# NCERT Solutions for Exercise 4.5 Class 12 Maths Chapter 4 - Determinants

In this article, you will get NCERT solutions for Class 12 Maths chapter 4 Determinants Exercise 4.5 consists of questions related to adjoint, cofactor, and inverse of a matrix. The adjoint of a square matrix A is defined as the transpose of the matrix of the cofactors of matrix A. The adjoint of matrix A is denoted by adj (A) and the inverse of matrix A is defined by A-1. It is a very important concept used in the solving system of linear equations, statistics, and scientific research.

Exercise 4.5 Class 12 Maths is very important for the board exams as generally one question is asked from this exercise. You are advised to solve the questions from Class 12 Maths ch 4 ex 4.5 including examples given before the exercise. You can go through Class 12 Maths chapter 4 exercise 4.5 solutions to understand the concept. Also, check for NCERT Solutions here.

Also, see

• Determinants Exercise 4.1
• Determinants Exercise 4.2
• Determinants Exercise 4.3
• Determinants Exercise 4.4
• Determinants Exercise 4.6
• Determinants Miscellaneous Exercise

## Determinants Exercise:4.5

Question:1 Find adjoint of each of the matrices.

$\small \begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix}$

Given matrix: $\small \begin{bmatrix} 1 &2 \\ 3 & 4 \end{bmatrix}= A$

Then we have,

$A_{11} = 4, A_{12}=-(1)3, A_{21} = -(1)2,\ and\ A_{22}= 1$

Hence we get:

$adjA = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} &A_{22} \end{bmatrix}^T = \begin{bmatrix} A_{11} & A_{21} \\ A_{12} &A_{22} \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ -3 &1 \end{bmatrix}$

Question:2 Find adjoint of each of the matrices

$\small \begin{bmatrix} 1 &-1 &2 \\ 2 & 3 &5 \\ -2 & 0 &1 \end{bmatrix}$

Given the matrix: $\small A = \begin{bmatrix} 1 &-1 &2 \\ 2 & 3 &5 \\ -2 & 0 &1 \end{bmatrix}$

Then we have,

$A_{11} = (-1)^{1+1}\begin{vmatrix} 3 &5 \\ 0& 1 \end{vmatrix} =(3-0)= 3$

$A_{12} = (-1)^{1+2}\begin{vmatrix} 2 &5 \\ -2& 1 \end{vmatrix} =-(2+10)= -12$

$A_{13} = (-1)^{1+3}\begin{vmatrix} 2 &3 \\ -2& 0 \end{vmatrix} =0+6= 6$

$A_{21} = (-1)^{2+1}\begin{vmatrix} -1 &2 \\ 0& 1 \end{vmatrix} =-(-1-0)= 1$

$A_{22} = (-1)^{2+2}\begin{vmatrix} 1 &2 \\ -2& 1 \end{vmatrix} =(1+4)= 5$

$A_{23} = (-1)^{2+3}\begin{vmatrix} 1 &-1 \\-2& 0 \end{vmatrix} =-(0-2)= 2$

$A_{31} = (-1)^{3+1}\begin{vmatrix} -1 &2 \\ 3& 5 \end{vmatrix} =(-5-6)= -11$

$A_{32} = (-1)^{3+2}\begin{vmatrix} 1 &2 \\2& 5\end{vmatrix} =-(5-4)= -1$

$A_{33} = (-1)^{3+3}\begin{vmatrix} 1 &-1 \\ 2& 3 \end{vmatrix} =(3+2)= 5$

Hence we get:

$adjA = \begin{bmatrix} A_{11} &A_{21} &A_{31} \\ A_{12}&A_{22} &A_{32} \\ A_{13}&A_{23} &A_{33} \end{bmatrix} = \begin{bmatrix} 3 &1 &-11 \\ -12&5 &-1 \\ 6&2 &5 \end{bmatrix}$

Question:3 Verify $\small A (adj A)=(adj A)A=|A|I$.

$\small \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}$

Given the matrix: $\small \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}$

Let $\small A = \begin{bmatrix} 2 &3 \\ -4 & -6 \end{bmatrix}$

Calculating the cofactors;

$\small A_{11} = (-1)^{1+1}(-6) = -6$

$\small A_{12} = (-1)^{1+2}(-4) = 4$

$\small A_{21} = (-1)^{2+1}(3) = -3$

$\small A_{22} = (-1)^{2+2}(2) = 2$

Hence, $\small adjA = \begin{bmatrix} -6 &-3 \\ 4& 2 \end{bmatrix}$

Now,

$\small A (adj A) = \begin{bmatrix} 2 &3 \\ -4&-6 \end{bmatrix}\left ( \begin{bmatrix} -6 &-3 \\ 4 &2 \end{bmatrix} \right )$

$\small \begin{bmatrix} -12+12 &-6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}$

aslo,

$\small (adjA)A = \begin{bmatrix} -6 &-3 \\ 4 & 2 \end{bmatrix}\begin{bmatrix} 2 &3 \\ -4& -6 \end{bmatrix}$

$\small = \begin{bmatrix} -12+12 &-18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$

Now, calculating |A|;

$\small |A| = -12-(-12) = -12+12 = 0$

So, $\small |A|I = 0\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$

Hence we get

$\small A (adj A)=(adj A)A=|A|I$

Question:4 Verify $\small A (adj A)=(adjA)A=|A| I$.

$\small \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}$

Given matrix: $\small \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}$

Let $\small A= \begin{bmatrix} 1 &-1 & 2\\ 3 &0 &-2 \\ 1 & 0 &3 \end{bmatrix}$

Calculating the cofactors;

$\small A_{11} = (-1)^{1+1} \begin{vmatrix} 0 &-2 \\ 0& 3 \end{vmatrix} = 0$

$\small A_{12} = (-1)^{1+2} \begin{vmatrix} 3 &-2 \\1& 3 \end{vmatrix} = -(9+2) =-11$

$\small A_{13} = (-1)^{1+3} \begin{vmatrix} 3 &0 \\ 1& 0 \end{vmatrix} = 0$

$\small A_{21} = (-1)^{2+1} \begin{vmatrix} -1 &2 \\ 0& 3 \end{vmatrix} = -(-3-0)= 3$

$\small A_{22} = (-1)^{2+2} \begin{vmatrix} 1 &2 \\ 1& 3 \end{vmatrix} = 3-2=1$

$\small A_{23} = (-1)^{2+3} \begin{vmatrix} 1 &-1 \\ 1& 0 \end{vmatrix} = -(0+1) = -1$

$\small A_{31} = (-1)^{3+1} \begin{vmatrix} -1 &2 \\ 0& -2 \end{vmatrix} = 2$

$\small A_{32} = (-1)^{3+2} \begin{vmatrix} 1 &2 \\ 3& -2 \end{vmatrix} = -(-2-6) = 8$

$\small A_{33} = (-1)^{3+3} \begin{vmatrix} 1 &-1 \\ 3& 0 \end{vmatrix} = 0+3 =3$

Hence, $\small adjA = \begin{bmatrix} 0 &3 &2 \\ -11 & 1& 8\\ 0 &-1 & 3 \end{bmatrix}$

Now,

$\small A (adj A) =\begin{bmatrix} 1 &-1 &2 \\ 3& 0 & -2\\ 1 & 0 & 3 \end{bmatrix}\begin{bmatrix} 0 &3 &2 \\ -11& 1& 8\\ 0& -1 &3 \end{bmatrix}$

$\small =\begin{bmatrix} 0+11+0 &3-1-2 &2-8+6 \\ 0+0+0 & 9+0+2 & 6+0-6 \\ 0+0+0 &3+0-3 & 2+0+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 &0 \\ 0& 11&0 \\ 0 & 0 & 11 \end{bmatrix}$

also,

$\small A (adj A) =\begin{bmatrix} 0 &3 &2 \\ -11& 1& 8\\ 0& -1 &3 \end{bmatrix}\begin{bmatrix} 1 &-1 &2 \\ 3& 0 & -2\\ 1 & 0 & 3 \end{bmatrix}$

$\small =\begin{bmatrix} 0+9+2 &0+0+0 &0-6+6 \\ -11+3+8 & 11+0+0 & -22-2+24 \\ 0-3+3 &0+0+0 & 0+2+9 \end{bmatrix} = \begin{bmatrix} 11 & 0 &0 \\ 0& 11&0 \\ 0 & 0 & 11 \end{bmatrix}$

Now, calculating |A|;

$\small |A| = 1(0-0) +1(9+2) +2(0-0) = 11$

So, $\small |A|I = 11\begin{bmatrix} 1 &0&0 \\ 0& 1&0 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 11 &0&0 \\ 0& 11&0\\ 0&0&11 \end{bmatrix}$

Hence we get,

$\small A (adj A)=(adj A)A=|A|I$.

Question:5 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 2 &-2 \\ 4 & 3 \end{bmatrix}$

Given matrix : $\small \begin{bmatrix} 2 &-2 \\ 4 & 3 \end{bmatrix}$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

|A| = (6+8) = 14

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (3) = 3$

$A_{12} = (-1)^{1+2} (4) = -4$

$A_{21} = (-1)^{2+1} (-2) = 2$

$A_{22} = (-1)^{2+2} (2) = 2$

So, we have $adjA = \begin{bmatrix} 3 &2 \\ -4& 2 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{14}\begin{bmatrix} 3 &2 \\ -4& 2 \end{bmatrix} = \begin{bmatrix} \frac{3}{14} &\frac{1}{7} \\ \\ \frac{-2}{7} & \frac{1}{7} \end{bmatrix}$

Question:6 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} -1 &5 \\ -3 &2 \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} -1 &5 \\ -3 &2 \end{bmatrix} = A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

|A| = (-2+15) = 13

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (2) = 2$

$A_{12} = (-1)^{1+2} (-3) = 3$

$A_{21} = (-1)^{2+1} (5) =-5$

$A_{22} = (-1)^{2+2} (-1) = -1$

So, we have $adjA = \begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{13}\begin{bmatrix} 2 &-5 \\ 3& -1 \end{bmatrix} = \begin{bmatrix} \frac{2}{13} &\frac{-5}{13} \\ \\ \frac{3}{13} & \frac{-1}{13} \end{bmatrix}$

Question:7 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 1 &2 &3 \\ 0 &2 &4 \\ 0 &0 &5 \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} 1 &2 &3 \\ 0 &2 &4 \\ 0 &0 &5 \end{bmatrix}= A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(10-0)-2(0-0)+3(0-0) = 10$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (10) = 10$ $A_{12} = (-1)^{1+2} (0) = 0$

$A_{13} = (-1)^{1+3} (0) =0$ $A_{21} = (-1)^{2+1} (10) = -10$

$A_{22} = (-1)^{2+2} (5-0) = 5$ $A_{23} = (-1)^{2+1} (0-0) = 0$

$A_{31} = (-1)^{3+1} (8-6) = 2$ $A_{32} = (-1)^{3+2} (4-0) =-4$

$A_{33} = (-1)^{3+3} (2-0) = 2$

So, we have $adjA = \begin{bmatrix} 10 &-10 &2 \\ 0& 5 &-4 \\ 0& 0 &2 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{1}{10}\begin{bmatrix} 10 &-10 &2 \\ 0 & 5& -4\\ 0 &0 &2 \end{bmatrix}$

Question:8 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 1 &0 &0 \\ 3 &3 &0 \\ 5 &2 &-1 \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} 1 &0 &0 \\ 3 &3 &0 \\ 5 &2 &-1 \end{bmatrix} = A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(-3-0)-0(-3-0)+0(6-15) = -3$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (-3-0) = -3$ $A_{12} = (-1)^{1+2} (-3-0) = 3$

$A_{13} = (-1)^{1+3} (6-15) =-9$ $A_{21} = (-1)^{2+1} (0-0) = 0$

$A_{22} = (-1)^{2+2} (-1-0) = -1$ $A_{23} = (-1)^{2+1} (2-0) = -2$

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

$A_{33} = (-1)^{3+3} (3-0) = 3$

So, we have $adjA = \begin{bmatrix} -3 &0 &0 \\ 3& -1 &0 \\ -9& -2 &3 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$= \frac{-1}{3}\begin{bmatrix} -3 &0 &0 \\ 3 & -1& 0\\ -9 &-2 &3 \end{bmatrix}$

Question:9 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 2 &1 &3 \\ 4 &-1 &0 \\ -7 &2 &1 \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} 2 &1 &3 \\ 4 &-1 &0 \\ -7 &2 &1 \end{bmatrix} =A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 2(-1-0)-1(4-0)+3(8-7) =-2-4+3 = -3$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (-1-0) = -1$ $A_{12} = (-1)^{1+2} (4-0) = -4$

$A_{13} = (-1)^{1+3} (8-7) =1$ $A_{21} = (-1)^{2+1} (1-6) = 5$

$A_{22} = (-1)^{2+2} (2+21) = 23$ $A_{23} = (-1)^{2+1} (4+7) = -11$

$A_{31} = (-1)^{3+1} (0+3) = 3$ $A_{32} = (-1)^{3+2} (0-12) =12$

$A_{33} = (-1)^{3+3} (-2-4) = -6$

So, we have $adjA = \begin{bmatrix} -1 &5 &3 \\ -4& 23 &12 \\ 1& -11 &-6 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$A^{-1} = \frac{1}{-3} \begin{bmatrix} -1 &5 &3 \\ -4& 23 &12 \\ 1& -11 &-6 \end{bmatrix}$

Question:10 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 1 & -1 & 2\\ 0 & 2 &-3 \\ 3 &-2 &4 \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} 1 & -1 & 2\\ 0 & 2 &-3 \\ 3 &-2 &4 \end{bmatrix} = A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(8-6)+1(0+9)+2(0-6) =2+9-12 = -1$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (8-6) = 2$ $A_{12} = (-1)^{1+2} (0+9) = -9$

$A_{13} = (-1)^{1+3} (0-6) =-6$ $A_{21} = (-1)^{2+1} (-4+4) = 0$

$A_{22} = (-1)^{2+2} (4-6) = -2$ $A_{23} = (-1)^{2+1} (-2+3) = -1$

$A_{31} = (-1)^{3+1} (3-4) = -1$ $A_{32} = (-1)^{3+2} (-3-0) =3$

$A_{33} = (-1)^{3+3} (2-0) = 2$

So, we have $adjA = \begin{bmatrix} 2 &0 &-1 \\ -9& -2 &3 \\ -6& -1 &2 \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 &0 &-1 \\ -9& -2 &3 \\ -6& -1 &2 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} -2 &0 &1 \\ 9& 2 &-3 \\ 6& 1 &-2 \end{bmatrix}$

Question:11 Find the inverse of each of the matrices (if it exists).

$\small \begin{bmatrix} 1 & 0&0 \\ 0 &\cos \alpha &\sin \alpha \\ 0 &\sin \alpha &-\cos \alpha \end{bmatrix}$

Given the matrix : $\small \begin{bmatrix} 1 & 0&0 \\ 0 &\cos \alpha &\sin \alpha \\ 0 &\sin \alpha &-\cos \alpha \end{bmatrix} =A$

To find the inverse we have to first find adjA then as we know the relation:

$A^{-1} = \frac{1}{|A|}adjA$

So, calculating |A| :

$|A| = 1(-\cos^2 \alpha-\sin^2 \alpha)+0(0-0)+0(0-0)$

$=-(\cos^2 \alpha + \sin^2 \alpha) = -1$

Now, calculating the cofactors terms and then adjA.

$A_{11} = (-1)^{1+1} (-\cos^2 \alpha - \sin^2 \alpha) = -1$ $A_{12} = (-1)^{1+2} (0-0) = 0$

$A_{13} = (-1)^{1+3} (0-0) =0$ $A_{21} = (-1)^{2+1} (0-0) = 0$

$A_{22} = (-1)^{2+2} (-\cos \alpha-0) = -\cos \alpha$ $A_{23} = (-1)^{2+1} (\sin \alpha-0) = -\sin \alpha$

$A_{31} = (-1)^{3+1} (0-0) = 0$ $A_{32} = (-1)^{3+2} (\sin \alpha-0) =-\sin \alpha$

$A_{33} = (-1)^{3+3} (\cos \alpha - 0) = \cos \alpha$

So, we have $adjA = \begin{bmatrix} -1 &0 &0 \\ 0& -\cos \alpha &-\sin \alpha \\ 0& -\sin \alpha &\cos \alpha \end{bmatrix}$

Therefore inverse of A will be:

$A^{-1} = \frac{1}{|A|}adjA$

$A^{-1} = \frac{1}{-1}\begin{bmatrix} -1 &0 &0 \\ 0& -\cos \alpha &-\sin \alpha \\ 0& -\sin \alpha &\cos \alpha \end{bmatrix} = \begin{bmatrix}1 &0 &0 \\ 0&\cos \alpha &\sin \alpha \\ 0& \sin \alpha &-\cos \alpha \end{bmatrix}$

Question:12 Let $\small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix}$ and $\small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix}$. Verify that$\small (AB)^-^1=B^{-1}A^{-1}$.

We have $\small A=\begin{bmatrix} 3 &7 \\ 2 & 5 \end{bmatrix}$ and $\small B=\begin{bmatrix} 6 &8 \\ 7 & 9 \end{bmatrix}$.

then calculating;

$AB = \begin{bmatrix} 3 &7 \\ 2& 5 \end{bmatrix}\begin{bmatrix} 6 &8 \\ 7& 9 \end{bmatrix}$

$=\begin{bmatrix} 18+49 &24+63 \\ 12+35 & 16+45 \end{bmatrix} = \begin{bmatrix} 67 &87 \\ 47& 61 \end{bmatrix}$

Finding the inverse of AB.

Calculating the cofactors fo AB:

$AB_{11}=(-1)^{1+1}(61) = 61$ $AB_{12}=(-1)^{1+2}(47) = -47$

$AB_{21}=(-1)^{2+1}(87) = -87$ $AB_{22}=(-1)^{2+2}(67) = 67$

$adj(AB) = \begin{bmatrix} 61 &-87 \\ -47& 67 \end{bmatrix}$

and |AB| = 61(67) - (-87)(-47) = 4087-4089 = -2

Therefore we have inverse:

$(AB)^{-1}=\frac{1}{|AB|}adj(AB) = -\frac{1}{2} \begin{bmatrix} 61 &-87 \\ -47 & 67 \end{bmatrix}$

$= \begin{bmatrix} \frac{-61}{2} &\frac{87}{2} \\ \\ \frac{47}{2} & \frac{-67}{2} \end{bmatrix}$ .....................................(1)

Now, calculating inverses of A and B.

|A| = 15-14 = 1 and |B| = 54- 56 = -2

$adjA = \begin{bmatrix} 5 &-7 \\ -2 & 3 \end{bmatrix}$ and $adjB = \begin{bmatrix} 9 &-8 \\ -7 & 6 \end{bmatrix}$

therefore we have

$A^{-1} = \frac{1}{|A|}adjA= \frac{1}{1} \begin{bmatrix} 5&-7 \\ -2& 3 \end{bmatrix}$ and $B^{-1} = \frac{1}{|B|}adjB= \frac{1}{-2} \begin{bmatrix} 9&-8 \\ -7& 6 \end{bmatrix}= \begin{bmatrix} \frac{-9}{2} & 4 \\ \\ \frac{7}{2} & -3 \end{bmatrix}$

Now calculating$B^{-1}A^{-1}$.

$B^{-1}A^{-1} =\begin{bmatrix} \frac{-9}{2} & 4 \\ \\ \frac{7}{2} & -3 \end{bmatrix}\begin{bmatrix} 5&-7 \\ -2& 3 \end{bmatrix}$

$=\begin{bmatrix} \frac{-45}{2}-8 && \frac{63}{2}+12 \\ \\ \frac{35}{2}+6 && \frac{-49}{2}-9 \end{bmatrix} = \begin{bmatrix} \frac{-61}{2} && \frac{87}{2} \\ \\ \frac{47}{2} && \frac{-67}{2} \end{bmatrix}$........................(2)

From (1) and (2) we get

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

Hence proved.

Question:13 If $\small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix}$? , show that $A^2-5A+7I=O$. Hence find $\small A^-^1$

Given $\small A=\begin{bmatrix} 3 &1 \\ -1 &2 \end{bmatrix}$ then we have to show the relation $A^2-5A+7I=0$

So, calculating each term;

$A^2 = \begin{bmatrix} 3& 1\\ -1& 2 \end{bmatrix}\begin{bmatrix} 3&1 \\ -1& 2 \end{bmatrix} = \begin{bmatrix} 9-1 &3+2 \\ -3-2&-1+4 \end{bmatrix} = \begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix}$

therefore $A^2-5A+7I$;

$=\begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix} - 5\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix} + 7 \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}$

$=\begin{bmatrix} 8 &5 \\ -5& 3 \end{bmatrix} - \begin{bmatrix} 15 &5 \\ -5& 10 \end{bmatrix} + \begin{bmatrix} 7 &0 \\ 0 & 7 \end{bmatrix}$

$\begin{bmatrix} 8-15+7 &&5-5+0 \\ -5+5+0 && 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 &&0 \\ 0 && 0 \end{bmatrix}$

Hence $A^2-5A+7I = 0$.

$\therefore A.A -5A = -7I$

$\Rightarrow A.A(A^{-1}) - 5AA^{-1} = -7IA^{-1}$

[Post multiplying by $A^{-1}$, also $|A| \neq 0$]

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

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

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

$\therefore A^{-1} = \frac{1}{7}(5\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}-\begin{bmatrix} 3 &1 \\ -1& 2 \end{bmatrix}) = \frac{1}{7}\begin{bmatrix} 2 &-1 \\ 1& 3 \end{bmatrix}$

Question:14 For the matrix $\small A=\begin{bmatrix} 3 &2 \\ 1 & 1 \end{bmatrix}$ , find the numbers $\small a$ and $\small b$ such that $A^2+aA+bI=0$.

Given $\small A=\begin{bmatrix} 3 &2 \\ 1 & 1 \end{bmatrix}$ then we have the relation $A^2+aA+bI=O$

So, calculating each term;

$A^2 = \begin{bmatrix} 3& 2\\ 1& 1 \end{bmatrix}\begin{bmatrix} 3&2 \\ 1& 1 \end{bmatrix} = \begin{bmatrix} 9+2 &6+2 \\ 3+1&2+1 \end{bmatrix} = \begin{bmatrix} 11 &8 \\ 4& 3 \end{bmatrix}$

therefore $A^2+aA+bI=O$;

$=\begin{bmatrix}11 &8 \\ 4& 3 \end{bmatrix} + a\begin{bmatrix} 3 &2 \\ 1& 1 \end{bmatrix} + b \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$

$\begin{bmatrix} 11+3a+b & 8+2a \\ 4+a & 3+a+b \end{bmatrix} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}$

So, we have equations;

$11+3a+b = 0,\ 8+2a = 0$ and $4+a = 0,and\ \ 3+a+b = 0$

We get $a = -4\ and\ b= 1$.

Question:15 For the matrix $\small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}$ Show that $\small A^3-6A^2+5A+11I=O$ Hence, find $\small A^-^1$.

Given matrix: $\small A=\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}$;

To show: $\small A^3-6A^2+5A+11I=O$

Finding each term:

$A^{2} = \begin{bmatrix} 1 & 1& 1\\ 1 & 2& -3\\ 2& -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1& 1\\ 1 & 2& -3\\ 2& -1 & 3 \end{bmatrix}$

$= \begin{bmatrix} 1+1+2 &&1+2-1 &&1-3+3 \\ 1+2-6 &&1+4+3 &&1-6-9 \\ 2-1+6 &&2-2-3 && 2+3+9 \end{bmatrix}$

$= \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}$

$A^{3} = \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}$

$= \begin{bmatrix} 4+2+2 &4+4-1 &4-6+3 \\ -3+8-28 &-3+16+14 & -3-24-42 \\ 7-3+28&7-6-14 &7+9+42 \end{bmatrix}$

$= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}$

So now we have, $\small A^3-6A^2+5A+11I$

$= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}-6\begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}+5\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}+11\begin{bmatrix} 1 &0 &0 \\ 0 &1 & 0\\ 0& 0& 1 \end{bmatrix}$

$= \begin{bmatrix} 8 &7 &1 \\ -23 &27 & -69 \\ 32&-13 &58 \end{bmatrix}-\begin{bmatrix} 24 &&12 &&6 \\ -18 &&48 &&-84 \\ 42 &&-18 && 84 \end{bmatrix}+\begin{bmatrix} 5 &5 &5 \\ 5 &10 &-15 \\ 10 &-5 &15 \end{bmatrix}+\begin{bmatrix} 11 &0 &0 \\ 0 &11 & 0\\ 0& 0& 11 \end{bmatrix}$

$= \begin{bmatrix} 8-24+5+11 &7-12+5 &1-6+5 \\ -23+18+5&27-48+10+11 &-69+84-15 \\ 32-42+10&-13+18-5 & 58-84+15+11 \end{bmatrix}$

$= \begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0&0 & 0 \end{bmatrix} = 0$

Now finding the inverse of A;

Post-multiplying by $A^{-1}$ as, $|A| \neq 0$

$\Rightarrow (AAA)A^{-1}-6(AA)A^{-1} +5AA^{-1}+11IA^{-1} = 0$

$\Rightarrow AA(AA^{-1})-6A(AA^{-1}) +5(AA^{-1})=- 11IA^{-1}$

$\Rightarrow A^{2}-6A +5I=- 11A^{-1}$

$A^{-1} = \frac{-1}{11}(A^{2}-6A+5I)$ ...................(1)

Now,

From equation (1) we get;

$A^{-1} = \frac{-1}{11}( \begin{bmatrix} 4 &&2 &&1 \\ -3 &&8 &&-14 \\ 7 &&-3 && 14 \end{bmatrix}-6\begin{bmatrix} 1 &1 &1 \\ 1 &2 &-3 \\ 2 &-1 &3 \end{bmatrix}+5\begin{bmatrix} 1 & 0& 0\\ 0&1 &0 \\ 0& 0&1 \end{bmatrix})$

$A^{-1} = \frac{-1}{11}( \begin{bmatrix} 4-6+5 &&2-6 &&1-6 \\ -3-6 &&8-12+5 &&-14+18 \\ 7-12 &&-3+6 && 14-18+5 \end{bmatrix}$

$A^{-1} = \frac{-1}{11}( \begin{bmatrix} 3 &&-4 &&-5 \\ -9 &&1 &&4 \\ -5 &&3 && 1 \end{bmatrix}$

Question:16 If $\small A=\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}$ , verify that $\small A^3-6A^2+9A-4I=O$. Hence find $\small A^-^1$.

Given matrix: $\small A=\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}$;

To show: $\small A^3-6A^2+9A-4I$

Finding each term:

$A^{2} = \begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}$

$= \begin{bmatrix} 4+1+1 &&-2-2-1 &&2+1+2 \\ -2-2-1 &&1+4+1 &&-1-2-2 \\ 2+1+2 &&-1-2-2 && 1+1+4 \end{bmatrix}$

$= \begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}$

$A^{3} =\begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}$

$= \begin{bmatrix} 12+5+5 &-6-10-5 &6+5+10 \\ -10-6-5 &5+12+5 & -5-6-10 \\ 10+5+6&-5-10-6 &5+5+12 \end{bmatrix}$

$= \begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}$

So now we have, $\small A^3-6A^2+9A-4I$

$=\begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}-6 \begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}+9\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}-4\begin{bmatrix} 1 &0 &0 \\ 0 &1 & 0\\ 0& 0& 1 \end{bmatrix}$

$=\begin{bmatrix} 22 &-21 &21 \\ -21 &22 & -21 \\ 21&-21 &22 \end{bmatrix}- \begin{bmatrix} 36 &&-30 &&30 \\ -30 &&36 &&-30 \\30 &&-30 && 36 \end{bmatrix}+\begin{bmatrix} 18 &-9 &9 \\ -9 &18 &-9 \\ 9 &-9 &18 \end{bmatrix}-\begin{bmatrix} 4 &0 &0 \\ 0 &4 & 0\\ 0& 0& 4 \end{bmatrix}$

$= \begin{bmatrix} 22-36+18-4 &-21+30-9 &21-30+9 \\ -21+30-9&22-36+18-4 &-21+30-9 \\ 21-30+9&-21+30-9 & 22-36+18-4 \end{bmatrix}$

$= \begin{bmatrix} 0 &0 &0 \\ 0&0 &0 \\ 0&0 & 0 \end{bmatrix} = O$

Now finding the inverse of A;

Post-multiplying by $A^{-1}$ as, $|A| \neq 0$

$\Rightarrow (AAA)A^{-1}-6(AA)A^{-1} +9AA^{-1}-4IA^{-1} = 0$

$\Rightarrow AA(AA^{-1})-6A(AA^{-1}) +9(AA^{-1})=4IA^{-1}$

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

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

Now,

From equation (1) we get;

$A^{-1} = \frac{1}{4}(\begin{bmatrix} 6 &&-5 &&5 \\ -5 &&6 &&-5 \\ 5 &&-5 && 6 \end{bmatrix}-6\begin{bmatrix} 2 &-1 &1 \\ -1 &2 &-1 \\ 1 &-1 &2 \end{bmatrix}+9\begin{bmatrix} 1 & 0& 0\\ 0&1 &0 \\ 0& 0&1 \end{bmatrix})$

$A^{-1} = \frac{1}{4} \begin{bmatrix} 6-12+9 &&-5+6 &&5-6 \\ -5+6 &&6-12+9 &&-5+6 \\ 5-6 &&-5+6 && 6-12+9 \end{bmatrix}$

Hence inverse of A is :

$A^{-1} = \frac{1}{4} \begin{bmatrix} 3 &&1 &&-1 \\ 1 &&3 &&1 \\ -1 &&1 && 3 \end{bmatrix}$

Question:17 Let A be a nonsingular square matrix of order $\small 3\times 3$. Then $\small |adjA|$ is equal to

(A) $\small |A|$ (B) $\small |A|^2$ (C) $\small |A|^3$ (D) $\small 3|A|$

We know the identity $(adjA)A = |A| I$

Hence we can determine the value of $|(adjA)|$.

Taking both sides determinant value we get,

$|(adjA)A| = ||A| I|$ or $|(adjA)||A| = ||A||| I|$

or taking R.H.S.,

$||A||| I| = \begin{vmatrix} |A| & 0&0 \\ 0&|A| &0 \\ 0&0 &|A| \end{vmatrix}$

$= |A| (|A|^2) = |A|^3$

or, we have then $|(adjA)||A| = |A|^3$

Therefore $|(adjA)| = |A|^2$

Hence the correct answer is B.

Question:18 If A is an invertible matrix of order 2, then det $\small (A^-^1)$ is equal to

(A) $\small det(A)$ (B) $\small \frac{1}{det (A)}$ (C) $\small 1$ (D) $\small 0$

Given that the matrix is invertible hence $A^{-1}$ exists and $A^{-1} = \frac{1}{|A|}adjA$

Let us assume a matrix of the order of 2;

$A = \begin{bmatrix} a &b \\ c &d \end{bmatrix}$.

Then $|A| = ad-bc$.

$adjA = \begin{bmatrix} d &-b \\ -c & a \end{bmatrix}$ and $|adjA| = ad-bc$

Now,

$A^{-1} = \frac{1}{|A|}adjA$

Taking determinant both sides;

$|A^{-1}| = |\frac{1}{|A|}adjA| = \begin{bmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{bmatrix}$

$\therefore|A^{-1}| = \begin{vmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix} = \frac{1}{|A|^2}\begin{vmatrix} d &-b \\ -c& a \end{vmatrix} = \frac{1}{|A|^2}(ad-bc) =\frac{1}{|A|^2}.|A| = \frac{1}{|A|}$

Therefore we get;

$|A^{-1}| = \frac{1}{|A|}$

Hence the correct answer is B.

## More About NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5:-

In the NCERT Solutions for Class 12 Maths chapter 4 Exercise 4.5 there are 18 questions including two multiple choice type questions. These questions are related to the most important concept of determinant i.e finding adjoint and inverse of a matrix. There are four examples given in the NCERT book before the exercise 4.5 that you can solve. Solving these examples will help you get conceptual clarity and you will be able to solve exercise questions of NCERT syllabus very easily.

Also Read| Determinants Class 12 Chapter 4 Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5:-

• Class 12 Maths chapter 4 exercise 4.5 solutions are helpful for the students to get conceptual clarity as these solutions are designed in a detailed manner.
• Only knowing the answer is not enough to get good marks in the board exams, one must know how to write in the board exams in order to perform well.
• NCERT solutions for Class 12 Maths chapter 4 exercise 4.4 are prepared by experts who know how to write in board exams, so it will help you to perform better.
• You can take Class 12 Maths ch 4 ex 4.5 solutions for reference.

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