Chapter 4 Determinants

NCERT Solutions Class 12 Maths Chapter 4 Determinants

Determinants is an important algebra chapter of Class 12 Maths that helps students to understand how numerical values are associated with square matrices. The chapter includes determinants of different orders, their properties, and practical applications such as solving linear equations. A clear understanding of determinants is essential for learning matrices, linear equations, and higher mathematics concepts used later in the syllabus.

ALLEN provides NCERT Solutions for class 12 Maths chapter 4 which are prepared strictly according to the latest textbook released by NCERT and aligned with the current syllabus prescribed by CBSE. These solutions are written in simple language with step-by-step explanations to help Class 12 students prepare confidently for board exams and competitive tests.

1.0Key Concepts of Class 12 Maths Chapter 4 Determinants

Class 12 Maths Chapter 4, Determinants, introduces the concept of determinants and their use in solving mathematical problems. The main topics covered in NCERT Solutions for Class 12 Maths Chapter 4 include:

Determinant of a Matrix: Understand the meaning of a determinant and how determinants are defined for square matrices of different orders.
Determinants of Order Two and Three: Learn how to evaluate determinants of order two and three using standard methods.
Properties of Determinants: Study important properties that can simplify determinant calculations and help solve questions efficiently.
Minors and Cofactors: Understand minors and cofactors and learn how they are used to expand determinants along rows or columns.
Area of a Triangle: Learn how determinants can be used to find the area of a triangle when the coordinates of its vertices are given.
Applications in Solving Linear Equations: Understand how determinants are used to solve systems of linear equations using Cramer's Rule.

2.0NCERT Class 12 Maths Chapter 4 Determinants : Detailed Solutions

EXERCISE - 4.1

Evaluate the determinants in Q. Nos. 1 and 2

$2 - 5 4 - 1$ Sol. Let $A = 2 - 5 4 - 1$

$\Rightarrow ∣ A ∣ = 2 \times (- 1) - (- 5) \times 4 = - 2 + 20 = 18$

$(i) cos θ sin θ - sin θ cos θ$

$(ii) x^{2} - x + 1 x + 1 x - 1 x + 1$

Sol.

(i)

$∣ A ∣ = (cos θ) (cos θ) - (sin θ) (- sin θ) = cos^{2} θ + sin^{2} θ = 1 (∵ sin^{2} θ + cos^{2} θ = 1)$

(ii)

$x^{2} - x + 1 x + 1 x - 1 x + 1 = (x^{2} - x + 1) (x + 1) - (x + 1) (x - 1) = x^{3} - x^{2} + x + x^{2} - x + 1 - (x^{2} - 1) = x^{3} + 1 - (x^{2} - 1) = x^{3} - x^{2} + 2$

If $A = [1422]$ , then show that $∣2 A ∣ = 4∣ A ∣$ Sol. Given, $A = [1422]$

$∴ 2 A = 2 [1422] = [2 \times 1 2 \times 4 2 \times 2 2 \times 2] = [2844] LHS = ∣2 A ∣ = 2844 = (2 \times 4) - (8 \times 4) = - 24$

Now, $∣ A ∣ = 1422 = (1 \times 2) - (4 \times 2) = - 6$

$∴ RHS = 4∣ A ∣ = 4 \times (- 6) = - 24$

$∴ LHS = RHS$ Hence; $∣2 A ∣ = 4∣ A ∣$

If $A = 100010124$ , then show that $∣3 A ∣ = 27∣ A ∣$ Sol. Given, $A = 100010124$ then $3 A = 3000303612$ Now;

$∣ A ∣ = 100010124 = 1 (4 - 0) - 0 (0) + 1 (0) = 4$

$∣3 A ∣ = 3000303612 = 3 (36 - 0) - 0 (0) + 3 (0) = 108$

$∣3 A ∣ = 108 = 27 \times 4 = 27∣ A ∣,$

Hence proved.

Evaluate the determinants. (i) $303 - 1 0 - 5 - 2 - 1 0$ (ii) $312 - 4 13 5 - 2 1$ (iii) $0 - 1 - 2 103 2 - 3 0$ (iv) $203 - 1 2 - 5 - 2 - 1 0$

Sol.

(i) Let $∣ A ∣ = 303 - 1 0 - 5 - 2 - 1 0$ By expanding along $R_{2}$

$∣ A ∣ = - 0 - 1 - 5 - 2 0 + 0 33 - 2 0 - (- 1) 33 - 1 - 5 = {3 \times (- 5) - 3 \times (- 1)} = - 15 + 3 = - 12$

(ii) Let $∣ A ∣ = 312 - 4 13 5 - 2 1$ By expanding along $R_{1}$ (first row), we get

$∣ A ∣ = 3 13 - 2 1 - (- 4) 12 - 2 1 + 51213 = 3 (1 + 6) + 4 (1 + 4) + 5 (3 - 2) = 3 (7) + 4 (5) + 5 (1) = 21 + 20 + 5 = 46$

(iii) Let $∣ A ∣ = 0 - 1 - 2 103 2 - 3 0$ By expanding along $R_{1}$ (first row), we get

$∣ A ∣ = 0 03 - 3 0 - 1 - 1 - 2 - 3 0 + 2 - 1 - 2 03 = - 1 (0 - 6) + 2 (- 3 - 0) = - 1 (- 6) + 2 (- 3) = 6 - 6 = 0$

(iv) Let $∣ A ∣ = 203 - 1 2 - 5 - 2 - 1 0$ By expanding along $R_{2}$ (second row), we get

$∣ A ∣ = - 0 - 1 - 5 - 2 0 + 2 23 - 2 0 - (- 1) 23 - 1 - 5 = 0 + 2 (0 + 6) + (- 10 + 3) = 12 - 7 = 5$

If $A = 125114 - 2 - 3 - 9$ , find $∣ A ∣$ ? Sol. Given, $A = 125114 - 2 - 3 - 9$ By expanding along $R_{1}$ (first row), we get

$∣ A ∣ = 1 14 - 3 - 9 - 1 25 - 3 - 9 + (- 2) 2514 = 1 (- 9 + 12) - 1 (- 18 + 15) - 2 (8 - 5) = 1 (3) - 1 (- 3) - 2 (3) = 3 + 3 - 6 = 0$

Find the values of x , if: (i) $2541 = 2 x 6 4 x$ (ii) $2435 = x 2 x 35$

Sol.

(i) Given, $2541 = 2 x 6 4 x$ On expanding both determinants, we get

$(2 \times 1) - (5 \times 4) = (2 x \times x) - (6 \times 4)$

$\Rightarrow 2 - 20 = 2 x^{2} - 24$

$\Rightarrow 2 x^{2} = - 18 + 24 \Rightarrow x^{2} = \frac{6}{2} = 3 \Rightarrow x = \pm 3$

(ii) Given, $2435 = x 2 x 35$ On expanding both determinants, we get

$(2 \times 5) - (4 \times 3) = (5 \times x) - (3 \times 2 x)$

$\Rightarrow 10 - 12 = 5 x - 6 x \Rightarrow - 2 = - x \Rightarrow x = 2$

Choose the correct answer in the following

If $x 18 2 x = 61826$ , then x is equal to ? (A) 6 (B) $\pm 6$ (C) -6 (D) 0 Sol. (B) Given, $x 18 2 x = 61826$ On expanding both determinants, we get

$(x \times x) - (18 \times 2) = (6 \times 6) - (18 \times 2)$

$\Rightarrow x^{2} - 36 = 36 - 36$ $\Rightarrow x^{2} - 36 = 0 \Rightarrow x^{2} = 36 \Rightarrow x = \pm 6$ So, (B) is the correct option.

EXERCISE - 4.2

Find the area of the triangle with vertices at the points in each of the following: (i) (1, 0), (6, 0), $(4, 3)$ (ii) (2, 7), (1, 1), (10, 8) (iii) (-2,-3), (3,2), (-1,-8)

Sol. Area of triangle $= \frac{1}{2} x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111$ (i) Required area $= \frac{1}{2} 164003111$

$= \frac{1}{2} ∣1 (0 - 3) - 0 (6 - 4) + 1 (18 - 0) ∣ = \frac{1}{2} - 3 + 18) = \frac{15}{2} sq. units$

(ii) Required area $= \frac{1}{2} 2110718111$

$= \frac{1}{2} ∣2 (1 - 8) - 7 (1 - 10) + 1 (8 - 10) ∣ = \frac{1}{2} ∣2 (- 7) - 7 (- 9) + 1 (- 2) ∣ = \frac{1}{2} ∣ - 14 + 63 - 2∣ = \frac{47}{2} sq. units$

(iii) Required area $= \frac{1}{2} - 2 3 - 1 - 3 2 - 8 111$

$= \frac{1}{2} ∣ - 2 (2 + 8) + 3 (3 + 1) + 1 (- 24 + 2) ∣ = \frac{1}{2} ∣ - 20 + 12 - 22∣ = \frac{1}{2} ∣ - 30∣ = 15 sq. units$

(Since area of triangle is always positive)

Show that the points $A (a, b + c), B (b, c + a)$ , $C (c, a + b)$ are collinear. Sol. Area of $△ ABC = \frac{1}{2} a b c b + c c + a a + b 111$

$= \frac{1}{2} a {(c + a) \times 1 - (a + b) \times 1} - (b + c) {b \times 1 - 1 \times c} + 1 {b \times (a + b) - (c + a) \times c} ∣ = \frac{1}{2} a {(c + a) \times 1 - (a + b) \times 1} - (b + c) {b \times 1 - 1 \times c} + 1 {b \times (a + b) - (c + a) \times c} ∣$

$= \frac{1}{2} a (c + a - a - b) - (b + c) (b - c) + 1 (ab + b^{2} - c^{2} - a c) = \frac{1}{2} a c - ab - b^{2} + c^{2} + ab + b^{2} - c^{2} - a c = \frac{1}{2} \times 0 = 0$

Since, area of $Δ ABC = 0$ . Hence, points $A (a, b + c), .. C (c, a + b)$ are collinear.

Find the value of k, if area of triangle is 4 sq. units and the vertices are: (i) (k, 0), (4, 0), (0, 2) (ii) (-2, 0), (0, 4), (0, k) Sol. Area of triangle $= \frac{1}{2} x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111$ (i) Given, $\frac{1}{2} k 40 002111 = 4$

$\Rightarrow \Rightarrow ∣ k (0 - 2) + 1 (8 - 0) ∣ = 8 k (0 - 2) + 1 (8 - 0) = \pm 8$

On taking positive sign;

$- 2 k + 8 = 8 \Rightarrow - 2 k = 0 \Rightarrow k = 0$

On taking negative sign;

$- 2 k + 8 = - 8 \Rightarrow - 2 k = - 16 \Rightarrow k = 8$

$∴ k = 0, 8$ (ii) Given, $\frac{1}{2} - 2 00 04 k 111 = 4$

$\Rightarrow \Rightarrow \Rightarrow ∣ - 2 (4 - k) + 1 (0 - 0) ∣ = 8 - 2 (4 - k) + 1 (0 - 0) = \pm 8 (- 8 + 2 k) = \pm 8$

On taking positive sign,

$2 k - 8 = 8 \Rightarrow 2 k = 16 \Rightarrow k = 8$

On taking negative sign,

$2 k - 8 = - 8 \Rightarrow 2 k = 0 \Rightarrow k = 0$

$∴ k = 0, 8$

(i) Find the equation of the line joining $(1, 2)$ and $(3, 6)$ using determinants. (ii) Find the equation of the line joining $(3, 1)$ and $(9, 3)$ using determinants.

Sol.

(i) Let $P (x, y)$ be any point on the line joining A(1,2) and B(3,6) . Area of triangle $= \frac{1}{2} x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111 = 0$ $∴ \frac{1}{2} 13 x 26 y 111 = 0$ $\Rightarrow \frac{1}{2} [1 (6 - y) - 2 (3 - x) + 1 (3 y - 6 x)] = 0$ $\Rightarrow 6 - y - 6 + 2 x + 3 y - 6 x = 0$ $\Rightarrow 2 y - 4 x = 0$ $\Rightarrow y = 2 x$ Hence, the equation of the line joining the given points is $y = 2 x$ . (ii) Let $P (x, y)$ be any point on the line joining A(3,1) and B(9,3). Area of triangle $= \frac{1}{2} x_{1} x_{2} x_{3} y_{1} y_{2} y_{3} 111 = 0$ $∴ \frac{1}{2} 39 x 13 y 111 = 0$ $\Rightarrow \frac{1}{2} ∣3 (3 - y) - 1 (9 - x) + 1 (9 y - 3 x) ∣ = 0$ $\Rightarrow 9 - 3 y - 9 + x + 9 y - 3 x = 0$ $\Rightarrow 6 y - 2 x = 0$ $\Rightarrow x - 3 y = 0$ Hence, the equation of the line joining the given points is $x - 3 y = 0$ .

Choose the correct answer in the following

If area of a triangle is 35 sq. units with vertices (2, -6), (5, 4) and $(k, 4)$ , then k is. (A) 12 (B) -2 (C) -12, -2 (D) 12,-2

Sol. (D) Given, $\frac{1}{2} 25 k - 6 44 111 = 35$ $\Rightarrow ∣2 (4 - 4) + 6 (5 - k) + 1 (20 - 4 k) ∣ = 70$ $\Rightarrow 2 (4 - 4) + 6 (5 - k) + 1 (20 - 4 k) = \pm 70$ $\Rightarrow 30 - 6 k + 20 - 4 k = \pm 70$ On taking positive sign, $- 10 k + 50 = 70 \Rightarrow - 10 k = 20 \Rightarrow k = - 2$ On taking negative sign, $- 10 k + 50 = - 70 \Rightarrow - 10 k = - 120 \Rightarrow k = 12$ $∴ k = 12, - 2$ . Hence, the correct option is (D).

EXERCISE - 4.3

(Q. Nos. 1 and 2) Write minors and cofactors of elements of following determinants.

(i) $20 - 4 3$ (ii) $∣ ac ∣$

Sol.

(i) Here, $20 - 4 3$ is given ∴ Minors, $M_{11} = 3, M_{12} = 0, M_{21} = - 4$ and $M_{22} = 2$ Also cofactors, $A_{11} = (- 1)^{1 + 1} M_{11} = 1 \times 3 = 3$ . $A_{12} = (- 1)^{1 + 2} M_{12} = (- 1) \times 0 = 0$ $A_{21} = (- 1)^{2 + 1} M_{21} = (- 1) \times (- 4) = 4$ and $A_{22} = (- 1)^{2 + 2} M_{22} = 1 \times 2 = 2$

(ii) Here, $a b c d$ is given $∴$ Minors, $M_{11} = d, M_{12} = b, M_{21} = c$ and

$M_{22} = a$

Also cofactors, $A_{11} = (- 1)^{1 + 1} M_{11} = 1 \times d = d$

$A_{12} = (- 1)^{1 + 2} M_{12} = (- 1) \times b = - b A_{21} = (- 1)^{2 + 1} M_{21} = (- 1) \times c = - c$

and $A_{22} = (- 1)^{2 + 2} M_{22} = 1 \times a = a$

(i) $100010001$ (ii) $130051 4 - 1 2$

Sol.

(i) Here, $100010001$ is given Minors of elements of first row are:

$M_{11} = 1001 = 1 - 0 = 1, M_{12} = 0001 = 0 - 0 = 0$

and $M_{13} = 0010 = 0 - 0 = 0$ Minors of elements of second row are:

$M_{21} = 0001 = 0 - 0 = 0, M_{22} = 1001 = 1 - 0 = 1$

and $M_{23} = 1000 = 0 - 0 = 0$ Minors of elements of third row are:

$M_{31} = 0100 = 0 - 0 = 0, M_{32} = 1000 = 0 - 0 = 0$

and

$M_{33} = 1001 = 1 - 0 = 1$

Hence, cofactors of elements of first row are:

$A_{11} = (- 1)^{1 + 1} M_{11} = 1 \times 1 = 1, A_{12} = (- 1)^{1 + 2} M_{12} = (- 1) \times 0 = 0$

and

$A_{13} = (- 1)^{1 + 3} M_{13} = 1 \times 0 = 0$

Cofactors of elements of second row are:

$A_{21} = (- 1)^{2 + 1} M_{21} = (- 1) \times 0 = 0, A_{22} = (- 1)^{2 + 2} M_{22} = 1 \times 1 = 1,$

and

$A_{23} = (- 1)^{2 + 3} M_{23} = (- 1) \times 0 = 0$

Cofactors of elements of third row are:

$A_{31} = (- 1)^{3 + 1} M_{31} = 1 \times 0 = 0, A_{32} = (- 1)^{3 + 2} M_{32} = (- 1) \times 0 = 0$

and

$A_{33} = (- 1)^{3 + 3} M_{33} = 1 \times 1 = 1$

(ii) Here, $130051 4 - 1 2$ is given Minors of elements of first row are:

$M_{11} = 51 - 1 2 = 10 + 1 = 11, M_{12} = 30 - 1 2 = 6 - 0 = 6$

and $M_{13} = 3051 = 3 - 0 = 3$ Minors of elements of second row are:

$M_{21} = 0142 = 0 - 4 = - 4, M_{22} = 1042 = 2 - 0 = 2$

and $M_{23} = 1001 = 1 - 0 = 1$ Minors of elements of third row are:

$M_{31} = 05 4 - 1 = 0 - 20 = - 20, M_{32} = 13 4 - 1 = - 1 - 12 = - 13$

and $M_{33} = 1305 = 5 - 0 = 5$

Hence, cofactors of elements of first row are:

$A_{12} = (- 1)^{1 + 2} M_{12} = (- 1) \times 6 = - 6, A_{13} = (- 1)^{1 + 3} M_{13} = 1 \times 3 = 3$

Cofactors of elements of second row are:

$A_{11} = (- 1)^{1 + 1} M_{11} = 1 \times 11 = 11, A_{21} = (- 1)^{2 + 1} M_{21} = (- 1) \times (- 4) = 4, A_{22} = (- 1)^{2 + 2} M_{22} = 1 \times 2 = 2$

and

$A_{23} = (- 1)^{2 + 3} M_{23} = (- 1) \times 1 = - 1$

Cofactors of elements of third row are:

$A_{31} = (- 1)^{3 + 1} M_{31} = 1 \times (- 20) = - 20, A_{32} = (- 1)^{3 + 2} M_{32} = (- 1) \times (- 13) = 13$

and

$A_{33} = (- 1)^{3 + 3} M_{33} = 1 \times 5 = 5$

Using cofactors of the elements of second row,

$evaluate Δ = 521302813 .$

Sol. Given, $Δ = 521302813$ Cofactors of the elements of second row are:

$A_{21} = (- 1)^{2 + 1} 3283 = - (9 - 16) = 7, [∵ A_{ij} = (- 1)^{i + j} M_{ij}]$

$A_{22} = (- 1)^{2 + 2} 5183 = (15 - 8) = 7$

and $A_{23} = (- 1)^{2 + 3} 5132 = - (10 - 3) = - 7$ Now, expansion of $Δ$ using cofactors of elements of second row is given by

$Δ = a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23} = (2 \times 7) + (0 \times 7) + (1) \times (- 7) = 14 - 7 = 7$

Using cofactors of elements of third column, evaluate $Δ = 111 x y z yz zx xy$ . Sol. Given, $Δ = 111 x y z yz zx xy$ Cofactors of the elements of third column are:

$A_{13} = (- 1)^{1 + 3} 11 y z = 1 (z - y) = z - y A_{23} = (- 1)^{2 + 3} 11 x z = - 1 (z - x) = x - z$

and $A_{33} = (- 1)^{3 + 3} 11 x y = 1 (y - x) = y - x$ Now, expansion of $Δ$ using cofactors of elements of third column is given by

$Δ = a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33} = yz (z - y) + z x (x - z) + x y (y - x) = y z^{2} - y^{2} z + z x^{2} - z^{2} x + x y^{2} - x^{2} y = x^{2} (z - y) + x (y^{2} - z^{2}) + yz (z - y) = (z - y) {x^{2} - x (y + z) + yz} = (z - y) {x^{2} - x y - x z + yz} = (z - y) [x (x - y) - z (x - y)] = (y - z) (x - y) (z - x) = (x - y) (y - z) (z - x)$

Choose the correct answer in the following : 5. If $Δ = a_{11} a_{21} a_{31} a_{12} a_{22} a_{32} a_{13} a_{23} a_{33}$ and $A_{ij}$ is cofactor of $a_{ij}$ , then value of $Δ$ is given by ?

(A) $a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33}$ (B) $a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31}$ (C) $a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13}$ (D) $a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$

Sol. (D) $Δ$ is equal to the sum of the products of the elements of a row (or a column) with their corresponding cofactors.

$Δ = a_{11} A_{11} + a_{12} A_{12} + a_{13} A_{13}$

or $a_{21} A_{21} + a_{22} A_{22} + a_{23} A_{23}$ or $a_{31} A_{31} + a_{32} A_{32} + a_{33} A_{33}$ or $a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$ or $a_{12} A_{12} + a_{22} A_{22} + a_{32} A_{32}$ or $a_{13} A_{13} + a_{23} A_{23} + a_{33} A_{33}$ Hence, sum of the products of the elements of first column with their corresponding cofactors is

$Δ = a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31}$

Hence, correct option is (D).

EXERCISE - 4.4

Find the adjoint of each of the matrices in Q. 1 and Q.2.

$[1324]$ Sol. Let $A = [1324]$ $∴ A_{11} = 4, A_{12} = - 3, A_{21} = - 2$ and $A_{22} = 1$ $∴ adj A = [A_{ij}]^{'} = [A_{11} A_{21} A_{12} A_{22}]^{'}$

$= [4 - 2 - 3 1]^{'} = [4 - 3 - 2 1]$

$12 - 2 - 1 30 251$ Sol. Let $A = 12 - 2 - 1 30 251$ Cofactors of elements of first row are:

$A_{11} = 3051 = 3 - 0 = 3, A_{12} = - 2 - 2 51 = - (2 + 10) = - 12$

and

$A_{13} = 2 - 2 30 = 0 - (- 6) = 6$

Cofactors of elements of second row are:

$A_{21} = - - 1 0 21 = - (- 1 - 0) = 1, A_{22} = 1 - 2 21 = (1 + 4) = 5$

and

$A_{23} = - 1 - 2 - 1 0 = - (0 - 2) = 2$

Cofactors of elements of third row are:

$A_{31} = - 1 3 25 = (- 5 - 6) = - 11, A_{32} = - 1225 = - (5 - 4) = - 1$

and

$A_{33} = 12 - 1 3 = 3 + 2 = 5$

Hence, $adj (A) = [A_{ij}]^{'}$

$= 31 - 11 - 12 5 - 1 625^{'} = 3 - 12 6 152 - 11 - 1 5$

(Q. Nos. 3 and 4) Verify $A (adj A) = (adj A)$ $A = ∣ A ∣ I_{n}$

$[2 - 4 3 - 6]$ Sol. Let $A = [2 - 4 3 - 6]$ ,

$∣ A ∣ = 2 - 4 3 - 6 = - 12 - (- 12) = - 12 + 12 = 0$

$∴ ∣ A ∣ I = 0 [1001] = [0000] = O$ Cofactors of A are

$A_{11} = - 6, A_{12} = 4, A_{21} = - 3, A_{22} = 2 ∴ adj (A) = [A_{ij}]^{'} = [- 6 - 3 42]^{'} = [- 6 4 - 3 2]$

Now, $(adj A)$

$A = [- 6 4 - 3 2] [2 - 4 3 - 6] = [- 12 + 12 8 - 8 - 18 + 18 12 - 12] = [0000] = 0$

Also,

$A (adj A) = [2 - 4 3 - 6] [- 6 4 - 3 2] = [- 12 + 12 24 - 24 - 6 + 6 12 - 12] = [0000] = O$

Hence, $A (adj A) = (adj A) A = ∣ A ∣ I_{2}$ .

$131 - 1 00 2 - 2 3$

Sol. Let $A = 131 - 1 00 2 - 2 3$ Now, $∣ A ∣ = 131 - 1 00 2 - 2 3$

$= 1 (0 - 0) - (- 1) (9 + 2) + 2 (0 - 0) = 0 + 11 + 0 = 11$

$∴ ∣ A ∣ I = 11100010001 = 110001100011$ Cofactors of A are:

$A_{11} = 0, A_{13} = 0, A_{22} = 3 - 2 = 1, A_{31} = 2 - 0 = 2, A_{33} = 0 + 3 = 3 A_{12} = - (9 + 2) = - 11, A_{21} = - (- 3 - 0) = 3, A_{23} = - (0 + 1) = - 1, A_{32} = - (- 2 - 6) = 8,$

$∴ Now, A (adj A) = = adj (A) = [A_{ij}]^{'} = 032 - 11 18 0 - 1 3^{'} = 0 - 11 0 31 - 1 283 = 131 - 1 00 2 - 2 3 0 - 11 0 31 - 1 283 0 + 11 + 0 0 + 0 + 0 0 + 0 + 0 3 - 1 - 2 9 + 0 + 2 3 + 0 - 3 2 - 8 + 6 6 + 0 - 6 2 + 0 + 9 110001100011$

Also, $(adj A) A = 0 - 11 0 31 - 1 283 131 - 1 00 2 - 2 3$

$= 0 + 9 + 2 - 11 + 3 + 8 0 - 3 + 3 0 + 0 + 0 11 + 0 + 0 0 + 0 + 0 0 - 6 + 6 - 22 - 2 + 24 0 + 2 + 9 = 110001100011 = 11100010001$

Hence, $A (adj A) = (adj A) A = ∣ A ∣ I_{3}$ Find the inverse of each of the matrices (if it exists) given in Q. 5 to 11

$[24 - 2 3]$

Sol. Let $A = [24 - 2 3]$ . We have, $∣ A ∣ = 24 - 2 3 = 6 - (- 8) = 14 \neq = 0$

$∴ A^{- 1}$ exists Cofactors of A are $A_{11} = 3, A_{12} = - 4$ ,

$A_{21} = 2, A_{22} = 2$

$∴ adj (A) = [A_{ij}]^{'} = [32 - 4 2]^{'} = [3 - 4 22]$ Now, $A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{14} [3 - 4 22]$

$= [\frac{3}{14} - \frac{4}{14} \frac{2}{14} \frac{2}{14}] = [\frac{3}{14} - \frac{2}{7} \frac{1}{7} \frac{1}{7}]$

$[- 1 - 3 52]$ Sol. Let $A = [- 1 - 3 52]$ . We have, $∣ A ∣ == - 2 - (- 15) = 13 \neq = 0$ $∴ A^{- 1}$ exists

Now, cofactors of A are

$A_{11} = 2, A_{12} = 3, A_{21} = - 5, A_{22} = - 1$

$∴ adj (A) = [A_{ij}]^{'} = [2 - 5 3 - 1]^{'} = [23 - 5 - 1]$ Now, $A^{- 1} = \frac{1}{∣ A ∣} (adj A)$

$= \frac{1}{13} [23 - 5 - 1] = [\frac{2}{13} \frac{3}{13} - \frac{5}{13} - \frac{1}{13}]$

$100220345$ Sol. Let $A = 100220345$ We have,

$∣ A ∣ = 1 (10 - 0) - 2 (0 - 0) + 3 (0 - 0) = 10 \neq = 0$

$∴ A^{- 1}$ exists

Now, cofactors of A are

$A_{11} = 10 - 0 = 10, A_{12} = - (0 - 0) = 0, A_{13} = 0 - 0 = 0, A_{21} = - (10 - 0) = - 10, A_{22} = 5 - 0 = 5, A_{23} = - (0 - 0) = 0, A_{31} = 8 - 6 = 2, A_{33} = 2 - 0 = 2 A_{32} = - (4 - 0) = - 4, .$

$∴ adj (A) = [A_{ij}]^{'} = 10 - 10 2 05 - 4 002^{'} = 1000 - 10 50 2 - 4 2$

Now, $A^{- 1} = \frac{1}{∣ A ∣} (adj A)$

$= \frac{1}{10} 1000 - 10 50 2 - 4 2 = 100 - 1 \frac{1}{2} 0 \frac{1}{5} - \frac{2}{5} \frac{1}{5}$

$135032 00 - 1$

Sol. Let $A = 135032 00 - 1$

$∣ A ∣ = 1 (- 3 - 0) - 0 + 0 = - 3 \neq = 0$

Cofactors of A are:

$A_{11} = - 3 - 0 = - 3, A_{13} = 6 - 15 = - 9, A_{21} = - (0 - 0) = 0, A_{23} = - (2 - 0) = - 2, A_{31} = 0 - 0 = 0, A_{32} = - (0 - 0) = 0, A_{12} = - (- 3 - 0) = 3, A_{22} = - 1 - 0 = - 1, A_{33} = 3 - 0 = 3$

$∴ adj (A) = [A_{ij}]^{'}$

$= - 3 00 3 - 1 0 - 9 - 2 3^{'} = - 3 3 - 9 0 - 1 - 2 003$

Now, $A^{- 1} = \frac{1}{∣ A ∣} (adj A)$

$= \frac{1}{- 3} - 3 3 - 9 0 - 1 - 2 003 = 1 - 1 3 0 \frac{1}{3} \frac{2}{3} 00 - 1$

$24 - 7 1 - 1 2 301$ Sol. Let $A = 24 - 7 1 - 1 2 301$

$∣ A ∣ = 2 (- 1 - 0) - 1 (4 - 0) + 3 (8 - 7) = - 2 - 4 + 3 = - 3 \neq = 0$

$∴ A^{- 1}$ exists

Cofactors of A are:

$A_{11} = - 1 - 0 = - 1, A_{12} = - (4 - 0) = - 4, A_{13} = 8 - 7 = 1, A_{21} = - (1 - 6) = 5, A_{22} = 2 + 21 = 23, A_{23} = - (4 + 7) = - 11, A_{31} = 0 + 3 = 3, A_{32} = - (0 - 12), A_{33} = - 2 - 4 = - 6$

$∴ adj (A) = [A_{ij}]^{'}$

$= - 1 53 - 4 2312 1 - 11 - 6^{'} = - 1 - 4 1 523 - 11 312 - 6$

Now,

$A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{- 3} - 1 - 4 1 523 - 11 312 - 6 = \frac{1}{3} \frac{4}{3} - \frac{1}{3} - \frac{5}{3} - \frac{23}{3} \frac{11}{3} - 1 - 4 2$

$103 - 1 2 - 2 2 - 3 4$

Sol. Let $A = 103 - 1 2 - 2 2 - 3 4$ We have,

$∣ A ∣ = 103 - 1 2 - 2 2 - 3 4 = 1 (8 - 6) - (- 1) (0 + 9) + 2 (0 - 6) = 2 + 9 - 12 = - 1 \neq = 0 ∴ A^{- 1} exists$

Cofactors of A are:

$A_{11} = 8 - 6 = 2, A_{13} = 0 - 6 = - 6, A_{21} = - (- 4 + 4) = 0, A_{23} = - (- 2 + 3) = - 1 A_{31} = 3 - 4 = - 1, A_{33} = 2 - 0 = 2 A_{12} = - (0 + 9) = - 9, A_{22} = 4 - 6 = - 2, A_{32} = - (- 3 - 0) = 3,$

$∴ adj (A) = [A_{ij}]^{'} = 20 - 1 - 9 - 2 3 - 6 - 1 2^{'} = 2 - 9 - 6 0 - 2 - 1 - 1 32$

Now; $A^{- 1} = \frac{1}{∣ A ∣} (adj A)$

$= \frac{1}{- 1} 2 - 9 - 6 0 - 2 - 1 - 1 32 = - 2 96 021 1 - 3 - 2$

$100 0 cos α sin α 0 sin α - cos α$

Sol. Let $A = 100 0 cos α sin α 0 sin α - cos α$

$\Rightarrow ∣ A ∣ = 100 0 cos α sin α 0 sin α - cos α = 1 (- cos^{2} α - sin^{2} α) = - (cos^{2} α + sin^{2} α) = - 1 \neq = 0 [∵ cos^{2} θ + sin^{2} θ = 1] ∴ A^{- 1} exists$

Cofactors of A are:

$A_{11} A_{12} A_{13} A_{21} A_{22} A_{23} A_{31} A_{32} A_{33} adj (A) = - cos^{2} α - sin^{2} α = - 1, = - (0 - 0) = 0, = 0 - 0 = 0, = - (0 - 0) = 0, = - cos α - 0 = - cos α, = - (sin α - 0) = - sin α, = 0 - 0 = 0, = - (sin α - 0) = - sin α, = cos α - 0 = cos α = [A_{ij}]^{'} = - 1 00 0 - cos α - sin α 0 - sin α cos α^{'} = - 1 00 0 - cos α - sin α 0 - sin α cos α$

Now, $A^{- 1} = \frac{1}{∣ A ∣} (adj A)$

$= \frac{1}{- 1} - 1 00 0 - cos α - sin α 0 - sin α cos α = 100 0 cos α sin α 0 sin α - cos α$

Let $A = [3275]$ and $B = [6789]$ .

Verify that $(AB)^{- 1} = B^{- 1} A^{- 1}$ . Sol. Given, $A = [3275]$ ;

$∣ A ∣ = 3275 = 15 - 14 = 1 \neq = 0$

$∴ A^{- 1}$ exists Cofactors of A are

$A_{11} = 5, A_{12} = - 2, A_{21} = - 7, A_{22} = 3$

$∴ adj (A) = [A_{ij}]^{'}$

$= [5 - 7 - 2 3]^{'} = [5 - 2 - 7 3]$

Now, $A^{- 1} = \frac{1}{∣ A ∣} (adj A)$

$= \frac{1}{1} [5 - 2 - 7 3] = [5 - 2 - 7 3]$

Here, $B = [6789]$ $∴ ∣ B ∣ = 6789 = 54 - 56 = - 2 \neq = 0$ $∴ B^{- 1}$ exists Cofactors of B are

$B_{11} = 9, B_{12} = - 7, B_{21} = - 8, B_{22} = 6 adj (B) = [B_{ij}]^{'} = [9 - 8 - 7 6]^{'} = [9 - 7 - 8 6]$

$∴ B^{- 1} = \frac{1}{∣ B ∣} (adj B) = \frac{1}{- 2} [9 - 7 - 8 6]$

Now,

$B^{- 1} A^{- 1} = \frac{1}{- 2} [9 - 7 - 8 6] [5 - 2 - 7 3] = \frac{1}{- 2} [45 + 16 - 35 - 12 - 63 - 24 49 + 18] = \frac{1}{- 2} [61 - 47 - 87 67] = [- \frac{61}{2} \frac{47}{2} \frac{87}{2} - \frac{67}{2}]$

Now, let

$C = AB = [3275] [6789] = [18 + 49 12 + 35 24 + 63 16 + 45] = [67478761] ∴ ∣ AB ∣ = 67478761 = (67 \times 61) - (47 \times 87) = 4087 - 4089 = - 2 \neq = 0 ∴ C^{- 1} exists.$

Cofactors of C are

$C_{11} = 61, C_{12} = - 47, C_{21} = - 87, C_{22} = 67$

$∴ (AB)^{- 1} = \frac{1}{∣ AB ∣} (adj AB) = \frac{1}{- 2} [61 - 47 - 87 67] = [- \frac{61}{2} \frac{47}{2} \frac{87}{2} - \frac{67}{2}]$

From Eq. (1) and (2), we get

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

Hence, the given result is proved.

If $A = [3 - 1 12]$ , show that

$A^{2} - 5 A + 7 I = O . Hence, find A^{- 1} .$

Sol. Given, $A = [3 - 1 12]$

$∴ A^{2} = A . A = [3 - 1 12] [3 - 1 12] = [9 - 1 - 3 - 2 3 + 2 - 1 + 4] = [8 - 5 53]$

Now, $A^{2} - 5 A + 7 I$

$= [8 - 5 53] - 5 [3 - 1 12] + 7 [1001] = [8 - 5 53] - [15 - 5 510] + [7007] = [8 - 15 + 7 - 5 + 5 + 0 5 - 5 + 0 3 - 10 + 7] = [0000] = O$

$∴ A^{2} - 5 A + 7 I = 0$

$∵ ∣ A ∣ = 3 - 1 12 = 6 + 1 = 7 \neq = 0,$

$∴ A^{- 1}$ exists. Now, A.A - 5A = -7I Post multiplying by $A^{- 1}$ on both sides, we get

$A.A (A^{- 1}) - 5 A A^{- 1} = - 7 I A^{- 1}$

$\Rightarrow AI - 5 I = - 7 A^{- 1}$

(Using AA $^{- 1} = I$ and $IA^{- 1} = A^{- 1}$ )

$\Rightarrow A^{- 1} = - \frac{1}{7} (A - 5 I)$

$\Rightarrow A^{- 1} = \frac{1}{7} (5 I - A) = \frac{1}{7} ([5005] - [3 - 1 12]) = \frac{1}{7} [21 - 1 3]$

$∴ A^{- 1} = \frac{1}{7} [21 - 1 3]$

For the matrix $A = [3121]$ , find the numbers ' a ' and ' b ' such that $A^{2} + aA + bI = O$ . Sol. Given, $A = [3121]$

$A^{2} = A \cdot A = [3121] [3121] = [9 + 2 3 + 1 6 + 2 2 + 1] = [11483]$

Given, $A^{2} + aA + bI = O$ On putting the values of $A^{2}, A$ and I ; we get

$\Rightarrow \Rightarrow [11483] + a [3121] + b [1001] = O [11 + 3 a + b 4 + a + 0 8 + 2 a + 0 3 + a + b] = O [11 + 3 a + b 4 + a 8 + 2 a 3 + a + b] = [0000]$

If two matrices are equal, then their corresponding elements are also equal.

$11 + 3 a + b = 0 8 + 2 a = 0 4 + a = 0$

and $3 + a + b = 0$ Solving Eq. (3) and (4), we get $4 + a = 0 \Rightarrow a = - 4$ And $3 + a + b = 0 \Rightarrow 3 - 4 + b = 0 \Rightarrow b = 1$ Thus, $a = - 4$ and $b = 1$

For the matrix $A = 112 12 - 1 1 - 3 3$ , show that

$A^{3} - 6 A^{2} + 5 A + 11 I = O . Hence, find A^{- 1} .$

Sol. Given, $A = 112 12 - 1 1 - 3 3$ ;

$∣ A ∣ = 1 (6 - 3) - 1 (3 + 6) + 1 (- 1 - 4) = 3 - 9 - 5 = - 11 \neq = 0$

$∴ A^{- 1}$ exists

Now, $A^{2} = A . A$

$= = 112 12 - 1 1 - 3 3 112 12 - 1 1 - 3 3 1 + 1 + 2 1 + 2 - 6 2 - 1 + 6 1 + 2 - 1 1 + 4 + 3 2 - 2 - 3 1 - 3 + 3 1 - 6 - 9 2 + 3 + 9 = 4 - 3 7 28 - 3 1 - 14 14$

And $A^{3} = A^{2} . A$

$= 4 - 3 7 28 - 3 1 - 14 14 112 12 - 1 1 - 3 3$

$= 4 + 2 + 2 - 3 + 8 - 28 7 - 3 + 28 4 + 4 - 1 - 3 + 16 + 14 7 - 6 - 14 4 - 6 + 3 - 3 - 24 - 42 7 + 9 + 42 = 8 - 23 32 727 - 13 1 - 69 58$

$∴ A^{3} - 6 A^{2} + 5 A + 11 I$

$= 8 - 23 32 727 - 13 1 - 69 58 - 6 4 - 3 7 28 - 3 1 - 14 14 + 5 112 12 - 1 1 - 3 3 + 11100010001$

$= 8 - 23 32 727 - 13 1 - 69 58 - 24 - 18 42 1248 - 18 6 - 84 84$

$= 8 - 24 + 5 + 11 - 23 + 18 + 5 + 0 32 - 42 + 10 + 0 7 - 12 + 5 + 0 27 - 48 + 10 + 11 - 13 + 18 - 5 + 0 1 - 6 + 5 + 0 - 69 + 84 - 15 + 0 58 - 84 + 15 + 11$

Now, $A^{3} - 6 A^{2} + 5 A + 11 I = O$

$\Rightarrow (AAA) A^{- 1} - 6 (AA) A^{- 1} + 5 AA^{- 1} + 111 A^{- 1} = O$

(Post-multiplying by $A^{- 1}$ as $∣ A ∣ \neq = 0$ )

$\Rightarrow AA (AA^{- 1}) - 6 A (AA^{- 1}) + 5 (AA^{- 1}) + 11 (IA^{- 1}) = 0 \Rightarrow AAI - 6 AI + 5 I + 11 A^{- 1} = O$

$\Rightarrow A^{- 1} = - \frac{1}{11} (A^{2} - 6 A + 5 I) A^{- 1} = \frac{1}{11} (- A^{2} + 6 A - 5 I) = \frac{1}{11} ⎩ ⎨ ⎧ - 4 - 3 7 28 - 3 1 - 14 14 + 6 112 12 - 1 1 - 3 3 - 5100010001 ⎭ ⎬ ⎫ = \frac{1}{11} ⎩ ⎨ ⎧ - 4 3 - 7 - 2 - 8 3 - 1 14 - 14 + 6612 612 - 6 6 - 18 18 - 500050005 ⎭ ⎬ ⎫ = \frac{1}{11} - 4 + 6 - 5 3 + 6 - 0 - 7 + 12 - 0 - 2 + 6 - 0 - 8 + 12 - 5 3 - 6 - 0 - 1 + 6 - 0 14 - 18 - 0 - 14 + 18 - 5 = 22 - 21 21 - 21 22 - 21 21 - 21 22 - 6 6 - 5 5 - 5 6 - 5 5 - 5 6$

If $A = 2 - 1 1 - 1 2 - 1 1 - 1 2$ , verify that

$A^{3} - 6 A^{2} + 9 A - 4 I = O and hence, find A^{- 1}$

Sol. Given, $A = 2 - 1 1 - 1 2 - 1 1 - 1 2$

$∴ A^{2} = A.A = 2 - 1 1 - 1 2 - 1 1 - 1 2 2 - 1 1 - 1 2 - 1 1 - 1 2$

and $A^{3} = A^{2} . A = 6 - 5 5 - 5 6 - 5 5 - 5 6 2 - 1 1 - 1 2 - 1 1 - 1 2$

$= 12 + 5 + 5 - 10 - 6 - 5 10 + 5 + 6 - 6 - 10 - 5 5 + 12 + 5 - 5 - 10 - 6 6 + 5 + 10 - 5 - 6 - 10 5 + 5 + 12 = 22 - 21 21 - 21 22 - 21 21 - 21 22$

$∴ A^{3} - 6 A^{2} + 9 A - 4 I = 22 - 21 21 - 21 22 - 21 21 - 21 22 - 6 6 - 5 5 - 5 6 - 5 5 - 5 6$

$+ 9 2 - 1 1 - 1 2 - 1 1 - 1 2 - 4100010001$

$= 22 - 21 21 - 21 22 - 21 21 - 21 22 - 36 - 30 30 - 30 36 - 30 30 - 30 36$

$+ 18 - 9 9 - 9 18 - 9 9 - 9 18 - 400040004$

$= 22 - 36 + 18 - 4 - 21 + 30 - 9 - 0 21 - 30 + 9 - 0 - 21 + 30 - 9 - 0 22 - 36 + 18 - 4 - 21 + 30 - 9 - 0 21 - 30 + 9 - 0 - 21 + 30 - 9 - 0 22 - 36 + 18 - 4$

$= 000000000 = O$

$∴ A^{3} - 6 A^{2} + 9 A - 4 I = O$ $\Rightarrow (AAA) A^{- 1} - 6 (AA) A^{- 1} + 9 AA^{- 1} - 4 IA^{- 1} = O$ (Post-multiplying by $A^{- 1}$ as $∣ A ∣ \neq = 0$ )

$[∵ ∣ A ∣ = 2 - 1 1 - 1 2 - 1 1 - 1 2 = 2 (4 - 1) + 1 (- 2 + 1) + 1 (1 - 2) = 6 - 1 - 1 = 4 \neq = 0]$

$AA (AA^{- 1}) - 6 A (AA^{- 1}) + 9 (AA^{- 1}) - 4 (IA^{- 1}) = O$ $\Rightarrow AAI - 6 AI + 9 I - 4 A^{- 1} = O$ (Using $AA^{- 1} = I$ and $IA^{- 1} = A^{- 1}$ )

$\Rightarrow A^{2} - 6 A + 9 I = 4 A^{- 1}$

(Using $A^{2} I = A^{2}$ and $AI = A$ )

$\Rightarrow A^{- 1} = \frac{1}{4} (A^{2} - 6 A + 9 I)$

$= \frac{1}{4} ⎩ ⎨ ⎧ 6 - 5 5 - 5 6 - 5 5 - 5 6 - 6 2 - 1 1 - 1 2 - 1 1 - 1 2 + 9100010001 ⎭ ⎬ ⎫$

$= \frac{1}{4} ⎩ ⎨ ⎧ 6 - 5 5 - 5 6 - 5 5 - 5 6 - 12 - 6 6 - 6 12 - 6 6 - 6 12 + 900090009 ⎭ ⎬ ⎫$

$= \frac{1}{4} 6 - 12 + 9 - 5 + 6 + 0 5 - 6 + 0 - 5 + 6 + 0 6 - 12 + 9 - 5 + 6 + 0 5 - 6 + 0 - 5 + 6 + 0 6 - 12 + 9$

$= \frac{1}{4} 31 - 1 131 - 1 13$

Choose the correct answer in the following questions from 17 & 18.

Let A be a non-singular square matrix of order $3 \times 3$ , then $∣ adj A ∣$ is equal to ? (A) $∣ A ∣$ (B) $∣ A ∣^{2}$ (C) $∣ A ∣^{3}$ (D) $3∣ A ∣$

Sol. (B) We know that $(adj A) A = ∣ A ∣ I$

$= ∣ A ∣ 100010001 = ∣ A ∣ 00 0 ∣ A ∣ 0 00 ∣ A ∣$

$\Rightarrow ∣ (adj A) A ∣ = A ∣ 00 0 ∣ A ∣ 0 00 ∣ A ∣ = ∣ A ∣^{3} 100010001$

$= ∣ A ∣^{3} ∣ I ∣$

$\Rightarrow ∣ adj A ∣∣ A ∣ = ∣ A ∣^{3} (∵ ∣ I ∣ = 1)$ $∴ ∣ adj A ∣ = ∣ A ∣^{2}$ Hence, the correct option is (B).

If A is an invertible matrix of order 2, then $det (A^{- 1})$ is equal to ? (A) $det (A)$ (B) $\frac{1}{det ( A )}$ (C) 1 (D) zero

Sol. (B) We know that $AA^{- 1} = I$ $∴ AA^{- 1} = ∣ I ∣ \Rightarrow ∣ A ∣ A^{- 1} = 1$ (Using $AA^{- 1} = ∣ A ∣ A^{- 1}$ and $∣ I ∣ = 1$ )

$\Rightarrow A^{- 1} = \frac{1}{∣ A ∣} = \frac{1}{det ( A )} .$

Hence, the correct option is (B).

EXERCISE - 4.5

(Q. Nos. 1 to 6) Examine the consistency of the following system of equations ?

$x + 2 y = 2, 2 x + 3 y = 3$ .

Sol. The given system can be written as $AX = B$ where

$A = [1223], X = [x y] and B = [23]$

Here, $∣ A ∣ = 1223 = 1 (3) - 2 (2) = 3 - 4 = - 1 \neq = 0$ ∴ A is non-singular. Therefore, $A^{- 1}$ exists. Hence, the given system of equations is consistent.

$2 x - y = 5, x + y = 4$ . Sol. The given system can be written as $AX = B$ , where

$A = [21 - 1 1], X = [x y] and B = [54]$

Here,

$∣ A ∣ = 21 - 1 1 = 2 (1) - (- 1) (1) = 2 + 1 = 3 \neq = 0$

∴ A is non-singular. Therefore, $A^{- 1}$ exists. Hence, the given system of equations is consistent.

$x + 3 y = 5, 2 x + 6 y = 8$ . Sol. The given system can be written as $AX = B$ , where

$A = [1236], X = [x y] and B = [58]$

Here, $∣ A ∣ = 1236 = 1 (6) - 3 (2) = 6 - 6 = 0$ ∴ A is singular matrix. Nothing can be said about consistency as yet. We compute

$(adj A) B = = [6 - 3 - 2 1]^{'} [58] [6 - 2 - 3 1] [58] = [30 - 24 - 10 + 8] = [6 - 2] \neq = O$

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

$x + y + z = 1, 2 x + 3 y + 2 z = 2$ ,

$a x + a y + 2 a z = 4$

Sol. The given system can be written as $AX = B$ , where

$A = 12 a 13 a 12 2 a, X = x y z and B = 124$

$Here, ∣ A ∣ = 12 a 13 a 12 2 a = 1 (6 a - 2 a) - 1 (4 a - 2 a) + 1 (2 a - 3 a) = 4 a - 2 a - a = 4 a - 3 a = a \neq = 0$

∴ A is non-singular. Therefore, $A^{- 1}$ exists.

Hence, the given system of equations is consistent.

$3 x - y - 2 z = 2, 2 y - z = - 1, 3 x - 5 y = 3$ . Sol. The given system is

$3 x - y - 2 z = 2 0 x + 2 y - z = - 1$

and $3 x - 5 y + 0 z = 3$ which can be written as $AX = B$ , where

$A = 303 - 1 2 - 5 - 2 - 1 0, X = x y z and B = 2 - 1 3$

Here, $∣ A ∣ = 303 - 1 2 - 5 - 2 - 1 0$

$= 3 (0 - 5) + 1 (0 + 3) - 2 (0 - 6) = - 15 + 3 + 12 = 0$

∴ A is singular matrix. Therefore, nothing can be said about consistency as yet. So, we compute (adj A)B. Cofactors of A are:

$A_{11} = - 5, A_{12} = - 3, A_{13} = - 6, A_{21} = 10, A_{22} = 6, A_{23} = 12, A_{31} = 5, A_{32} = 3, A_{33} = 6$

$adj (A) = [A_{ij}]^{'} = - 5 105 - 3 63 - 6 126^{'} = - 5 - 3 - 6 10612536$

$∴ (adj A) B = - 5 - 3 - 6 10612536 \cdot 2 - 1 3$

$= - 10 - 10 + 15 - 6 - 6 + 9 - 12 - 12 + 18 = - 5 - 3 - 6 \neq = 0$

Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.

$5 x - y + 4 z = 5, 2 x + 3 y + 5 z = 2$ ,

$5 x - 2 y + 6 z = - 1$

Sol. The given system can be written as $AX = B$ , where

$A = 525 - 1 3 - 2 456, X = x y z and B = 52 - 1$

Here, $∣ A ∣ = 525 - 1 3 - 2 456$

$= 5 (18 + 10) - (- 1) (12 - 25) + 4 (- 4 - 15) = 140 - 13 - 76 = 51 \neq = 0$

∴ A is non-singular. Therefore, $A^{- 1}$ exists. Hence, the given system of equations is consistent.

(Q. Nos. 7 to 14) Solve the following system of linear equations, using matrix method

$5 x + 2 y = 4, 7 x + 3 y = 5$ Sol. The given system can be written as $AX = B$ , where $A = [5723], X = [x y]$ and $B = [45]$ Here, $∣ A ∣ = 5723 = 15 - 14 = 1 \neq = 0$ Thus, A is non-singular. Therefore, its $A^{- 1}$ exists. Therefore, the given system is consistent and has a unique solution given by

$A^{- 1} (A X) = A^{- 1} B \Rightarrow X = A^{- 1} B$

Cofactors of A are,

$A_{11} = 3, A_{12} = - 7, A_{21} = - 2, A_{22} = 5$

$∴ adj (A) = [A_{ij}]^{'} = [3 - 2 - 7 5]^{'} = [3 - 7 - 2 5]$ Now,

$A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{1} [3 - 7 - 2 5] = [3 - 7 - 2 5]$

$\Rightarrow X = A^{- 1} B = [3 - 7 - 2 5] [45]$

$= [3 \times 4 + (- 2) \times 5 (- 7) \times 4 + 5 \times 5] = [2 - 3]$

$\Rightarrow [x y] = [2 - 3]$ Hence, $x = 2$ and $y = - 3$ .

$2 x - y = - 2, 3 x + 4 y = 3$ . Sol. The given system can be written as $AX = B$ , where

$A = [23 - 1 4], X = [x y] and B = [- 2 3]$

Here, $∣ A ∣ = 23 - 1 4 = 2 \times 4 - (- 3) = 11 \neq = 0$ Thus, A is non-singular. Therefore, its $A^{- 1}$ exists. Therefore, the given system is consistent and has a unique solution given by

$X = A^{- 1} B$

Cofactors of A are,

$A_{11} = 4, A_{12} = - 3, A_{21} = 1, A_{22} = 2$

$adj (A) = [A_{ij}]^{'} = [41 - 3 2]^{'} = [4 - 3 12]$

$∴ A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{11} [4 - 3 12]$

Now, $X = A^{- 1} B = \frac{1}{11} [4 - 3 12] [- 2 3]$

$= \frac{1}{11} [- 8 + 3 6 + 6] = \frac{1}{11} [- 5 12] = [- \frac{5}{11} \frac{12}{11}]$

$\Rightarrow [x y] = [\frac{- 5}{11} \frac{12}{11}]$ Hence, $x = \frac{- 5}{11}$ and $y = \frac{12}{11}$ .

$4 x - 3 y = 3, 3 x - 5 y = 7$ .

Sol. The given system can be written as $AX = B$ , where

$A = [43 - 3 - 5], X = [x y] and B = [37]$

Here,

$∣ A ∣ = 43 - 3 - 5 = 4 (- 5) - 3 (- 3) = - 20 + 9 = - 11 \neq = 0$

Thus, A is non-singular. Therefore, its $A^{- 1}$ exists. Therefore, the given system is consistent and has a unique solution given by

$X = A^{- 1} B$

Cofactors of A are,

$A_{11} = - 5, A_{12} = - 3, A_{21} = 3, A_{22} = 4$

$adj (A) = [A_{ij}]^{'} = [- 5 3 - 3 4]^{'} = [- 5 - 3 34]$

$∴ A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{- 11} [- 5 - 3 34]$

Now, $X = A^{- 1} B = \frac{1}{- 11} [- 5 - 3 34] [37]$

$= \frac{1}{- 11} [- 15 + 21 - 9 + 28] = \frac{1}{- 11} [619]$

$\Rightarrow [x y] = [\frac{- 6}{11} \frac{- 19}{11}]$ Hence, $x = - \frac{6}{11}$ and $y = - \frac{19}{11}$

$5 x + 2 y = 3, 3 x + 2 y = 5$ .

Sol. The given system can be written as $AX = B$ , where $A = [5322], X = [x y]$ and $B = [35]$ Here, $∣ A ∣ = 5322 = 10 - 6 = 4 \neq = 0$ Thus, A is non-singular. Therefore, its $A^{- 1}$ exists. Therefore, the given system is consistent and has a unique solution given by $X = A^{- 1} B$ Cofactors of A are,

$A_{11} = 2, A_{12} = - 3, A_{21} = - 2, A_{22} = 5$

$adj (A) = [A_{ij}]^{'} = [2 - 2 - 3 5]^{'} = [2 - 3 - 2 5]$ $∴ A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{4} [2 - 3 - 2 5]$ Now, $X = A^{- 1} B = \frac{1}{4} [2 - 3 - 2 5] [35]$

$= \frac{1}{4} [6 - 10 - 9 + 25] = \frac{1}{4} [- 4 16] = [- 1 4]$

$\Rightarrow [x y] = [- 1 4]$ Hence, $x = - 1$ and $y = 4$

$2 x + y + z = 1, x - 2 y - z = \frac{3}{2}, 3 y - 5 z = 9$

Sol. The given system can be written as $AX = B$ , where

$A = 220 1 - 4 3 1 - 2 - 5, X = x y z and B = 139$

Here, $∣ A ∣ = 220 1 - 4 3 1 - 2 - 5$

$= 2 (20 + 6) - 1 (- 10 - 0) + 1 (6 - 0) = 52 + 10 + 6 = 68 \neq = 0$

Thus, A is non-singular.

Therefore, its $A^{- 1}$ exists. Therefore, the given system is consistent and has a unique solution given by

$X = A^{- 1} B$

Cofactors of A are,

$A_{11} = 20 + 6 = 26, A_{12} = - (- 10 - 0) = 10, A_{13} = 6 - 0 = 6 A_{21} = - (- 5 - 3) = 8, A_{22} = - 10 - 0 = - 10, A_{23} = - (6 - 0) = - 6 A_{31} = (- 2 + 4) = 2, A_{32} = - (- 4 - 2) = 6, A_{33} = - 8 - 2 = - 10$

$adj (A) = [A_{ij}]^{'} = 2682 10 - 10 6 6 - 6 - 10^{'}$

$= 26106 8 - 10 - 6 26 - 10$

$∴ A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{68} 26106 8 - 10 - 6 26 - 10$ Now, $X = A^{- 1} B$

$\Rightarrow x y z = \frac{1}{68} 26106 8 - 10 - 6 26 - 10 \cdot 139 x y z = \frac{1}{68} 26 + 24 + 18 10 - 30 + 54 6 - 18 - 90 = \frac{1}{68} 6834 - 102 \Rightarrow x y z = 1 \frac{1}{2} \frac{- 3}{2}$

Hence; $x = 1, y = \frac{1}{2}$ and $z = \frac{- 3}{2}$

$x - y + z = 4, 2 x + y - 3 z = 0, x + y + z = 2$

Sol. The given system can be written as $AX = B$ , where $A = 121 - 1 11 1 - 3 1, X = x y z$ and $B = 402$ Here, $∣ A ∣ = 121 - 1 11 1 - 3 1$

$= 1 (1 + 3) - (- 1) (2 + 3) + 1 (2 - 1) = 4 + 5 + 1 = 10 \neq = 0$

Thus, A is non-singular. Therefore, its $A^{- 1}$ exists. Therefore, the given system is consistent and has a unique solution given by

$X = A^{- 1} B$

Cofactors of A are,

$A_{11} = 1 + 3 = 4, A_{13} = 2 - 1 = 1, A_{21} = - (- 1 - 1) = 2, A_{23} = - (1 + 1) = - 2 A_{31} = 3 - 1 = 2, A_{33} = 1 + 2 = 3 A_{12} = - (2 + 3) = - 5, A_{22} = 1 - 1 = 0, A_{32} = - (- 3 - 2) = 5,$

$∴ adj (A) = [A_{ij}]^{'} = 422 - 5 05 1 - 2 3^{'}$

$= 4 - 5 1 20 - 2 253$

$∴ A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{10} 4 - 5 1 20 - 2 253$ Now, $X = A^{- 1} B$

$\Rightarrow x y z = \frac{1}{10} 4 - 5 1 20 - 2 253 \cdot 402$

$\Rightarrow x y z = \frac{1}{10} 16 + 0 + 4 - 20 + 0 + 10 4 + 0 + 6 = \frac{1}{10} 20 - 10 10 = 2 - 1 1$ Hence; $x = 2, y = - 1$ and $z = 1$

$2 x + 3 y + 3 z = 5, x - 2 y + z = - 4$ ,

$3 x - y - 2 z = 3.$

Sol. The given system can be written as $AX = B$ , where

$A = 213 3 - 2 - 1 31 - 2, X = x y z, and B = 5 - 4 3$

Here, $∣ A ∣ = 213 3 - 2 - 1 31 - 2$

$= 2 (4 + 1) - 3 (- 2 - 3) + 3 (- 1 + 6) = 10 + 15 + 15 = 40 \neq = 0$

Thus, A is non-singular. Therefore, its $A^{- 1}$ exists. Therefore, the given system is consistent and has a unique solution given by $X = A^{- 1} B$ Cofactors of A are,

$A_{11} = 4 + 1 = 5, A_{13} = (- 1 + 6) = 5 A_{21} = - (- 6 + 3) = 3, A_{23} = - (- 2 - 9) = 11 A_{31} = 3 + 6 = 9, A_{33} = - 4 - 3 = - 7 A_{12} = - (- 2 - 3) = 5, A_{22} = (- 4 - 9) = - 13, A_{32} = - (2 - 3) = 1,$

$adj (A) = [A_{ij}]^{'} = 539 5 - 13 1 511 - 7^{'}$

$= 555 3 - 13 11 91 - 7$

$∴ A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{40} 555 3 - 13 11 91 - 7$ Now, $X = A^{- 1} B$ $\Rightarrow x y z = \frac{1}{40} 555 3 - 13 11 91 - 7 \cdot 5 - 4 3$

$= \frac{1}{40} 25 - 12 + 27 25 + 52 + 3 25 - 44 - 21 = \frac{1}{40} 4080 - 40 = 12 - 1$

Hence, $x = 1, y = 2$ and $z = - 1$

$x - y + 2 z = 7, 3 x + 4 y - 5 z = - 5$ ,

$2 x - y + 3 z = 12.$

Sol. The given system can be written as $AX = B$ , where

$A = 132 - 1 4 - 1 2 - 5 3, X = x y z, and B = 7 - 5 12$

Here,

$∣ A ∣ = 132 - 1 4 - 1 2 - 5 3 = 1 (12 - 5) - (- 1) (9 + 10) + 2 (- 3 - 8) = 7 + 19 - 22 = 4 \neq = 0$

Thus, A is non-singular. Therefore, its $A^{- 1}$ exists. Therefore, the given system is consistent and has a unique solution given by

$X = A^{- 1} B$

Cofactors of A are,

$A_{11} = 12 - 5 = 7, A_{12} = - (9 + 10) = - 19, A_{13} = - 3 - 8 = - 11 A_{21} = - (- 3 + 2) = 1, A_{22} = 3 - 4 = - 1, A_{23} = - (- 1 + 2) = - 1 A_{31} = 5 - 8 = - 3, A_{32} = - (- 5 - 6) = 11, A_{33} = 4 + 3 = 7$

$adj (A) = [A_{ij}]^{'} = 71 - 3 - 19 - 1 11 - 11 - 1 7^{'}$

$= 7 - 19 - 11 1 - 1 - 1 - 3 117$

$∴ A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{4} 7 - 19 - 11 1 - 1 - 1 - 3 117$ Now, $X = A^{- 1} B$

$x y z = \frac{1}{4} 7 - 19 - 11 1 - 1 - 1 - 3 117 \cdot 7 - 5 12 = \frac{1}{4} 49 - 5 - 36 - 133 + 5 + 132 - 77 + 5 + 84$

Hence, $x = 2, y = 1$ and $z = 3$

If $A = 231 - 3 21 5 - 4 - 2$ , find $A^{- 1}$ . Using $A^{- 1}$ , solve the system of equations

$2 x - 3 y + 5 z = 11 3 x + 2 y - 4 z = - 5 x + y - 2 z = - 3$

Sol. The given system can be written as $AX = B$ , where $A = 231 - 3 21 5 - 4 - 2, X = x y z$ and

$B = 11 - 5 - 3$

Here, $∣ A ∣ = 231 - 3 21 5 - 4 - 2$

$= 2 (- 4 + 4) - (- 3) (- 6 + 4) + 5 (3 - 2) = 0 - 6 + 5 = - 1 \neq = 0$

Thus, A is non-singular. Therefore, its $A^{- 1}$ exists. Therefore, the given system is consistent and has a unique solution given by $X = A^{- 1} B$ Cofactors of A are,

$A_{11} = - 4 + 4 = 0, A_{12} = - (- 6 + 4) = 2, A_{13} = 3 - 2 = 1 A_{21} = - (6 - 5) = - 1, A_{22} = - 4 - 5 = - 9, A_{23} = - (2 + 3) = - 5 A_{31} = (12 - 10) = 2, A_{32} = - (- 8 - 15) = 23, A_{33} = 4 + 9 = 13$

$adj (A) = [A_{ij}]^{'}$

$= 0 - 1 2 2 - 9 23 1 - 5 13^{'} = 021 - 1 - 9 - 5 22313$

$∴ A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{- 1} 021 - 1 - 9 - 5 22313$

$= 0 - 2 - 1 195 - 2 - 23 - 13$

Now, $X = A^{- 1} B$

$\Rightarrow x y z = 0 - 2 - 1 195 - 2 - 23 - 13 \cdot 11 - 5 - 3 \Rightarrow x y z = 0 - 5 + 6 - 22 - 45 + 69 - 11 - 25 + 39 = 123$

Hence, $x = 1, y = 2$ and $z = 3$

The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion, 2kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method. Sol. Let the prices (per kg) of onion, wheat and rice be ₹x, ₹y and ₹ $z$ , respectively. Then

$4 x + 3 y + 2 z = 60, 2 x + 4 y + 6 z = 90 6 x + 2 y + 3 z = 70$

The given system can be written as $AX = B$ , where

$A = 426342263, X = x y z and B = 609070$

Here,

$∣ A ∣ = 426342263 = 4 (12 - 12) - 3 (6 - 36) + 2 (4 - 24) = 0 + 90 - 40 = 50 \neq = 0$

Thus, A is non-singular. Therefore, its $A^{- 1}$ exists. Therefore, the given system is consistent and has a unique solution given by

$X = A^{- 1} B$

Cofactors of A are,

$A_{11} = 12 - 12 = 0, A_{12} = - (6 - 36) = 30, A_{13} = 4 - 24 = - 20, A_{21} = - (9 - 4) = - 5, A_{22} = 12 - 12 = 0, A_{23} = - (8 - 18) = 10, A_{31} = (18 - 8) = 10, A_{32} = - (24 - 4) = - 20, A_{33} = 16 - 6 = 10$

$adj (A) = [A_{ij}]^{'} = 0 - 5 10 300 - 20 - 20 1010^{'} = 030 - 20 - 5 010 10 - 20 10$

$∴ A^{- 1} = \frac{1}{∣ A ∣} (adj A) = \frac{1}{50} 030 - 20 - 5 010 10 - 20 10$

Now, $X = A^{- 1} B$

$x y z = \frac{1}{50} 030 - 20 - 5 010 10 - 20 10 \cdot 609070 = \frac{1}{50} 0 - 450 + 700 1800 + 0 - 1400 - 1200 + 900 + 700$

$\Rightarrow x y z = \frac{1}{50} 250400400 = 588$ $∴ x = 5, y = 8$ and $z = 8$ . Hence, price of onion per kg is ₹5, price of wheat per kg is ₹8 and that of rice per kg is ₹8.

MISCELLANEOUS EXERCISE

Prove that the determinant

$x - sin θ cos θ sin θ - x 1 cos θ 1 x is independent of θ .$

Sol. Let $∣ A ∣ = x - sin θ cos θ sin θ - x 1 cos θ 1 x$

Expanding to corresponding first row, we get

$∣ A ∣ = x - x 1 1 x - sin θ - sin θ cos θ 1 x + cos θ - sin θ cos θ - x 1 = x (- x^{2} - 1) - sin θ (- x sin θ - cos θ) + cos θ (- sin θ + x cos θ) = - x^{3} - x + x sin^{2} θ + sin θ cos θ - sin θ cos θ + x cos^{2} θ = - x^{3} - x + x (sin^{2} θ + cos^{2} θ) = - x^{3} - x + x (∵ sin^{2} θ + cos^{2} θ = 1) = - x^{3}$

Hence, A is independent of $θ$ .

Evaluate $cos α cos β - sin β sin α cos β cos α sin β cos β sin α sin β - sin α 0 cos α$ .

Sol. Given, $∣ A ∣ = cos α cos β - sin β sin α cos β cos α sin β cos β sin α sin β - sin α 0 cos α$ Expanding corresponding to $R_{1}$ , we get

$∣ A ∣ = = = = cos α cos β (cos α cos β - 0) - cos α sin β (- cos α sin β - 0) - sin α (- sin^{2} β sin α - cos^{2} β sin α) cos^{2} α (cos^{2} β + sin^{2} β) + sin^{2} α (sin^{2} β + cos^{2} β) [∵ sin^{2} θ + cos^{2} θ = 1] cos^{2} α (1) + sin^{2} α (1) cos^{2} α + sin^{2} α = 1$

If $A^{- 1} = 3 - 15 5 - 1 6 - 2 1 - 5 2$ and

$B = 1 - 1 0 23 - 2 - 2 01, find (A B)^{- 1} .$

Sol. We know that, $(AB)^{- 1} = B^{- 1} A^{- 1}$ and $A^{- 1}$ is known, therefore we proceed to find $B^{- 1}$ .

Here,

$∣ B ∣ = 1 - 1 0 23 - 2 - 2 01 = 1 (3 - 0) - 2 (- 1 - 0) - 2 (2 - 0) = 3 + 2 - 4 = 1 \neq = 0$

$∴ B^{- 1}$ exists. Cofactors of B are:

$B_{11} = (3 - 0) = 3, B_{13} = (2 - 0) = 2, B_{21} = - (2 - 4) = 2, B_{23} = - (- 2 - 0) = 2, B_{31} = (0 + 6) = 6, B_{33} = (3 + 2) = 5 B_{12} = - (- 1 - 0) = 1, B_{22} = (1 - 0) = 1, B_{32} = - (0 - 2) = 2,$

$∴ adj (B) = [B_{ij}]^{'} = 326112225^{'} = 312212625$ $∴ B^{- 1} = \frac{1}{∣ B ∣} adj (B) = \frac{1}{1} 312212625 = 312212625$ Now, $(AB)^{- 1} = B^{- 1} A^{- 1}$

$= 312212625 3 - 15 5 - 1 6 - 2 1 - 5 2 = 9 - 30 + 30 3 - 15 + 10 6 - 30 + 25 - 3 + 12 - 12 - 1 + 6 - 4 - 2 + 12 - 10 3 - 10 + 12 1 - 5 + 4 2 - 10 + 10 = 9 - 2 1 - 3 10 502$

Let $A = 1 - 2 1 - 2 31 115$ , verify that (a)

$[adj A]^{- 1} = adj (A^{- 1}), (b) (A^{- 1})^{- 1} = A$

Sol. Given, $A = 1 - 2 1 - 2 31 115$

$\Rightarrow ∣ A ∣ = 1 - 2 1 - 2 31 115 = 1 (15 - 1) - (- 2) (- 10 - 1) + 1 (- 2 - 3) = 14 - 22 - 5 = - 13 \neq = 0$

$∴ A^{- 1}$ exists Cofactors of A are:

$A_{11} = 15 - 1 = 14, A_{12} = - (- 10 - 1) = 11, A_{13} = - 2 - 3 = - 5, A_{21} = - (- 10 - 1) = 11, A_{22} = 5 - 1 = 4, A_{23} = - (1 + 2) = - 3 A_{31} = (- 2 - 3) = - 5, A_{32} = - (1 + 2) = - 3, A_{33} = 3 - 4 = - 1$

$adj (A) = [A_{ij}]^{'} = 1411 - 5 114 - 3 - 5 - 3 - 1^{'}$

$= 1411 - 5 114 - 3 - 5 - 3 - 1$

$∴ A^{- 1} = \frac{1}{∣ A ∣} adj (A) = \frac{1}{- 13} 1411 - 5 114 - 3 - 5 - 3 - 1 = - \frac{14}{13} - \frac{11}{13} \frac{5}{13} - \frac{11}{13} - \frac{4}{13} \frac{3}{13} \frac{5}{13} \frac{3}{13} \frac{1}{13}$

(a) $Here, adj (A) = 1411 - 5 114 - 3 - 5 - 3 - 1 = B ($ let $)$

$∴ ∣ B ∣ = ∣ adj A ∣ = 1411 - 5 114 - 3 - 5 - 3 - 1 = 14 (- 4 - 9) - 11 (- 11 - 15) - 5 (- 33 + 20) = - 182 + 286 + 65 = 169 \neq = 0$

$∴ B^{- 1}$ exists. Cofactors of B are:

$B_{11} = (- 4 - 9) = - 13, B_{12} = - (- 11 - 15) = 26, B_{13} = (- 33 + 20) = - 13, B_{21} = - (- 11 - 15) = 26, B_{22} = (- 14 - 25) = - 39, B_{23} = - (- 42 + 55) = - 13, B_{31} = (- 33 + 20) = - 13, B_{32} = - (- 42 + 55) = - 13, B_{33} = (56 - 121) = - 65$

$\Rightarrow adj (B) = adj (adj A)$

$= [B_{ij}]^{'} = - 13 26 - 13 26 - 39 - 13 - 13 - 13 - 65^{'} = - 13 26 - 13 26 - 39 - 13 - 13 - 13 - 65$

$∴ B^{- 1} = [adj A]^{- 1} = \frac{1}{∣ adj A ∣} {adj (adj A)}$ $\Rightarrow [adj A]^{- 1} = \frac{1}{169} - 13 26 - 13 26 - 39 - 13 - 13 - 13 - 65$

$= \frac{1}{13} - 1 2 - 1 2 - 3 - 1 - 1 - 1 - 5$

Cofactors of $A^{- 1}$ are:

$A_{11} = \frac{- 13}{169}, A_{21} = \frac{26}{169}, A_{31} = \frac{- 13}{169}, A_{12} = \frac{26}{169}, A_{22} = \frac{- 39}{169}, A_{32} = - \frac{13}{169}, A_{13} = \frac{- 13}{169}, A_{23} = \frac{- 13}{169}, A_{33} = - \frac{65}{169}$

$N o w, adj (A^{- 1}) = - \frac{13}{169} \frac{26}{169} - \frac{13}{169} \frac{26}{169} - \frac{39}{169} - \frac{13}{169} - \frac{13}{169} - \frac{13}{169} - \frac{65}{169}^{t}$

$= - \frac{13}{169} \frac{26}{169} - \frac{13}{169} \frac{26}{169} - \frac{39}{169} - \frac{13}{169} - \frac{13}{169} - \frac{13}{169} - \frac{65}{169}$

⇒

$adj (A^{- 1}) = \frac{1}{13} - 1 2 - 1 2 - 3 - 1 - 1 - 1 - 5$

From Eq. (1) and (2), we get that $adj (A^{- 1}) = (adj A)^{- 1}$

(b)

$A^{- 1} = = = = = - \frac{14}{13} (- \frac{4}{169} - \frac{9}{169}) + \frac{11}{13} (- \frac{11}{169} - \frac{15}{169}) + \frac{5}{13} (- \frac{33}{169} + \frac{20}{169}) - \frac{14}{13} (- \frac{13}{169}) + \frac{11}{13} (- \frac{26}{169}) + \frac{5}{13} (- \frac{13}{169}) \frac{14}{169} - \frac{22}{169} - \frac{5}{169} - \frac{13}{169} - \frac{1}{13}$

and $adj (A^{- 1}) = \frac{1}{13} - 1 2 - 1 2 - 3 - 1 - 1 - 1 - 5$ $∴ (A^{- 1})^{- 1} = \frac{1}{∣ A ^{- 1} ∣} (adjA^{- 1})$

$= \frac{1}{( - \frac{1}{13} )} \times \frac{1}{13} - 1 2 - 1 2 - 3 - 1 - 1 - 1 - 5 = 1 - 2 1 - 2 31 115 = A$

Evaluate $x y x + y y x + y x x + y x y$ .

Sol. Let $Δ = x y x + y y x + y x x + y x y$ Expand along $R_{1}$

$Δ = x (x y + y^{2} - x^{2}) - y (y^{2} - x^{2} - x y) + (x + y) (x y - x^{2} - 2 x y - y^{2}) Δ = x^{2} y + x y^{2} - x^{3} - y^{3} + x^{2} y + x y^{2} + x^{2} y - x^{3} - 2 x^{2} y - x y^{2} + x y^{2} - x^{2} y - 2 x y^{2} - y^{3} Δ = - 2 (x^{3} + y^{3})$

Evaluate $111 x x + y x y y x + y$ .

Sol. Let $Δ = 111 x x + y x y y x + y$ Expand along $C_{1}$

$Δ = 1 (x^{2} + 2 x y + y^{2} - x y) - 1 (x^{2} + x y - x y) + 1 (x y - x y - y^{2}) Δ = x^{2} + x y + y^{2} - x^{2} - y^{2} = x y$

Solve the following system of equations:

$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4; \frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1 and \frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2.$

Sol. Let $\frac{1}{x} = p, \frac{1}{y} = q$ and $\frac{1}{z} = r$ , then the given equations become $2 p + 3 q + 10 r = 4$ ,

$4 p - 6 q + 5 r = 1, 6 p + 9 q - 20 r = 2$

This system can be written as $AX = B$ , where

$A = 246 3 - 6 9 105 - 20, X = p q r, B = 412$

Here,

$∣ A ∣ = 246 3 - 6 9 105 - 20 = 2 (120 - 45) - 3 (- 80 - 30) + 10 (36 + 36) = 150 + 330 + 720 = 1200$

Thus, A is non-singular. Therefore, its $A^{- 1}$ exists.

Therefore, the given system is consistent and has a unique solution given by $X = A^{- 1} B$ Cofactors of A are:

$A_{11} = 120 - 45 = 75, A_{12} = - (- 80 - 30) = 110, A_{13} = (36 + 36) = 72, A_{21} = - (- 60 - 90) = 150, A_{22} = (- 40 - 60) = - 100, A_{23} = - (18 - 18) = 0, A_{31} = 15 + 60 = 75, A_{32} = - (10 - 40) = 30, A_{33} = - 12 - 12 = - 24$

$\Rightarrow adj (A) = [A_{ij}]^{'} = 7515075 110 - 100 30 720 - 24^{'} = 7511072 150 - 100 0 7530 - 24$

Now;

$A^{- 1} = \frac{1}{∣ A ∣} adj (A) = \frac{1}{1200} 7511072 150 - 100 0 7530 - 24$

$∵ \Rightarrow X = A^{- 1} B p q r = \frac{1}{1200} 7511072 150 - 100 0 7530 - 24 412 p q r = \frac{1}{1200} 300 + 150 + 150 440 - 100 + 60 288 + 0 - 48 = \frac{1}{1200} 600400240 = \frac{1}{2} \frac{1}{3} \frac{1}{5}$

$\Rightarrow \Rightarrow \Rightarrow p = \frac{1}{2}, q = \frac{1}{3}, r = \frac{1}{5} \frac{1}{x} = \frac{1}{2}, \frac{1}{y} = \frac{1}{3}, \frac{1}{z} = \frac{1}{5} x = 2, y = 3 and z = 5$

Choose the correct answer in the following Exercises from 8 and 9. 8. If x,y,z are non-zero real numbers, then the inverse of matrix $A = x 00 0 y 0 00 z$ is ? (A) $x^{- 1} 00 0 y^{- 1} 0 00 z^{- 1}$ (B) $xyz x^{- 1} 00 0 y^{- 1} 0 00 z^{- 1}$ (C) $\frac{1}{xyz} x 00 0 y 0 00 z$ (C) $\frac{1}{xyz} x 00 0 y 0 00 z$ (D) $\frac{1}{xyz} 100010001$

Sol. (A) Given, $A = x 00 0 y 0 00 z$

$\Rightarrow ∣ A ∣ = x 00 0 y 0 00 z = x (yz - 0) = xyz \neq = 0$

( $∵ x, y$ and z are non-zero) $∴ A^{- 1}$ exists Cofactors of A are:

$A_{11} = (yz - 0) = yz, A_{13} = 0 - 0 = 0, A_{21} = - (0 - 0) = 0, A_{23} = - (0 - 0) = 0, A_{31} = 0 - 0 = 0, A_{33} = (x y - 0) = x y A_{12} = - (0 - 0) = 0, A_{22} = x z - 0 = x z, A_{32} = - (0 - 0) = 0,$

$∴ adj (A) = [A_{ij}]^{'} = yz 00 0 xz 0 00 xy^{'}$

$= yz 00 0 xz 0 00 xy$

Now, $A^{- 1} = \frac{1}{∣ A ∣} (adj A)$

$\Rightarrow A^{- 1} = \frac{1}{xyz} yz 00 0 xz 0 00 xy = \frac{1}{x} 00 0 \frac{1}{y} 0 00 \frac{1}{z} = x^{- 1} 00 0 y^{- 1} 0 00 z^{- 1}$

Hence, the correct option is (A).

Let $A = 1 - sin θ - 1 sin θ 1 - sin θ 1 sin θ 1$ , where $0 \leq θ \leq 2 π$ , then: (A) $det A = 0$ (B) $det A \in (2, \infty)$ (C) $det A \in (2, 4)$ (D) $det A \in [2, 4]$

Sol. (D) Given, $A = 1 - sin θ - 1 sin θ 1 - sin θ 1 sin θ 1$

$\Rightarrow ∣ A ∣ = 1 - sin θ - 1 sin θ 1 - sin θ 1 sin θ 1 = 1 (1 + sin^{2} θ) - sin θ (- sin θ + sin θ) + 1 (sin^{2} θ + 1)$

$∴ ∣ A ∣ = 2 + 2 sin^{2} θ$ For $0 \leq θ \leq 2 π, - 1 \leq sin θ \leq 1$

$\Rightarrow 0 \leq sin^{2} θ \leq 1$

$\Rightarrow 1 \leq 1 + sin^{2} θ \leq 2 \Rightarrow 2 \leq 2 (1 + sin^{2} θ) \leq 4$ $∴ det (A) \in [2, 4]$ Hence, the correct option is (D).

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

Check NCERT Solutions for Class 12 Maths chapter-wise, covering exercise questions, useful formulas, and step-by-step answers for effective preparation.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Relations and Functions
Chapter 2	Inverse Trigonometric Functions
Chapter 3	Matrices
Chapter 5	Continuity and Differentiability
Chapter 6	Application of Derivatives
Chapter 7	Integrals
Chapter 8	Application of Integrals
Chapter 9	Differential Equations
Chapter 10	Vector Algebra
Chapter 11	Three Dimensional Geometry
Chapter 12	Linear Programming
Chapter 13	Probability

4.0Class 12 Maths Chapter 4 Determinants Exercise-wise Solutions

Exercise	Number of Questions	Important Topics
Exercise 4.1 Solutions	8 Questions	Determinants of 2 × 2 and 3 × 3 matrices and their evaluation
Exercise 4.2 Solutions	5 Questions	Solving linear equations using determinants and Cramer's rule
Exercise 4.3 Solutions	5 Questions	Inverse of a matrix using determinants and applications of determinants
Exercise 4.4 Solutions	18 Questions	Minors, cofactors and their use in finding the inverse of a matrix
Exercise 4.5 Solutions	16 Questions	Adjoint of a matrix, its properties and inverse of a matrix
MISCELLANEOUS EXERCISE	9 Questions

5.0Key Features and Benefits of NCERT Solutions Class 12 Maths Chapter 4 Determinants

Step-by-step solutions explain the properties of determinants clearly and help students avoid common calculation errors.
Detailed methods make it easier to understand the expansion of determinants and present solutions properly in board examinations.
The exercises are based on the NCERT Class 12 Maths syllabus and cover important questions from the textbook.
Regular practice with NCERT questions can help students improve their speed and accuracy while solving determinant-based problems.
A strong understanding of determinants can support preparation for mathematics olympiads and other entrance examinations.
Clear concepts in determinants can help students understand linear equations and other advanced topics in algebra.

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NCERT Solutions Class 12 Maths Chapter 4 Determinants

1.0Key Concepts of Class 12 Maths Chapter 4 Determinants

2.0NCERT Class 12 Maths Chapter 4 Determinants : Detailed Solutions

EXERCISE - 4.1

EXERCISE - 4.2

EXERCISE - 4.3

EXERCISE - 4.4

EXERCISE - 4.5

MISCELLANEOUS EXERCISE

3.0Class 12 Maths NCERT Solutions – Chapter-wise Links

4.0Class 12 Maths Chapter 4 Determinants Exercise-wise Solutions

5.0Key Features and Benefits of NCERT Solutions Class 12 Maths Chapter 4 Determinants

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