}{ccc} 11 & 5 & 3 \\ 5 & - 1 & 5 \\ 3 & 5 & 11 \end{array}]$ . Entry by entry, $A^{3} - 3 A^{2} - 2 A + 8 I = O$ (for instance the top-left entry is $11 - 15 - 4 + 8 = 0$ ).
$∣ A ∣ = 2 (- 3) - 1 (2) + 0 = - 8 \neq 0$ . Multiplying by $A^{- 1}$ : $A^{2} - 3 A - 2 I + 8 A^{- 1} = O$ , so $A^{- 1} = - \frac{1}{8} (A^{2} - 3 A - 2 I) = - \frac{1}{8} [\begin{array}{ccc} - 3 & - 2 & 1 \\ - 2 & 4 & - 2 \\ 1 & - 2 & - 3 \end{array}] = \frac{1}{8} [\begin{array}{ccc} 3 & 2 & - 1 \\ 2 & - 4 & 2 \\ - 1 & 2 & 3 \end{array}]$ .
Short answerHardNCERT Exercise 4.4, Q16
For $A = [\begin{array}{ccc} 3 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 3 \end{array}]$ , verify that $A^{3} - 9 A^{2} + 24 A - 20 I = O$ and hence find $A^{- 1}$ .
Show solution
Answer $A^{- 1} = \frac{1}{10} [\begin{array}{ccc} 4 & - 1 & - 1 \\ - 1 & 4 & - 1 \\ - 1 & - 1 & 4 \end{array}]$
$A^{2}$ has $11$ on the diagonal and $7$ elsewhere; $A^{3}$ has $47$ on the diagonal and $39$ elsewhere. Diagonal check: $47 - 99 + 72 - 20 = 0$ ; off-diagonal check: $39 - 63 + 24 = 0$ .
$∣ A ∣ = 3 (8) - 1 (2) + 1 (- 2) = 20 \neq 0$ . Multiplying by $A^{- 1}$ : $A^{2} - 9 A + 24 I - 20 A^{- 1} = O$ , so $A^{- 1} = \frac{1}{20} (A^{2} - 9 A + 24 I)$ .
$A^{2} - 9 A + 24 I$ has $11 - 27 + 24 = 8$ on the diagonal and $7 - 9 = - 2$ elsewhere, so $A^{- 1} = \frac{1}{20} [\begin{array}{ccc} 8 & - 2 & - 2 \\ - 2 & 8 & - 2 \\ - 2 & - 2 & 8 \end{array}] = \frac{1}{10} [\begin{array}{ccc} 4 & - 1 & - 1 \\ - 1 & 4 & - 1 \\ - 1 & - 1 & 4 \end{array}]$ .
Quick answerEasyNCERT Exercise 4.4, Q17
$A$ is a non-singular $3 \times 3$ matrix with $∣ A ∣ = 5$ . Find $∣ adj A ∣$ .
Show answer
Answer 25
$∣ adj A ∣ = 5^{2} = 25$ .
Quick answerMediumNCERT Exercise 4.4, Q18
$A$ is an invertible $3 \times 3$ matrix with $∣ A ∣ = - 4$ . Then $∣ 2 A^{- 1} ∣$ equals
$- 2$ $- \frac{1}{2}$ $- 8$ $\frac{1}{2}$
Show answer
Answer (A) $- 2$
$∣ 2 A^{- 1} ∣ = 2^{3} ∣ A^{- 1} ∣ = 8 \times (- \frac{1}{4}) = - 2$ . The answer $- \frac{1}{2}$ forgets the cube.

4.6 Applications of Determinants and Matrices

A shopkeeper pricing mixed packs, an engineer balancing the forces at a joint and a dietitian matching nutrients all end up with several linear equations in several unknowns. Matrices let us write such a system as a single equation, and the inverse then solves it in one move. First, two words for what can happen.

Definition

Consistent system

A system of equations is consistent if it has at least one solution.

Definition

Inconsistent system

A system of equations is inconsistent if it has no solution.

A consistent system may have exactly one solution or infinitely many. In this chapter we solve only systems with exactly one solution; for the others we only decide whether solutions exist.

4.6.1 Solution of system of linear equations using inverse of a matrix

Collect the coefficients, the unknowns and the right-hand sides of $a_{1} x + b_{1} y + c_{1} z = d_{1}$ , $a_{2} x + b_{2} y + c_{2} z = d_{2}$ , $a_{3} x + b_{3} y + c_{3} z = d_{3}$ into three matrices. Matrix multiplication then turns the three equations into one:

Matrix form AX = B

\underset{A}{\underset{⏟}{[\begin{array}{ccc} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}]}} \underset{X}{\underset{⏟}{[\begin{array}{c} x \\ y \\ z \end{array}]}} = \underset{B}{\underset{⏟}{[\begin{array}{c} d_{1} \\ d_{2} \\ d_{3} \end{array}]}}

DerivationCase I: the coefficient matrix is non-singular

If $∣ A ∣ \neq 0$ , then $A^{- 1}$ exists. Multiply both sides of $A X = B$ on the left by $A^{- 1}$ and regroup (matrix multiplication is associative):

\begin{aligned} A^{- 1} (A X) & = A^{- 1} B \\ (A^{- 1} A) X & = A^{- 1} B \\ I X & = A^{- 1} B \end{aligned}

Matrix method

X = A^{- 1} B

Inverses are unique, so no other $X$ can satisfy the equation. This recipe, invert the coefficient matrix and multiply, is the matrix method.

Case II: $∣ A ∣ = 0$ . Now there is no inverse, but Theorem 1 still helps. Multiply $A X = B$ on the left by $adj A$ : since $(adj A) A = ∣ A ∣ I = O$ , any solution $X$ would force $(adj A) B = O$ . So if $(adj A) B \neq O$ , no solution can exist and the system is inconsistent. If $(adj A) B = O$ , the test is silent: the system may have infinitely many solutions (then it is consistent) or none, and you decide by looking at the equations themselves (Fig. 4.14).

3/3Compute $(adj A) B$ . If it is not $O$ , there is no solution. If it is $O$ , check the equations: infinitely many solutions or none.

Fig. 4.14First look at

∣ A ∣

; only if it is

0

do you need

(adj A) B

A flow chart. At the top a box AX = B leads to a diamond asking whether |A| = 0. In the second step the no branch is shown: it leads to a green box reading X = A inverse B, exactly one solution. In the third step the yes branch is added beside it: it leads to a box find (adj A)B, which splits into an orange box not O, no solution, and another box equals O, test further.

Example

Two singular systems

Decide whether each system is consistent. (a) $x + y + z = 6$ , $x + 2 y + 3 z = 14$ , $2 x + 3 y + 4 z = 20$ . (b) The same left-hand sides with right-hand sides $6$ , $14$ , $21$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 16NCERT Example 16 · p. 95

Two equations in two unknowns

Solve $4 x + 3 y = 11$ , $3 x - 2 y = 4$ by the matrix method.

Try it yourself first. Then reveal the solution one step at a time.

Example 17NCERT Example 17 · p. 95

Three equations in three unknowns

Solve by the matrix method: $2 x + y - z = 0$ , $x - 3 y + 2 z = 11$ , $3 x + 2 y + z = 7$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 18NCERT Example 18 · p. 96

Ticket prices at a mela

At a village mela, one family pays ₹270 for 2 adult, 3 child and 1 senior ticket. A second family pays ₹200 for 1 adult, 2 child and 2 senior tickets, and a third pays ₹280 for 3 adult, 1 child and 2 senior tickets. Find the price of each kind of ticket by the matrix method.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

For

A X = B

, suppose

∣ A ∣ = 0

and

(adj A) B \neq O

. The system

has exactly one solutionhas infinitely many solutionshas no solutionmay or may not have a solution

Show answer

Answer (C) has no solution

Any solution would force

(adj A) B = ∣ A ∣ X = O

. Since it is not

O

, there is no solution. “May or may not” is the verdict only when

(adj A) B = O

Miscellaneous example

Example 19NCERT Example 19 · p. 98

Reading an inverse off a product

Find the product $P Q$ for $P = [\begin{array}{ccc} 1 & 0 & - 1 \\ 1 & 2 & - 2 \\ 1 & - 1 & 0 \end{array}]$ and $Q = [\begin{array}{ccc} - 2 & 1 & 2 \\ - 2 & 1 & 1 \\ - 3 & 1 & 2 \end{array}]$ . Use it to solve $x - z = 1$ , $x + 2 y - 2 z = 1$ , $x - y = 2$ .

Try it yourself first. Then reveal the solution one step at a time.

Practice · 13 questions

Practice questions with answers (NCERT “Miscellaneous exercises”)

ProofMediumNCERT Miscellaneous Exercises, Q1
Show that $∣ \begin{array}{ccc} 1 & x & \sin θ \\ - x & 1 & \cos θ \\ - \sin θ & - \cos θ & 1 \end{array} ∣$ does not depend on $θ$ .
Show solution
Along $R_{1}$ : $1 (1 + \cos^{2} θ) - x (- x + \sin θ \cos θ) + \sin θ (x \cos θ + \sin θ)$ .
$= 1 + \cos^{2} θ + x^{2} - x \sin θ \cos θ + x \sin θ \cos θ + \sin^{2} θ = x^{2} + 2$ , which has no $θ$ in it.
Short answerMediumNCERT Miscellaneous Exercises, Q2
Evaluate $∣ \begin{array}{ccc} \cos θ & - \sin θ \cos φ & \sin θ \sin φ \\ \sin θ & \cos θ \cos φ & - \cos θ \sin φ \\ 0 & \sin φ & \cos φ \end{array} ∣$ .
Show solution
Along $R_{3}$ : the $(3, 2)$ term is $- \sin φ (- \cos^{2} θ \sin φ - \sin^{2} θ \sin φ) = \sin^{2} φ$ .
The $(3, 3)$ term is $\cos φ (\cos^{2} θ \cos φ + \sin^{2} θ \cos φ) = \cos^{2} φ$ . The total is $\sin^{2} φ + \cos^{2} φ = 1$ .
Short answerMediumNCERT Miscellaneous Exercises, Q3
Given $A = [\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{array}]$ and $B^{- 1} = [\begin{array}{ccc} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 1 \end{array}]$ , find $(A B)^{- 1}$ .
Show solution
Answer $\frac{1}{4} [\begin{array}{ccc} 1 & - 2 & 3 \\ 5 & - 6 & 3 \\ 0 & 4 & 0 \end{array}]$
$∣ A ∣ = 1 (2) - 2 (0 - 1) + 0 = 4$ and $A^{- 1} = \frac{1}{4} [\begin{array}{ccc} 2 & - 4 & 2 \\ 1 & 2 & - 1 \\ - 1 & 2 & 1 \end{array}]$ .
$(A B)^{- 1} = B^{- 1} A^{- 1} = \frac{1}{4} [\begin{array}{ccc} 2 - 1 & - 4 + 2 & 2 + 1 \\ 4 + 1 & - 8 + 2 & 4 - 1 \\ 1 - 1 & 2 + 2 & - 1 + 1 \end{array}] = \frac{1}{4} [\begin{array}{ccc} 1 & - 2 & 3 \\ 5 & - 6 & 3 \\ 0 & 4 & 0 \end{array}]$ .
ProofHardNCERT Miscellaneous Exercises, Q4
Let $A = [\begin{array}{ccc} 1 & 0 & 2 \\ - 1 & 1 & 1 \\ 2 & 1 & 1 \end{array}]$ . Verify that (a) $(adj A)^{- 1} = adj (A^{- 1})$ and (b) $(A^{- 1})^{- 1} = A$ .
Show solution
$∣ A ∣ = 1 (1 - 1) - 0 + 2 (- 1 - 2) = - 6$ and $adj A = [\begin{array}{ccc} 0 & 2 & - 2 \\ 3 & - 3 & - 3 \\ - 3 & - 1 & 1 \end{array}]$ , so $A^{- 1} = - \frac{1}{6} adj A$ .
(a) $adj A = ∣ A ∣ A^{- 1}$ , so $(adj A)^{- 1} = \frac{1}{∣ A ∣} A = - \frac{1}{6} A$ . Also $adj (A^{- 1}) = ∣ A^{- 1} ∣ (A^{- 1})^{- 1} = \frac{1}{∣ A ∣} A = - \frac{1}{6} A$ . Both equal $[\begin{array}{ccc} - \frac{1}{6} & 0 & - \frac{1}{3} \\ \frac{1}{6} & - \frac{1}{6} & - \frac{1}{6} \\ - \frac{1}{3} & - \frac{1}{6} & - \frac{1}{6} \end{array}]$ ; multiplying this by $adj A$ gives $I$ , which confirms it.
(b) $A^{- 1} A = A A^{- 1} = I$ says that $A$ is an inverse of $A^{- 1}$ ; inverses are unique, so $(A^{- 1})^{- 1} = A$ . Numerically, $A^{- 1} \cdot A = - \frac{1}{6} (adj A) A = - \frac{1}{6} (- 6 I) = I$ .
Short answerMediumNCERT Miscellaneous Exercises, Q5
Evaluate $∣ \begin{array}{ccc} p & q & q \\ q & p & q \\ q & q & p \end{array} ∣$ and write the answer as a product of factors.
Show solution
Answer $(p - q)^{2} (p + 2 q)$
Along $R_{1}$ : $p (p^{2} - q^{2}) - q (p q - q^{2}) + q (q^{2} - p q) = p^{3} - 3 p q^{2} + 2 q^{3}$ .
This vanishes at $p = q$ , and dividing gives $p^{3} - 3 p q^{2} + 2 q^{3} = (p - q)^{2} (p + 2 q)$ (expand to check).
Short answerMediumNCERT Miscellaneous Exercises, Q6
Evaluate $∣ \begin{array}{ccc} x + 1 & x & x \\ x & x + 2 & x \\ x & x & x + 3 \end{array} ∣$ .
Show solution
Answer $11 x + 6$
Along $R_{1}$ : $(x + 1) [(x + 2) (x + 3) - x^{2}] - x [x (x + 3) - x^{2}] + x [x^{2} - x (x + 2)]$ .
$= (x + 1) (5 x + 6) - x (3 x) + x (- 2 x) = 5 x^{2} + 11 x + 6 - 3 x^{2} - 2 x^{2} = 11 x + 6$ .
Short answerHardNCERT Miscellaneous Exercises, Q7
Solve $\frac{4}{x} + \frac{3}{y} + \frac{8}{z} = 5$ , $\frac{2}{x} - \frac{6}{y} + \frac{4}{z} = 0$ , $\frac{6}{x} + \frac{9}{y} - \frac{4}{z} = 5$ .
Show solution
Answer $x = 2$ , $y = 3$ , $z = 4$
With $u = \frac{1}{x}$ , $v = \frac{1}{y}$ , $w = \frac{1}{z}$ : $4 u + 3 v + 8 w = 5$ , $2 u - 6 v + 4 w = 0$ , $6 u + 9 v - 4 w = 5$ .
$∣ A ∣ = 4 (24 - 36) - 3 (- 8 - 24) + 8 (18 + 36) = - 48 + 96 + 432 = 480$ and $adj A = [\begin{array}{ccc} - 12 & 84 & 60 \\ 32 & - 64 & 0 \\ 54 & - 18 & - 30 \end{array}]$ .
$[\begin{array}{c} u \\ v \\ w \end{array}] = \frac{1}{480} [\begin{array}{c} - 60 + 0 + 300 \\ 160 + 0 + 0 \\ 270 + 0 - 150 \end{array}] = [\begin{array}{c} \frac{1}{2} \\ \frac{1}{3} \\ \frac{1}{4} \end{array}]$ , so $x = 2$ , $y = 3$ , $z = 4$ .
Quick answerMediumNCERT Miscellaneous Exercises, Q8
If $x$ , $y$ , $z$ are non-zero, the inverse of $A = [\begin{array}{ccc} 0 & 0 & x \\ 0 & y & 0 \\ z & 0 & 0 \end{array}]$ is
$[\begin{array}{ccc} 0 & 0 & x^{- 1} \\ 0 & y^{- 1} & 0 \\ z^{- 1} & 0 & 0 \end{array}]$ $\frac{1}{x y z} [\begin{array}{ccc} 0 & 0 & x \\ 0 & y & 0 \\ z & 0 & 0 \end{array}]$ $[\begin{array}{ccc} 0 & 0 & - z^{- 1} \\ 0 & y^{- 1} & 0 \\ - x^{- 1} & 0 & 0 \end{array}]$ $[\begin{array}{ccc} 0 & 0 & z^{- 1} \\ 0 & y^{- 1} & 0 \\ x^{- 1} & 0 & 0 \end{array}]$
Show answer
Answer (D) $[\begin{array}{ccc} 0 & 0 & z^{- 1} \\ 0 & y^{- 1} & 0 \\ x^{- 1} & 0 & 0 \end{array}]$
Row 1 of $A$ , $(0, 0, x)$ , must meet column 1 of the inverse and give 1, so the bottom-left entry is $x^{- 1}$ ; similarly the top-right entry is $z^{- 1}$ . Option A swaps them.
Quick answerHardNCERT Miscellaneous Exercises, Q9
Let $A = [\begin{array}{ccc} 2 & \sin θ & 0 \\ - \sin θ & 2 & \sin θ \\ 0 & - \sin θ & 2 \end{array}]$ . As $θ$ varies over all real numbers, what is the largest value of $\det A$ ?
Show answer
Answer 12
Along $R_{1}$ : $2 (4 + \sin^{2} θ) - \sin θ (- 2 \sin θ) = 8 + 4 \sin^{2} θ$ . Since $0 \leq \sin^{2} θ \leq 1$ , $\det A$ takes every value from $8$ (at $θ = 0$ ) to $12$ (at $θ = \frac{π}{2}$ ); the largest is $12$ .
Quick answerMedium
$A$ and $B$ are $3 \times 3$ matrices with $∣ A ∣ = 2$ and $∣ B ∣ = - 3$ . Find $∣ 2 A B ∣$ .
Show answer
Answer -48
$∣ 2 A B ∣ = 2^{3} ∣ A ∣ ∣ B ∣ = 8 \times 2 \times (- 3) = - 48$ .
11
Quick answerEasy
Match each quantity (for a non-singular $3 \times 3$ matrix $A$ ) with its value.
Tap an item on the left, then its match on the right.
Show answer
Answer
$∣ A^{- 1} ∣$ : $1 / ∣ A ∣$
$∣ adj A ∣$ : $∣ A ∣^{2}$
$∣ 5 A ∣$ : $125 ∣ A ∣$
$∣ A^{T} ∣$ : $∣ A ∣$
$∣ A ∣ ∣ A^{- 1} ∣ = 1$ ; $∣ adj A ∣ = ∣ A ∣^{n - 1}$ with $n = 3$ ; $∣ k A ∣ = k^{3} ∣ A ∣$ ; and a determinant can be expanded along rows or columns alike, so transposing does not change it.
Quick answerEasy
For what value of $k$ are the points $(1, 1)$ , $(2, 3)$ and $(4, k)$ collinear?
Show answer
Answer 7
$1 (3 - k) - 1 (2 - 4) + (2 k - 12) = k - 7 = 0$ , so $k = 7$ .
Quick answerEasy
A square matrix whose determinant is zero is called a matrix.
Show answer
Answer singular
A matrix with $∣ A ∣ = 0$ is singular; it has no inverse.

Flash cards · 10

Formulas at a glance

1 / 10

Recap

Determinants: key points and summary

Every square matrix has a determinant: $∣ [a] ∣ = a$ and $∣ \begin{array}{cc} a & b \\ c & d \end{array} ∣ = a d - b c$ (Section 4.2.2).
A $3 \times 3$ determinant can be expanded along any row or column with the signs $(- 1)^{i + j}$ ; all six expansions agree, so choose the line with the most zeros (Section 4.2.3).
For a square matrix of order $n$ , $∣ k A ∣ = k^{n} ∣ A ∣$ (Property).
The area of a triangle is $∣ Δ ∣$ , where $Δ = \frac{1}{2} \det$ of the rows $(x_{i}, y_{i}, 1)$ ; its sign shows whether the vertices run anticlockwise or clockwise (Section 4.3).
Three points are collinear exactly when $Δ = 0$ , which also gives the equation of the line through two points (line equation).
The minor $M_{i j}$ is the determinant left after deleting row $i$ and column $j$ ; the cofactor is $A_{i j} = (- 1)^{i + j} M_{i j}$ (Section 4.4).
$∣ A ∣$ equals the sum of the entries of any row or column times their own cofactors; with the cofactors of a different line the sum is $0$ (Property).
$adj A$ is the transpose of the matrix of cofactors; for order 2, swap the diagonal and negate the rest (Section 4.5.1).
$A (adj A)$ and $(adj A) A$ both equal $∣ A ∣ I$ (Theorem 1).
$A$ is singular if $∣ A ∣ = 0$ and non-singular otherwise; $∣ A B ∣ = ∣ A ∣ ∣ B ∣$ , $∣ A^{- 1} ∣ = 1 / ∣ A ∣$ and $∣ adj A ∣ = ∣ A ∣^{n - 1}$ (Singular matrices and the inverse).
$A$ has an inverse exactly when $∣ A ∣ \neq 0$ , and then $A^{- 1} = \frac{1}{∣ A ∣} adj A$ (Theorem 4).
If $∣ A ∣ \neq 0$ , the system $A X = B$ has the unique solution $X = A^{- 1} B$ (Section 4.6.1).
If $∣ A ∣ = 0$ : when $(adj A) B \neq O$ the system is inconsistent; when $(adj A) B = O$ it has infinitely many solutions or none, and you check directly (Fig. 4.14).

Timeline

Historical note: how determinants grew

c. 200 BCE–100 CE
Counting-rod tables in China
The Nine Chapters on the Mathematical Art lays out the coefficients of several equations as a grid of rods and eliminates unknowns by combining columns.
1683
Seki Takakazu
The Japanese mathematician writes down determinant-like expansions to eliminate an unknown from equations.
1693
Leibniz
In a letter he labels coefficients with two indices and states when a system can be solved.
1750
Cramer's rule
Gabriel Cramer publishes a rule that solves n equations in n unknowns as ratios of such expressions.
1771–1773
Vandermonde, Laplace, Lagrange
Vandermonde studies determinants for their own sake, Laplace expands them by minors, and Lagrange uses them for areas and volumes.
1801
Gauss
He uses the word determinant, though for a quantity attached to quadratic forms.
1812
Binet and Cauchy
Both prove the product rule $∣ A B ∣ = ∣ A ∣ ∣ B ∣$ ; Cauchy gives the word its modern meaning.
1841
Jacobi
His papers make determinants a standard tool across mathematics.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 4.1, Q1	Practice 4.1 · Q1
NCERT Exercise 4.1, Q2	Practice 4.1 · Q2
NCERT Exercise 4.1, Q3	Practice 4.1 · Q3
NCERT Exercise 4.1, Q4	Practice 4.1 · Q4
NCERT Exercise 4.1, Q5	Practice 4.1 · Q5
NCERT Exercise 4.1, Q6	Practice 4.1 · Q6
NCERT Exercise 4.1, Q7	Practice 4.1 · Q7
NCERT Exercise 4.1, Q8	Practice 4.1 · Q8
NCERT Exercise 4.2, Q1	Practice 4.2 · Q1
NCERT Exercise 4.2, Q2	Practice 4.2 · Q2
NCERT Exercise 4.2, Q3	Practice 4.2 · Q3
NCERT Exercise 4.2, Q4	Practice 4.2 · Q4
NCERT Exercise 4.2, Q5	Practice 4.2 · Q5
NCERT Exercise 4.3, Q1	Practice 4.3 · Q1
NCERT Exercise 4.3, Q2	Practice 4.3 · Q2
NCERT Exercise 4.3, Q3	Practice 4.3 · Q3
NCERT Exercise 4.3, Q4	Practice 4.3 · Q4
NCERT Exercise 4.3, Q5	Practice 4.3 · Q5
NCERT Exercise 4.4, Q1	Practice 4.4 · Q1
NCERT Exercise 4.4, Q2	Practice 4.4 · Q2
NCERT Exercise 4.4, Q3	Practice 4.4 · Q3
NCERT Exercise 4.4, Q4	Practice 4.4 · Q4
NCERT Exercise 4.4, Q5	Practice 4.4 · Q5
NCERT Exercise 4.4, Q6	Practice 4.4 · Q6
NCERT Exercise 4.4, Q7	Practice 4.4 · Q7
NCERT Exercise 4.4, Q8	Practice 4.4 · Q8
NCERT Exercise 4.4, Q9	Practice 4.4 · Q9
NCERT Exercise 4.4, Q10	Practice 4.4 · Q10
NCERT Exercise 4.4, Q11	Practice 4.4 · Q11
NCERT Exercise 4.4, Q12	Practice 4.4 · Q12
NCERT Exercise 4.4, Q13	Practice 4.4 · Q13
NCERT Exercise 4.4, Q14	Practice 4.4 · Q14
NCERT Exercise 4.4, Q15	Practice 4.4 · Q15
NCERT Exercise 4.4, Q16	Practice 4.4 · Q16
NCERT Exercise 4.4, Q17	Practice 4.4 · Q17
NCERT Exercise 4.4, Q18	Practice 4.4 · Q18
NCERT Exercise 4.5, Q1	Practice 4.5 · Q1
NCERT Exercise 4.5, Q2	Practice 4.5 · Q2
NCERT Exercise 4.5, Q3	Practice 4.5 · Q3
NCERT Exercise 4.5, Q4	Practice 4.5 · Q4
NCERT Exercise 4.5, Q5	Practice 4.5 · Q5
NCERT Exercise 4.5, Q6	Practice 4.5 · Q6
NCERT Exercise 4.5, Q7	Practice 4.5 · Q7
NCERT Exercise 4.5, Q8	Practice 4.5 · Q8
NCERT Exercise 4.5, Q9	Practice 4.5 · Q9
NCERT Exercise 4.5, Q10	Practice 4.5 · Q10
NCERT Exercise 4.5, Q11	Practice 4.5 · Q11
NCERT Exercise 4.5, Q12	Practice 4.5 · Q12
NCERT Exercise 4.5, Q13	Practice 4.5 · Q13
NCERT Exercise 4.5, Q14	Practice 4.5 · Q14
NCERT Exercise 4.5, Q15	Practice 4.5 · Q15
NCERT Exercise 4.5, Q16	Practice 4.5 · Q16
NCERT Miscellaneous Exercises, Q1	Miscellaneous exercises · Q1
NCERT Miscellaneous Exercises, Q2	Miscellaneous exercises · Q2
NCERT Miscellaneous Exercises, Q3	Miscellaneous exercises · Q3
NCERT Miscellaneous Exercises, Q4	Miscellaneous exercises · Q4
NCERT Miscellaneous Exercises, Q5	Miscellaneous exercises · Q5
NCERT Miscellaneous Exercises, Q6	Miscellaneous exercises · Q6
NCERT Miscellaneous Exercises, Q7	Miscellaneous exercises · Q7
NCERT Miscellaneous Exercises, Q8	Miscellaneous exercises · Q8
NCERT Miscellaneous Exercises, Q9	Miscellaneous exercises · Q9

Quick check: MCQs on Determinants

4.1 Introduction2 questions
4.2.2 Determinant of a matrix of order two2 questions
4.2.3 Determinant of a matrix of order 3 × 34 questions
4.3 Area of a Triangle4 questions
4.4 Minors and Cofactors4 questions
Singular matrices and the inverse4 questions
4.6.1 Solution of system of linear equations using inverse of a matrix4 questions

Glossary · 12

Key terms (English–Hindi)

adjointसहखंडजsahkhandaj: The transpose of the matrix of cofactors of a square matrix $A$ , written $adj A$ .

cofactorसहखंडsahkhand: $A_{i j} = (- 1)^{i + j} M_{i j}$ : the minor of $a_{i j}$ with its checkerboard sign attached.
collinear pointsसंरेख बिंदुsanrekh bindu: Points that lie on one straight line.First seen in the chapter →
consistent systemसंगत निकायsangat nikaay: A system of equations that has at least one solution.First seen in the chapter →

determinantसारणिकsaarnik: The number assigned to a square matrix $A$ , written $∣ A ∣$ or $\det A$ . For a $2 \times 2$ matrix it is $a d - b c$ .First seen in the chapter →

expansion of a determinantसारणिक का प्रसारsaarnik ka prasaar: Writing a determinant as the sum of the entries of one row or column, each multiplied by its cofactor.First seen in the chapter →

inconsistent systemअसंगत निकायasangat nikaay: A system of equations that has no solution.First seen in the chapter →
inverse of a matrixव्युत्क्रमvyutkram: The matrix $A^{- 1}$ with $A A^{- 1} = A^{- 1} A = I$ . When $∣ A ∣ \neq 0$ it equals $\frac{1}{∣ A ∣} adj A$ .First seen in the chapter →

matrix methodआव्यूह विधिaavyooh vidhi: Solving a system written as $A X = B$ by computing $X = A^{- 1} B$ , which works when $∣ A ∣ \neq 0$ .First seen in the chapter →
minorउपसारणिकupsaarnik: The minor $M_{i j}$ of $a_{i j}$ is the determinant left after deleting row $i$ and column $j$ .

non-singular matrixव्युत्क्रमणीय आव्यूहvyutkramaniya aavyooh: A square matrix whose determinant is not 0. It always has an inverse.First seen in the chapter →

singular matrixअव्युत्क्रमणीय आव्यूहavyutkramaniya aavyooh: A square matrix whose determinant is 0. It has no inverse.First seen in the chapter →

Determinants: Class 12 Maths · Chapter 4

What you will learn in Determinants

4.1 Introduction

4.2 Determinant

4.2.1 Determinant of a matrix of order one

4.2.2 Determinant of a matrix of order two

4.2.3 Determinant of a matrix of order 3 × 3

Practice 4.1: questions with answers

4.3 Area of a Triangle

Practice 4.2: questions with answers

4.4 Minors and Cofactors

Practice 4.3: questions with answers

4.5 Adjoint and Inverse of a Matrix

4.5.1 Adjoint of a matrix

Singular matrices and the inverse

Practice 4.4: questions with answers

4.6 Applications of Determinants and Matrices

4.6.1 Solution of system of linear equations using inverse of a matrix

Practice 4.5: questions with answers

Miscellaneous example

Practice questions with answers (NCERT “Miscellaneous exercises”)

Determinants: key points and summary

Find your NCERT exercise question

Quick check: MCQs on Determinants

Key terms (English–Hindi)

You reached the end