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

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

Edited By Ramraj Saini | Updated on Dec 03, 2023 04:11 PM IST | #CBSE Class 12th
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NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.5

NCERT Solutions for Exercise 4.5 Class 12 Maths Chapter 4 Determinants are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. 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. 12th class Maths exercise 4.5 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Determinants Exercise: 4.5

Question:1 Find adjoint of each of the matrices.

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

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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}.

Answer:

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

Then we have adj(AB):

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 calculatingB^{-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

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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|

Answer:

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

Answer:

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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Key Features Of NCERT Solutions for Exercise 4.5 Class 12 Maths Chapter 4

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 4.5 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 4.5, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 4.5 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 4.5 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 4.5 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 4.5 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. Does singular matrices are invertible ?

No, singular matrices are not invertible.

2. Does non-singular matrices are invertible ?

Yes, non-singular matrices are invertible.

3. what is an invertible matrix ?

If an inverse of a square matrix exists then it is called an invertible matrix.

4. If |A| = 5 and order of A is 2 then |3A| = ?​

|3A| = 3^2|A| = 45

5. If A is a symmetric matrix then the transpose of A is?

If A is a symmetric matrix then the transpose of A is A.

6. If A is a skew-symmetric matrix then the transpose of A is?

If A is a skew-symmetric matrix then the transpose of A is -A.

7. If A is a matrix and A' is the transpose of matrix A then what is |A|?

If A is a matrix and A' is the transpose of matrix A then |A| = |A'|.

8. Does square diagonal marix is a symmetric matrix ?

Yes, every square diagonal matrix is a symmetric matrix.

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Can I change my board from CBSE to CHSE Odisha in Class 12th?

Hello,

Yes, you can switch from CBSE to CHSE Odisha in Class 12th, but there are a few conditions:

  1. Eligibility : Check if CHSE Odisha allows lateral entry into Class 12th. Generally, students can transfer from CBSE to state boards with prior approval.
  2. Document Requirements : You may need to provide transcripts, mark sheets, and a transfer certificate.
  3. Subject Alignment : Ensure that the subjects you studied in CBSE align with the CHSE curriculum.
  4. Application Process : Contact the respective CHSE Odisha authority or school for specific transfer procedures and deadlines.

Hope it helps !

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Prachi Kumari 30 December,2024
quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

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Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
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Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

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Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024
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