RD Sharma Solutions Class 12 Mathematics Chapter 6 MCQ

# RD Sharma Solutions Class 12 Mathematics Chapter 6 MCQ

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 02:47 PM IST

RD Sharma is considered as the best material for maths as it contains in detail explanation for questions and concepts. It is widely used by many CBSE schools in the country. Many teachers use this book as a reference for lectures and setting up question papers. This is why it is the best choice for students to prepare for their exams.

Also Read - RD Sharma Solution for Class 9 to 12 Maths

## RD Sharma Class 12 Solutions Chapter 6 MCQ Adjoint & Inverse of Matrix - Other Exercise

Adjoint and inverse of a matrix exercise multiple choice questions question

(c) BA = CA
Given: A is an invertible matrix of order 3
Hints: A inverse is possible only when $\left | A \right |\neq 0$
Solution:
$\left | adj A \right | = \left | A \right |^{2}$ $\left [ \therefore \left | adj A \right | = \left | A \right |^{n-1} \right ]$
∴ Option A is correct
Now option B,
We know that in of invertible matrix
$\left ( A^{-1} \right )^{-1}=A$
∴ Option B is also correct
Now option D
$\left ( AB \right )=B^{-1}A^{-1}$ (∴ A & B two invertible matrices of the same order)
Now, option C
B A = C A
⇒B=C
But it is given B ≠ C. Hence Option C is not true.

Adjoint and inverse of a matrix exercise multiple choice questions question 3

Answer: option (d) none of these
Given: $A=\left[\begin{array}{ll} 3 & 4 \\ 2 & 4 \end{array}\right] \& B=\left[\begin{array}{cc} -2 & -2 \\ 0 & -1 \end{array}\right]$
Hint: add the matrices A and B then find the inverse
Solution:
\begin{aligned} &A+B=\left[\begin{array}{ll} 3 & 4 \\ 2 & 4 \end{array}\right]+\left[\begin{array}{cc} -2 & -2 \\ 0 & -1 \end{array}\right] \\ &=\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right] \\ &|A+B|=3-4=-1 \end{aligned}
Now, $(A+B)^{-1}=\frac{\operatorname{adj}(A+B)}{|A+B|}$

\begin{aligned} &C_{11}=(-1)^{1+1} 3=3 \\ &C_{12}=(-1)^{1+2} 2=-2 \\ &C_{21}=(-1)^{2+1} 2=-2 \\ &C_{22}=(-1)^{2+2} 1=1 \\ &\operatorname{Adj}(A+B)=\left[\begin{array}{cc} 3 & -2 \\ -2 & 1 \end{array}\right]^{T}=\left[\begin{array}{cc} 3 & -2 \\ -2 & 1 \end{array}\right] \end{aligned}
Hence, $(A+B)^{-1}=\frac{\operatorname{adj}(A+B)}{|A+B|}$
\begin{aligned} &=\frac{1}{-1}\left[\begin{array}{cc} 3 & -2 \\ -2 & 1 \end{array}\right] \\ &=-1\left[\begin{array}{cc} 3 & -2 \\ -2 & 1 \end{array}\right] \\ &=\left[\begin{array}{cc} -3 & 2 \\ 2 & -1 \end{array}\right] \end{aligned}

We can clearly see that $(A+B)^{-1}$ exists. So, option (c) is incorrect.

Now we will check the other options.
\begin{aligned} &\mathrm{A}^{-1}=\frac{1}{4}\left[\begin{array}{cc} 4 & -2 \\ -4 & 3 \end{array}\right]^{T}=\frac{1}{4}\left[\begin{array}{cc} 4 & -4 \\ -2 & 3 \end{array}\right] \\ &\mathrm{B}^{-1}=\frac{1}{2}\left[\begin{array}{cc} -1 & 0 \\ 2 & -2 \end{array}\right]^{T}=\frac{1}{2}\left[\begin{array}{cc} -1 & 2 \\ 0 & -2 \end{array}\right] \\ &\mathrm{A}^{-1}+\mathrm{B}^{-1}=\frac{1}{4}\left[\begin{array}{cc} 4 & -2 \\ -4 & 3 \end{array}\right]+\frac{1}{2}\left[\begin{array}{cc} -1 & 2 \\ 0 & -2 \end{array}\right] \\ &=\left[\begin{array}{cc} 1 & -\frac{1}{2} \\ -1 & \frac{3}{4} \end{array}\right]+\left[\begin{array}{rc} -\frac{1}{2} & 0 \\ 1 & -1 \end{array}\right] \\ &=\left[\begin{array}{cc} \frac{1}{2} & \frac{-1}{2} \\ 0 & -\frac{1}{4} \end{array}\right] \\ &\Rightarrow(A+B)^{-1} \neq \mathrm{A}^{-1}+\mathrm{B}^{-1} \end{aligned}

This shows option (b) is incorrect.

It is not skew symmetric matrix and for this we will Show $\left ( \left ( A+B \right )^{-1} \right )^{T}\neq -\left ( A+B \right )^{-1}$
From the above calculation
\begin{aligned} &\text { L.H.S }=\left((A+B)^{-1}\right)^{T}=\left[\begin{array}{cc} -3 & 2 \\ 2 & -1 \end{array}\right]^{T}=\left[\begin{array}{cc} -3 & 2 \\ 2 & -1 \end{array}\right] \\ &\text { R.H.S }=-(A+B)^{-1}=-\left[\begin{array}{cc} -3 & 2 \\ 2 & -1 \end{array}\right]=\left[\begin{array}{cc} 3 & -2 \\ -2 & 1 \end{array}\right] \\ &\text { L.H.S } \neq \text { R.H.S } \end{aligned}

Hence option (d) is correct.

Adjoint and inverse of a matrix exercise multiple choice questions question 4

$\text { (b) }\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]$
Given: $s=\left[\begin{array}{cc} a & b \\ c & d \end{array}\right]$
Hint: find adjoint S by finding cofactors and transposing it.
Solution :
\begin{aligned} &S=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right] \\ &C_{11}=(-1)^{1+1} d=d \\ &C_{12}=(-1)^{1+2} c=-c \\ &C_{21}=(-1)^{2+1} b=-b \\ &C_{22}=(-1)^{2+2} a=a \\ &\text { Adj } S=\left[\begin{array}{cc} d & -c \\ -b & a \end{array}\right]^{T}=\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right] \end{aligned}

Adjoint and inverse of a matrix exercise multiple choice questions question 5

Option (b) Singular
Given: A is a singular matrix
Hint: $\left | A \right |=0$ when matrix is singular
Solution:
We know that
\begin{aligned} &A(a d j A)=|A| \times 1=0 \times 1=0 \\ &\Rightarrow A(\operatorname{adj} A)=0 \end{aligned}
Hence, matrix is singular then adjoint is also singular. So, option b is correct
option (a) AB is non-singular
Given: A & B are non-singular matrix
Hint: A & B are singular matrices only when the$\left | A \right |\& \left | B \right |$ is equal to Zero
Solution:
If A & B are non-singular matrices of order nxn

\begin{aligned} &\therefore|\mathrm{A}| \neq 0 \text { and }|\mathrm{B}| \neq 0 \\ &|\mathrm{AB}|=|\mathrm{A}||\mathrm{B}| \\ &\Rightarrow \therefore|\mathrm{AB}| \neq 0 \end{aligned}
∴ AB is non-singular
Hence option (a) is correct.

Adjoint and inverse of a matrix exercise multiple choice questions question 7

: option (c) a6
Given: $A=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]$
Hint: Find $\left | adj A \right |$ , find adj A
Solution:
\begin{aligned} &A=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]\\ &\operatorname{Adj} A=\left[\begin{array}{ccc} a^{2} & 0 & 0 \\ 0 & a^{2} & 0 \\ 0 & 0 & a^{2} \end{array}\right]^{T}\\ &=\left[\begin{array}{ccc} a^{2} & 0 & 0 \\ 0 & a^{2} & 0 \\ 0 & 0 & a^{2} \end{array}\right]\\ &|\operatorname{adjA}|=a^{2}\left(a^{4}-0\right)=a^{2} \cdot a^{4}=a^{6} \end{aligned}
Hence, option c is correct.

Adjoint and inverse of a matrix exercise multiple choice questions question 8

: option (a) 144
Given: $A= \begin{bmatrix} 1& 2 & -1 \\ -1& 1& 2\\ 2& -1& 1 \end{bmatrix}$
Hint: calculate adjoint of A and put the value in it.
Solution:
\begin{aligned} &A=\left[\begin{array}{ccc} 1 & 2 & -1 \\ -1 & 1 & 2 \\ 2 & -1 & 1 \end{array}\right] \\ &|A|=1(1+2)-2(-1-4)-1(1-2) \\ &=3+10+1=14 \\ &|A| \neq 0 \\ &\Rightarrow \operatorname{adj}(\operatorname{adj} A)=|A|^{(n-1)^{2}} \mid \end{aligned}
Where n=3 because of 3×3 matrices
\begin{aligned} &=|A|^{(3-1)^{2}} \\ &=|A|^{2^{2}} \\ &=|A|^{4} \\ &=14^{4} \quad(\therefore|\mathrm{A}|=14) \\ \end{aligned}
Hence, option (a) is correct.

Adjoint and inverse of a matrix exercise multiple choice questions question 9

option (c) Det (A)
Given: B is non-singular matrix and A is square matrix
Hint: Use the property B-1B=I
Solution:
\begin{aligned} &\operatorname{det}\left(B^{-1} A B\right) \\ &=\operatorname{det}\left(B^{-1} B A\right) \\ &=\operatorname{det}(I A) \quad\left(\text { since }, B^{-1} B=I\right) \\ &=\operatorname{det}(A) \end{aligned}
Hence, option (c) is correct.

Adjoint and inverse of a matrix exercise multiple choice question question 10

option (c) 10
Give: $A\left ( adj A \right )=\begin{bmatrix} 10 &0 \\ 0 & 10 \end{bmatrix}$
Hint: calculating determinant of A i.e; |A|
Solution:
$A\left ( adj A \right )=\begin{bmatrix} 10 &0 \\ 0 & 10 \end{bmatrix}$
$A (adj A) = 10 I$
We know that
$A^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})$
Pre- multiplying by ‘A’
\begin{aligned} &A A^{-1}=\frac{A(\operatorname{adjA})}{|A|} \\ &I=\frac{A(\operatorname{adjA})}{|A|} \\ &A(\operatorname{adj} A)=|A|I \\ &\Rightarrow|A|=10 \end{aligned}
Hence option (c) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 11

option (d) none of these
Given: $A^{5}=0 \text { such that } A^{n} \neq \mathrm{I} \text { for } 1 \leq n \leq 4$
Hint: zero matrix is that whose every element is zero
Solution:
\begin{aligned} &\mathrm{I}-\mathrm{A}^{5}=(\mathrm{I}-\mathrm{A})\left(\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\mathrm{A}^{3}+\mathrm{A}^{4}\right) \\ &\mathrm{I}=(\mathrm{I}-\mathrm{A})\left(\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\mathrm{A}^{3}+\mathrm{A}^{4}\right) \quad\left(\therefore \mathrm{A}^{5}=0\right) \\ &\frac{I}{(I-A)}=\left(\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\mathrm{A}^{3}+\mathrm{A}^{4}\right) \\ &(\mathrm{I}-\mathrm{A})^{-1}=\left(\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\mathrm{A}^{3}+\mathrm{A}^{4}\right) \end{aligned}
Hence, option (d) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 12

option (d)$\lambda \neq 0$
Given: A satisfies the equation $x^{3}-5x^{2}+4x+ \lambda =0$
Hint: since A satisfies the equation so put x = A
Solution :
Since A satisfies the equation ∴ $A^{3}-5A^{2}+4A+ \lambda =0$
A-1 Exist if $\lambda \neq 0$
since if $\lambda =0$ then the above equation gives A = 0 and in that case A-1 will not exist
Hence option (d) is correct.

Adjoint and inverse of a matrix exercise multiple choice question question 13

: option (a) A2
Given: $A^{3}=I$
Hint: to find A-1, multiple both side by A-1
Solutions:
$A^{3}=I$
Post multiplying by A-1 both sides
\begin{aligned} &A^{3} \cdot A^{-1}=\mathrm{I} \cdot A^{-1} \\ &A^{2} \cdot A \cdot A^{-1}=\mathrm{I} \cdot A^{-1} \\ &A^{2} \cdot \mathrm{I}=A^{-1} \quad\left(\therefore A A^{-1}=\mathrm{I}\right) \\ &\Rightarrow A^{-1}=A^{2} \end{aligned}
Hence, option (a) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 14

option (b) $A^{2}+B^{2}$
Given: A & B are square matrices and $B = - A^{-1} BA$
Hint: to solve the question you must know the formula of $\left (A+B \right )^{2}$
Solution:
$B = - A^{-1} BA$
Multiply both sides with A
\begin{aligned} &A B=A\left(-A^{-1} B A\right) \\ &A B=-\left(A A^{-1}\right) B A \\ &A B=-B A \quad\left(\therefore A A^{-1}=I\right) \\ &\text { Now, }(A+B)^{2}=(A+B)(A+B) \\ &=A^{2}+A B+B A+B^{2} \\ &=A^{2}-B A+B A+B^{2} \quad(\text { since } A B=-\text { BA }) \\ &(A+B)^{2}=A^{2}+B^{2} \end{aligned}
Hence, option b is correct

Adjoint and inverse of a matrix exercise multiple choice question question15

option (c) 16A
Given: $A=\begin{bmatrix} 2 &0 &0 \\ 0 & 2& 0\\ 0&0 &2 \end{bmatrix}$
Hint: you must know the multiplication concept of matrices
Solution:

\begin{aligned} &\mathrm{A}^{2}=\mathrm{A} \cdot \mathrm{A}\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right] \\ &=\left[\begin{array}{lll} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right] \\ &A^{5}= A^{2} \cdot A^{2} \cdot A \\ &A^{5}=\left[\begin{array}{lll} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right]\left[\begin{array}{lll} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right] \text { A } \end{aligned}
\begin{aligned} &=\left[\begin{array}{ccc} 16 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 16 \end{array}\right] \mathrm{A} \\ &=16\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \text { A } \\ &= 16 \mathrm{I} \mathrm{A} \\ &\Rightarrow \mathrm{A}^{5}=16 \mathrm{~A} \end{aligned}

Hence, option (c) is correct
option (d) $C^{-1}BA^{-1}$
Given: A, B & C are non-singular square matrices of same orders
Hint: to find the answer you must know the properties of inverse
Solution: $\left (AB^{-1}C \right )^{-1}$
\begin{aligned} &=C^{-1}\left(B^{-1}\right)^{-1} A^{-1} \\ &=C^{-1} B A^{-1} \\ &\Rightarrow\left(A B^{-1} C\right)^{-1}=C^{-1} B A^{-1} \end{aligned}
Hence, option (d) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 17

option (d) non-existent
Given: $\begin{bmatrix} 5 &10 &3 \\ -2&-4 &6 \\ -1& -2 &b \end{bmatrix}$ is a singular matrices
Hint: for any singular matrix, the value of the determinant is 0
Solution:
\begin{aligned} &A=\left[\begin{array}{ccc} 5 & 10 & 3 \\ -2 & -4 & 6 \\ -1 & -2 & b \end{array}\right] \\ &|A|=5(-4 b+12)-10(-2 b+6)+3(4-4) \\ &=-20 b+60+20 b-60 \\ &=0 \end{aligned}
Hence, b is non-exist

∴ option (d) is correct
option (b) $d^{n-1}$
Given: d is the determinant of a square matrix A of order n
Hint: you must know the formula of $|\operatorname{adj} A|$
Solution:
\begin{aligned} &|\operatorname{adj} A|=|A|^{n-1} \\ &|\operatorname{adj} A|=d^{n-1} \quad(\therefore|A|=d) \end{aligned}

Adjoint and inverse of a matrix exercise multiple choice question question 19

option (d)$2^{6}$
Given:$n=3 \& |A|=8$
Hint: to find the answer, put the given value in the formula $|adj A|$
Solution:\begin{aligned} &|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{n-1} \\ &|\operatorname{adj} \mathrm{A} \mid=(8)^{3-1} \quad(\therefore n=3 \&|A|=8) \\ &=8^{2} \\ &=\left(2^{3}\right)^{2} \\ & |adj A|=2^{6} \end{aligned}

Hence, option (d) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 20

option (c) $I-A$
Given: $A^{2}-A+I=0$
Hint: use multiplication formula to find the answer
Solution: $A^{2}-A+I=0$
$I=A-A^{2}$
$I=A-A.A$
Post – multiply both sides by A-1
\begin{aligned} &\mathrm{I} \mathrm{A}^{-1}=(\mathrm{A}-\mathrm{A} \cdot \mathrm{A}) \mathrm{A}^{-1} \\ &\mathrm{I} \mathrm{A}^{-1}=\mathrm{AA}^{-1}-\mathrm{A} \cdot\left(\mathrm{AA}^{-1}\right) \\ &\mathrm{A}^{-1}=\mathrm{I}-\mathrm{A} \end{aligned}

Hence , option (c) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 21

option (c) $(A+B)^{-1} = A^{-1}+ B^{-1}$
Given: A & B are invertible matrices
Hint: correct option is that whose right hand side is not equal to left hand side.
Solution :
We know that $A^{-1}=\frac{adj A}{|A|}$
$\Rightarrow adj.A=|A|A^{-1}$
Option (a) is correct
\begin{aligned} &\text { Also A. } \mathrm{A}^{-1}=\mathrm{I} \\ &\left|\mathrm{AA}^{-1}\right|=|\mathrm{I}| \\ &\Rightarrow|\mathrm{A}|\left|\mathrm{A}^{-1}\right|=\mathrm{I} \\ &\Rightarrow\left|\mathrm{A}^{-1}\right|=\frac{\mathrm{I}}{|\mathrm{A}|} \\ &\Rightarrow \operatorname{det}\left(\mathrm{A}^{-1}\right)=[\operatorname{det}(\mathrm{A})]^{-1} \end{aligned}
Option (b) is also correct

\begin{aligned} &\text { Now, }(A+B)^{-1}=\frac{\operatorname{adj}(A+B)}{|A+B|} \\ &\& A^{-1}+B^{-1}=\frac{a d j A}{|A|}+\frac{a d j B}{|B|} \\ &\Rightarrow(A+B)^{-1} \neq A^{-1}+B^{-1} \end{aligned}

Adjoint and inverse of a matrix exercise multiple choice question question 22

option (b) A
Given: A is a square matrix such that $A^{2}=I$
Hint: you must know about the concept of square matrix
Solution: $A^{2}=I$
$\Rightarrow A.A=I$
Multiply both sides by $A^{-1}$
\begin{aligned} &\text { A. }\left(\mathrm{AA}^{-1}\right)=\mathrm{I} \mathrm{A}^{-1}\\ &\mathrm{A} . \mathrm{I}=\mathrm{A}^{-1}\\ &\Rightarrow \mathrm{A}^{-1}=\mathrm{A} \end{aligned}
Hence, option (b) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 23

option (a) $\frac{1}{2}\begin{bmatrix} 2 &4 \\ 3&-5 \end{bmatrix}$
Given: $A=\begin{bmatrix} 1 &2 \\ 3&-5 \end{bmatrix} \& A=\begin{bmatrix} 1 &0 \\ 0&2 \end{bmatrix}$ and
X be a matrix such that A = BX
Hint: put A & B in equation and square to find matrix X.
Solution: – A= BX
\begin{aligned} &\Rightarrow \mathrm{B}^{-1} \mathrm{~A}=\mathrm{X} \\ &\mathrm{B}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 2 \end{array}\right] \\ &|B|=2 \\ &B^{-1}=\frac{\mathrm{adj} \mathrm{B}}{|\mathrm{B}|} \\ &=\frac{1}{2}\left[\begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right] \\ &\text { Now, } \mathrm{B}^{-1} \mathrm{~A}=\mathrm{X} \\ &\frac{1}{2}\left[\begin{array}{cc} 2 & 0 \\ 0 & 1 \end{array}\right]\left[\begin{array}{cc} 1 & 2 \\ 3 & -5 \end{array}\right]=\mathrm{X} \\ &\frac{1}{2}\left[\begin{array}{cc} 2 & 4 \\ 3 & -5 \end{array}\right]=\mathrm{X} \end{aligned}
Hence, option (a) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 24

option (b) $\frac{1}{19}$
Given: $A=\begin{bmatrix} 2 &3 \\ 5& -2 \end{bmatrix} \& A^{-1}=KA$
Hint: simply multiply the matrices
\begin{aligned} &\text { Solution: } \mathrm{A}=\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]\\ &\therefore|\mathrm{A}|=-4-15=-19\\ &\text { So } \mathrm{A}^{-1} \text { exists }\\ &\text { We knows= that } A^{-1} \mathrm{~A}=\mathrm{I}\\ &\Rightarrow \mathrm{KA} \cdot \mathrm{A}=\mathrm{I} \quad\left(\therefore \mathrm{A}^{-1}=\mathrm{KA}\right)\\ &\Rightarrow \mathrm{K}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ &\Rightarrow \mathrm{K}\left[\begin{array}{cc} 19 & 0 \\ 0 & 19 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ &\Rightarrow \mathrm{K}(19 \mathrm{I})=\mathrm{I}\\ &\Rightarrow 19 \mathrm{~K}=1\\ &\Rightarrow \mathrm{K}=\frac{1}{19} \end{aligned}
Hence, option (b) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 26

option (d) none of these
Given: $A=\begin{bmatrix} 1 &0 &1 \\ 0& 0& 1\\ a&b &2 \end{bmatrix}$
Hint: identity matrix is that whose diagonal elements is 1 and rest and zero
Solution:
\begin{aligned} &A^{2}=\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 1 \\ 0 & 0 & 1 \\ a & b & 2 \end{array}\right] \\ &=\left[\begin{array}{ccc} 1+a & b & 3 \\ a & b & 2 \\ 3 a & 2 b & a+b+4 \end{array}\right] \\ &\text { Now, a I }+b A+2 A^{2} \\ &=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right]+\left[\begin{array}{ccc} b & 0 & b \\ 0 & 0 & b \\ a b & b^{2} & 2 b \end{array}\right]+\left[\begin{array}{ccc} 2+2 a & 2 b & 6 \\ 2 a & 2 b & 4 \\ 6 a & 4 b & 2 a+2 b+8 \end{array}\right] \\ &=\left[\begin{array}{ccc} 3 a+b+2 & 2 b & b+6 \\ 2 a & a+2 b & b+4 \\ a b+6 a & b^{2}+4 b & 3 a+4 b+8 \end{array}\right] \end{aligned}
Now, value of a & b is not given so it is not solved further
Hence, option (d) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 27

$\text { option (b) } A=\cos 2 \theta \& b=\sin 2 \theta$
$\text { Given: }\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array}\right]^{-1}=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]$
Hint: use simple matrix multiplication.
Solution :
\begin{aligned} &\text { Inverse of }\left[\begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array}\right]^{-1}=\frac{1}{1+\tan ^{2} \theta}\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right]\\ &\text { From the given part }\\ &\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right] \frac{1}{1+\tan ^{2} \theta}\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right]=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]\\ &\frac{1}{1+\tan ^{2} \theta}\left[\begin{array}{cc} 1-\tan ^{2} \theta & -2 \tan \theta \\ 2 \tan \theta & 1-\tan ^{2} \theta \end{array}\right]=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right] \end{aligned}
\begin{aligned} &\left[\begin{array}{cc} \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} & \frac{-2 \tan \theta}{1+\tan ^{2} \theta} \\ \frac{2 \tan \theta}{1+\tan ^{2} \theta} & \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} \end{array}\right]=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]\\ &\left[\begin{array}{cc} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array}\right]=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]\\ &\text { On comparing, we get }\\ &A=\cos 2 \theta \ \& \ b=\sin 2 \theta \end{aligned}
Hence, option (b) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 28

option (d) none of these
Given: $3 A^{3}+2A^{2}+5A+I=0$
Hint: use simple calculation to get the result.
\begin{aligned} &\text { Solution: } 3 A^{3}+2 A^{2}+5 A=-I \\ &A\left(3 A^{2}+2 A+5 I\right)=-I \\ &3 A^{2}+2 A+5 I=-A^{-1}\left(\text { multiply } A^{-1}\right. \text { on both sides) } \\ &\Rightarrow A^{-1}=-\left(3 A^{2}+2 A+5 I\right) \end{aligned}
Hence, option (d) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 29

option (b) $\frac{1}{det(A)}$
Given – A is an invertible matrix
Hint – you must know about invertible matrix concept
Solution –
We know that $AA^{-1}=I$
\begin{aligned} &\Rightarrow\left|\mathrm{AA}^{-1}\right|=|\mathrm{I}| \\ &\Rightarrow|\mathrm{A}|\left|\mathrm{A}^{-1}\right|=1 \\ &\Rightarrow\left|A^{-1}\right|=|A|^{-1} \end{aligned}
Hence, is option (b) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 30

option (a) $\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$,if n is an even natural number.
Given $-A=\begin{bmatrix} 2 &-1 \\ 3& -2 \end{bmatrix}$
Hint – find the value of A2
Solution$-A=\begin{bmatrix} 2 &-1 \\ 3& -2 \end{bmatrix}$
\begin{aligned} &A^{2}=\left[\begin{array}{ll} 2 & -1 \\ 3 & -2 \end{array}\right]\left[\begin{array}{ll} 2 & -1 \\ 3 & -2 \end{array}\right] \\ &A^{2}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &A^{2}=I \\ &\Rightarrow A^{-1}=A \end{aligned}

Therefore, when n is odd $A^{n}=A^{-1}$ and when n is even $A^{n}=I$

Hence, option (a) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 31

option (a) $\begin{bmatrix} x^{-1} &0 &0 \\ 0& y^{-1} & 0\\ 0&0 &z^{-1} \end{bmatrix}$
Given $-A=\begin{bmatrix} x &0 &0 \\ 0& y & 0\\ 0&0 &z \end{bmatrix}$
Hint – Find the inverse of matrix A using the formula: $A^{-1}=\frac{adj A}{\left | A \right |}$
Solution-
\begin{aligned} &A=\left[\begin{array}{lll} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{array}\right] \\ &|A|=x(y z-0)-0+0=x y z \\ &\operatorname{adj} A=\left[\begin{array}{ccc} y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y \end{array}\right] \\ &A^{-1}=\frac{a d j A}{|A|} \\ &=\frac{1}{x y z}\left[\begin{array}{ccc} y z & 0 & 0 \\ 0 & x z & 0 \\ 0 & 0 & x y \end{array}\right] \end{aligned}
\begin{aligned} &=\left[\begin{array}{ccc} 1 / x & 0 & 0 \\ 0 & 1 / y & 0 \\ 0 & 0 & 1 / z \end{array}\right] \\ &\mathrm{A}^{-1}=\left[\begin{array}{ccc} x^{-1} & 0 & 0 \\ 0 & y^{-1} & 0 \\ 0 & 0 & \mathrm{z}^{-1} \end{array}\right] \end{aligned}
Hence, option (a) is correct.

Adjoint and inverse of a matrix exercise multiple choice question question 32

option (d)$(A+B)^{-1}=B^{-1}+A^{-1}$
Given – A & B are invertible matrices.
Hint – answers is that whose left have side is not equal to right hand side.
\begin{aligned} &\text { Solution }-\text { adj } A=|A| \cdot A^{-1} \quad\left(\therefore A^{-1}=\frac{a d j A}{|A|}\right) \\ &\text { Also } A \cdot A^{-1}=I \\ &\Rightarrow|A|\left|A^{-1}\right|=|I| \\ &\Rightarrow\left|A^{-1}\right|=\frac{I}{|A|} \\ &\Rightarrow \operatorname{det}\left|A^{-1}\right|=[\operatorname{det}|A|]^{-1} \\ &\text { Now }(A+B)^{-1}=\frac{\operatorname{adj}(A+B)}{|A+B|} \\ &\& B^{-1}+A^{-1}==\frac{a d j B}{|B|}+\frac{a d j A}{|A|} \\ &\Rightarrow(A+B)^{-1} \neq B^{-1}+A^{-1} \end{aligned}

Hence option (d) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 33

option (d) none of these
Given $-A=\begin{bmatrix} 2 & \lambda &-3 \\ 0&2 &5 \\ 1&1 &3 \end{bmatrix}$
Hint- find the value of |A|
Solution –
\begin{aligned} &A=\left[\begin{array}{ccc} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{array}\right] \\ &|A|=2(6-5)-\lambda(0-5)-3(0-2) \\ &|A|=2+5 \lambda+6 \\ &A^{-1} \text { exist if }|A| \neq 0 \\ &\Rightarrow 8+5 \lambda \neq 0 \\ &\Rightarrow \lambda \neq \frac{-8}{5} \end{aligned}
Hence, option (d) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 34

option (a) $I$
Given – A is a square matrix &$A^{2}=A$
Hint –Use the concept of square of identity matrix $I^{2}=I$
\begin{aligned} &\text { Solutions }-(\mathrm{I}-\mathrm{A})^{3}+\mathrm{A} \\ &=(\mathrm{I}-\mathrm{A})^{2}(\mathrm{I}-\mathrm{A})+\mathrm{A} \\ &=\left(\mathrm{I}^{2}-2 \mathrm{AI}+\mathrm{A}^{2}\right)(\mathrm{I}-\mathrm{A})+\mathrm{A} \\ &=(\mathrm{I}-2 \mathrm{~A}+\mathrm{A})(\mathrm{I}-\mathrm{A})+\mathrm{A} \quad \quad\left(\therefore \mathrm{A}^{2}=\mathrm{A} \text { and } \mathrm{I}^{2}=\mathrm{I}\right) \end{aligned}
\begin{aligned} &=(I-A)(I-A)+A \\ &=\left(I^{2}-2 A I+A^{2}\right)+A \\ &=(I-2 A+A)+A \quad \quad\left(\therefore A^{2}=A \text { and } I^{2}=I\right) \\ &=I-A+A=I \\ &\Rightarrow(I-A)^{3}+A=I \end{aligned}
Hence, option (a) is correct

Adjoint and inverse of a matrix exercise multiple choice question question 35

option (c) $\lambda =1$
Given - $\begin{bmatrix} 2 & -1 &3 \\ \lambda &0 &7 \\ -1& 1 &4 \end{bmatrix}$
Hint – Find the determinant of A i.e, $\left | A \right |$
Solution – since matrix is not invertible\begin{aligned} &\therefore|\mathrm{A}|=0 \\ &\mathrm{~A}=\left[\begin{array}{ccc} 2 & -1 & 3 \\ \lambda & 0 & 7 \\ -1 & 1 & 4 \end{array}\right] \\ &|\mathrm{A}|=2(0-7)+1(4 \lambda+7)+3(\lambda+0) \\ &0=-14+4 \lambda+7+3 \lambda \quad(\therefore|\mathrm{A}|=0) \\ &0=-7+7 \boldsymbol{\lambda} \\ &7=7 \boldsymbol{\lambda} \\ &\Rightarrow \lambda=1 \end{aligned}

MCQs to get a good understanding of all the concepts in this chapter.

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## RD Sharma Chapter wise Solutions

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4. What is the inverse of a Matrix?

A Matrix, when multiplied by its inverse, should result in an identity Matrix. That is called the inverse value of a Matrix. For more info, check RD Sharma Class 12th Chapter 6 MCQ.

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