RD Sharma Solutions Class 12 Mathematics Chapter 6 MCQ

Schools
RD Sharma Solutions
RD Sharma Solutions Class 12 Mathematics Chapter 6 MCQ

RD Sharma Solutions Class 12 Mathematics Chapter 6 MCQ

Kuldeep MauryaUpdated on 20 Jan 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

Answer:

(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:

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

Answer:

$\text { (b) }\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]$
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

Answer:

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

Adjoint and inverse of a matrix exercise multiple choice questions question 6
Answer:

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

Answer:

: 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

Answer:

: 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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

: 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

Answer:

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

Answer:

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

Adjoint and inverse of a matrix exercise multiple choice question question 16
Answer:

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

Answer:

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

Adjoint and inverse of a matrix exercise multiple choice question question 18
Answer:

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

Answer:

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

Answer:

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

Answer:

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

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

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

$\text { option (b) } A=\cos 2 \theta \& b=\sin 2 \theta$
$\text { Given: }\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array}\right]^{-1}=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]$
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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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

Answer:

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.

Career360 has prepared RD Sharma Class 12th Chapter 6 MCQ specifically for students to save time and prepare for their exams. As exercise problems can be time consuming, MCQs are basically a distraction from those problems. In order to complete these quickly students can refer to this material.

As the board exams have 20 marks reserved for MCQs it is essential that students get each and everyone right. RD Sharma Class 12th Chapter 6 MCQ solutions is designed to help students prepare well for the MCQs. Solving all the MCQs is a tiring task and will use up a lot of time.

A majority of students spend a lot of time-solving MCQs and then panic when there’s no time left for main questions. RD Sharma Class 12th Chapter 6 MCQ material helps to get a clear understanding with the help of solutions and can help students divide the work to study systematically. If students stay in line with their class lectures and practice this material, there is nothing else they will need to score well in exams. As RD Sharma books are widely used in CBSE schools, it is possible that questions from this might appear in the exams.

RD Sharma books come up with a new version occasionally, it gets hard for students to keep track of the latest version materials. This means that as soon as the new version is released, the older materials become less helpful. RD Sharma Class 12 Chapter 6 MCQ by Career360 is updated to the latest version which means that students won’t have to worry about missing or different questions.

RD Sharma Class 12 Solutions Chapter 6 MCQ material is prepared by experts who have years of experience and insight on exam patterns. These solutions are meant to provide students with the best source of preparation and get them a unified material for all their chapters.

The best part about it is that it is free to download and available on Career360’s website. Thousands of students have already joined the team and started preparing for their exams using this material. If students didn’t know about this, they should definitely give it a try.