RD Sharma Solutions Class 12 Mathematics Chapter 6 MCQ

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

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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.

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RD Sharma Chapter wise Solutions

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