RD Sharma Solutions Class 12 Mathematics Chapter 4 MCQ
MCQs are a great way to test students' knowledge of an entire chapter by combining questions from all sections. This is why RD Sharma class 12th exercise MCQ can be highly beneficial to students preparing for board exams. The RD Sharma class 12 chapter 4 exercise MCQ is based on the chapter Algebra of Matrices, which covers addition, multiplication, division, and subtraction of Matrices. You will also learn about skew-symmetric matrices, null matrices, symmetric matrices, and so on. RD Sharma class 12th exercise MCQ part of the textbook will have 46 questions that you will answer.
Also Read - RD Sharma Solution for Class 9 to 12 Maths
RD Sharma Class 12 Solutions Chapter 4 MCQ Algebra of Matrices-Other Exercise
- Chapter 4 - Algebra of Matrices Ex 4.1
- Chapter 4 - Algebra of Matrices Ex 4.2
- Chapter 4 - Algebra of Matrices Ex 4.3
- Chapter 4 - Algebra of Matrices Ex 4.4
- Chapter 4 - Algebra of Matrices Ex 4.5
- Chapter 4 - Algebra Matrices Ex FBQ
- Chapter 4 - Algebra Matrices Ex VSA
Algebra of Matrices Excercise: MCQ
Answer: The option (B) is correct. As the main diagonal elements are 1 except others which are 0, so it is a Unit Matrix.Given: $A=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{array}\right]$
Solution:
$\begin{aligned} A^{2} &=A \times A \\ &=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{array}\right] \times\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{array}\right] \\ &=\left[\begin{array}{lll} 1+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+1+0 & 0+0+0 \\ a+0-a & 0+b-b & 0+0+1 \end{array}\right] \\ &=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}$
Answer: The correct option is $\text { (C), } A^{4}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
Given:$A=\left[\begin{array}{ll} i & 0 \\ 0 & i \end{array}\right]$
Solution:
Now,$A^{2}=A \times A$
$\begin{aligned} &=\left[\begin{array}{ll} i & 0 \\ 0 & i \end{array}\right] \times\left[\begin{array}{cc} i & 0 \\ 0 & i \end{array}\right] \\ &\quad=\left[\begin{array}{cc} i^{2} & 0 \\ 0 & i^{2} \end{array}\right] \\ &\quad=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] \\ &A^{3}=A^{2} \times A \\ &=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] \times\left[\begin{array}{cc} i & 0 \\ 0 & i \end{array}\right] \\ &=\left[\begin{array}{cc} -i & 0 \\ 0 & -i \end{array}\right] \\ &A^{4}=A^{3} \times A \\ &=\left[\begin{array}{cc} -i & 0 \\ 0 & -i \end{array}\right] \times\left[\begin{array}{cc} i & 0 \\ 0 & i \end{array}\right] \\ &=\left[\begin{array}{cc} -i^{2} & 0 \\ 0 & -i^{2} \end{array}\right] \\ &=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}$
So,$A^{4}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
Given:$A=\left[\begin{array}{ll} i & 0 \\ 0 & i \end{array}\right]$
Solution:
Now,$A^{2}=A \times A$
$\begin{aligned} &=\left[\begin{array}{ll} i & 0 \\ 0 & i \end{array}\right] \times\left[\begin{array}{cc} i & 0 \\ 0 & i \end{array}\right] \\ &\quad=\left[\begin{array}{cc} i^{2} & 0 \\ 0 & i^{2} \end{array}\right] \\ &\quad=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] \\ &A^{3}=A^{2} \times A \\ &=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] \times\left[\begin{array}{cc} i & 0 \\ 0 & i \end{array}\right] \\ &=\left[\begin{array}{cc} -i & 0 \\ 0 & -i \end{array}\right] \\ &A^{4}=A^{3} \times A \\ &=\left[\begin{array}{cc} -i & 0 \\ 0 & -i \end{array}\right] \times\left[\begin{array}{cc} i & 0 \\ 0 & i \end{array}\right] \\ &=\left[\begin{array}{cc} -i^{2} & 0 \\ 0 & -i^{2} \end{array}\right] \\ &=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}$
So,$A^{4}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
Answer: The correct option is $\text { (A), } B^{2}=B$
Given:$AB=A,BA=B$
Solution:
It is Given that,
$\mathrm{AB}=\mathrm{A}$ ....(1)
$\mathrm{BA}=\mathrm{B}$ ....(2)
Now from Equation (2),$\begin{aligned} &B \times(A B)=B \\ &B^{2} A=B \end{aligned}$
Again from Equation (2),
$\begin{aligned} &\mathrm{B}^{2} \mathrm{~A}=\mathrm{BA} \\ &\mathrm{B}^{2}=\mathrm{B} \end{aligned}$
So, the option (B) is correct.
4 Exercise MCQs Question 4
Answer: The correct option is $\text { (B), } \mathrm{B}^{2}=\mathrm{B} \text { and } \mathrm{A}^{2}=\mathrm{A}$Given:$A B=A, B A=B$
Solution:
$A B=A$ ....(1)
$BA=B$ ....(2)
Now from Equation (2),
$\begin{aligned} &B \times(A B)=B \\ &B^{2} A=B \end{aligned}$
Again from Equation (2),
$\begin{aligned} &\mathrm{B}^{2} \mathrm{~A}=\mathrm{BA} \\ &\mathrm{B}^{2}=\mathrm{B} \end{aligned}$
Now from Equation (1),
$\begin{aligned} &A \times(B A)=A \\ &A^{2} B=A \end{aligned}$
Again from Equation (1),
$\begin{aligned} &\mathrm{A}^{2} \mathrm{~B}=\mathrm{AB} \\ &\mathrm{A}^{2}=\mathrm{A} \end{aligned}$
So,$\mathrm{A}^{2}=\mathrm{A} \text { and } \mathrm{B}^{2}=\mathrm{B}$
Hence, the correct option is (B).
4 Algebra of Matrices Exercise MCQs Question 5
Answer: The correct option is $\text { (C), } A^{2}+B^{2}=A+B$Given:$A B=A, B A=B$
Solution:
$\mathrm{AB}=\mathrm{A}$ .....(1)
$\mathrm{BA}=\mathrm{B}$ .....(2)
Now from Equation (2),
$\begin{aligned} &B \times(A B)=B \\ &\mathrm{~B}^{2} \mathrm{~A}=\mathrm{B} \end{aligned}$
Again from Equation (2),
$\mathrm{B}^{2} \mathrm{~A}=\mathrm{BA}$
$\mathrm{B}^{2}=\mathrm{B}$
Now from Equation (1),
$\begin{aligned} &A \times(B A)=\mathrm{A} \\ &\mathrm{A}^{2} \mathrm{~B}=\mathrm{A} \end{aligned}$
Again from Equation (1),
$\begin{aligned} &A^{2} B=A B \\ &A^{2}=A \end{aligned}$
So,$\mathrm{A}^{2}=\mathrm{A} \text { and } \mathrm{B}^{2}=\mathrm{B}$
So, the option (C) is the correct option.
4 Algebra of Matrices Exercise MCQs Question 6
Answer: The correct option is (D), k = 7Given: $\left[\begin{array}{cc} \cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7} \end{array}\right]^{\hat{k}}=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
Solution:
Let $A=\left[\begin{array}{cc} \cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\ \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7} \end{array}\right]$
Then $\begin{aligned} |\mathrm{A}| &=\left(\cos \frac{2 \pi}{7}\right)\left(\cos \frac{2 \pi}{7}\right)-\left(\sin \frac{2 \pi}{7}\right)\left(-\sin \frac{2 \pi}{7}\right) \\ &=\left(\cos ^{2} \frac{2 \pi}{7}\right)+\left(\sin ^{2} \frac{2 \pi}{7}\right) \\ &=1 \end{aligned}$
$I=1$
$I^{k}=1$ {k cab be anything}
$\text { Let } \theta=\frac{2 \pi}{7}$
$\begin{aligned} \mathrm{A}^{2} &=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \times\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \\ &=\left[\begin{array}{cc} \cos ^{2} \theta-\sin ^{2} \theta & -\sin \theta \cos \theta-\sin \theta \cos \theta \\ \sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^{2} \theta-\sin ^{2} \theta \end{array}\right] \\ &=\left[\begin{array}{cc} \cos ^{2} \theta-\sin ^{2} \theta & -2 \sin \theta \cos \theta \\ 2 \sin \theta \cos \theta & \cos ^{2} \theta-\sin ^{2} \theta \end{array}\right] \\ &=\left[\begin{array}{cc} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array}\right] \quad\left[\because \cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta, 2 \sin \theta \cos \theta=\sin 2 \theta\right] \end{aligned}$
Now,$\begin{aligned} \mathrm{A}^{4} &=\left[\begin{array}{cc} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array}\right] \times\left[\begin{array}{cc} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{array}\right] \\ &=\left[\begin{array}{cc} \cos 4 \theta & -\sin 4 \theta \\ \sin 4 \theta & \cos 4 \theta \end{array}\right] \end{aligned}$
Similarly,$A^{7}=\left[\begin{array}{cc} \cos 7 \theta & -\sin 7 \theta \\ \sin 7 \theta & \cos 7 \theta \end{array}\right]$
Hence, $\theta=\frac{2 \pi}{7}$
Now, multiplying $\cos \Theta$and $\sin \Theta$ to LHS and RHS,
Then,
$\begin{aligned} &\cos 7 \theta=\cos 2 \pi=1 \\ &\sin 7 \theta=\sin 2 \pi=0 \\ &{\left[\begin{array}{cc} \cos 7 \theta & -\sin 7 \theta \\ \sin 7 \theta & \cos 7 \theta \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\ &\mathrm{A}^{7}=1 \end{aligned}$
Hence,k=7
So, the option (D) is correct.
4 Algebra of Matrices Exercise MCQs Question 7
Answer: The correct option is (A), $A^{n}=\left[\begin{array}{ccc} a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a^{n} \end{array}\right]$Given: The matrix AB is zero.
Solution:
If the matrix AB is zero, then, it is not necessary that either A = 0 or B = 0.
So, the option (A) is correct
4 Algebra of Matrices Exercise MCQs Question 8
Answer: The correct option is (C).Given: $\left[\begin{array}{ccc} a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a^{n} \end{array}\right]$
Solution:
$\begin{aligned} &\text { Let } A=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right] \\ &\Rightarrow A^{n}=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right] \times\left[\begin{array}{lll} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right] \ldots \text { n times } \\ &\Rightarrow A^{n}=\left[\begin{array}{ccc} a^{n} & 0 & 0 \\ 0 & a^{n} & 0 \\ 0 & 0 & a^{n} \end{array}\right] \end{aligned}$
So, the correct option is option (C).
4 Algebra of Matrices Exercise MCQs Question 9
Answer: The correct option is (A), B is a null matrix.
Given: AB = 0
Solution:
As, AB = 0 and the order of the matrices is A and B is 3,
So, the matrix B has to be a null matrix.
Thus, the correct option is (A)
Answer: The correct option is (B), AB = nB
Given:$A=\left[\begin{array}{lll} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{array}\right] \text { and } B=\left[\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right]$
Solution:$\begin{aligned} A B &=\left[\begin{array}{lll} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{array}\right] \times\left[\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right] \\ &=\left[\begin{array}{lll} n a_{1}+0+0 & n a_{2}+0+0 & n a_{3}+0+0 \\ 0+n b_{1}+0 & 0+n b_{2}+0 & 0+n b_{3}+0 \\ 0+0+n c_{1} & 0+0+n c_{2} & 0+0+n c_{3} \end{array}\right] \\ &=\left[\begin{array}{lll} n a_{1} & n a_{2} & n a_{3} \\ n b_{1} & n b_{2} & n b_{3} \\ n c_{1} & n c_{2} & n c_{3} \end{array}\right] \end{aligned}$
Now $\begin{aligned} &=n\left[\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right] \\ &=\mathrm{nB} \end{aligned}$
So, the option (B) is correct.
Given:$A=\left[\begin{array}{lll} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{array}\right] \text { and } B=\left[\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right]$
Solution:$\begin{aligned} A B &=\left[\begin{array}{lll} n & 0 & 0 \\ 0 & n & 0 \\ 0 & 0 & n \end{array}\right] \times\left[\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right] \\ &=\left[\begin{array}{lll} n a_{1}+0+0 & n a_{2}+0+0 & n a_{3}+0+0 \\ 0+n b_{1}+0 & 0+n b_{2}+0 & 0+n b_{3}+0 \\ 0+0+n c_{1} & 0+0+n c_{2} & 0+0+n c_{3} \end{array}\right] \\ &=\left[\begin{array}{lll} n a_{1} & n a_{2} & n a_{3} \\ n b_{1} & n b_{2} & n b_{3} \\ n c_{1} & n c_{2} & n c_{3} \end{array}\right] \end{aligned}$
Now $\begin{aligned} &=n\left[\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right] \\ &=\mathrm{nB} \end{aligned}$
So, the option (B) is correct.
4 Algebra of Matrices Exercise Multiple Choice Questions Question 11
Answer: The correct option is (A).Given: $A=\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]$
Solution: $A=\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right]$
$A^{n}=\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right] \times\left[\begin{array}{cc} 1 & a \\ 0 & 1 \end{array}\right] \times\left[\begin{array}{ll} 1 & a \\ 0 & 1 \end{array}\right] \ldots \mathrm{n} \text { times }$ $\{\text { where, } n \in N\}$
$A^{n}=\left[\begin{array}{cc} 1 & n a \\ 0 & 1 \end{array}\right]$
So, the correct option is (A).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 12
Answer: The correct option is (A).Given:$A=\left[\begin{array}{lll} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \text { and } B=\left[\begin{array}{ccc} 1 & -2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
Solution: $A B=\left[\begin{array}{lll} 1 & 2 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \times\left[\begin{array}{ccc} 1 & -2 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
$=\left[\begin{array}{ccc} 1+0+0 & -2+2+0 & y+0+x \\ 0+0+0 & 0+1+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \end{array}\right]$
$=\left[\begin{array}{ccc} 1 & 0 & x+y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
$I_{3}=\left[\begin{array}{ccc} 1 & 0 & x+y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$
Hence,
x + y = 0
So, the correct option is (A).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 13
Answer: The correct option is (B).
Given:$A=\left[\begin{array}{rr} 1 & -1 \\ 2 & -1 \end{array}\right], B=\left[\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right] \text { and }(A+B)^{2}=A^{2}+B^{2}$
Solution: $\begin{aligned} A+B &=\left[\begin{array}{ll} 1 & -1 \\ 2 & -1 \end{array}\right]+\left[\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right] \\ &=\left[\begin{array}{cc} a+1 & 0 \\ b+2 & -2 \end{array}\right] \end{aligned}$
$(A+B)^{2}=\left[\begin{array}{cc} a+1 & 0 \\ b+2 & -2 \end{array}\right] \times\left[\begin{array}{cc} a+1 & 0 \\ b+2 & -2 \end{array}\right]$
$\begin{aligned} &=\left[\begin{array}{cc} (a+1)^{2}+0 & 0+0 \\ (b+2)(a+1)-4-b & 0+4 \end{array}\right] \\ &=\left[\begin{array}{cc} a^{2}+2 a+1 & 0 \\ 2+2 a+b+a b-4-b & 4 \end{array}\right] \\ &=\left[\begin{array}{cc} a^{2}+2 a+1 & 0 \\ 2 a+a b-2 & 4 \end{array}\right] \end{aligned}$
$\begin{aligned} A^{2} &=\left[\begin{array}{rr} 1 & -1 \\ 2 & -1 \end{array}\right] \times\left[\begin{array}{rr} 1 & -1 \\ 2 & -1 \end{array}\right] \\ &=\left[\begin{array}{rr} 1-2 & -1+1 \\ 2-2 & -2+1 \end{array}\right] \\ A^{2} &=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] \end{aligned}$
$\begin{aligned} &B^{2}=\left[\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right] \times\left[\begin{array}{cc} a & 1 \\ b & -1 \end{array}\right] \\ &B^{2}=\left[\begin{array}{ll} a^{2}+b & a-1 \\ a b-b & b+1 \end{array}\right] \\ &A^{2}+B^{2}=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right]+\left[\begin{array}{cc} a^{2}+b & a-1 \\ a b-b & b+1 \end{array}\right] \\ &A^{2}+B^{2}=\left[\begin{array}{cc} a^{2}+b-1 & a-1 \\ a b-b & b \end{array}\right] \end{aligned}$
As, $A^{2}+B^{2}=(A+B)^{2}$
$\begin{aligned} &{\left[\begin{array}{cc} a^{2}+b-1 & a-1 \\ a b-b & b \end{array}\right]=\left[\begin{array}{cc} a^{2}+2 a+1 & 0 \\ 2 a+a b-2 & 4 \end{array}\right]} \\ &a-1=0 \\ &a=1 \\ &b=4 \end{aligned}$
So, the correct option is (B).
Answer: The correct option is (C).
Given:$A=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \text { and } \mathrm{A}^{2}=\mathrm{I}$
Solution: $\begin{aligned} &A^{2}=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \times\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \\ &A^{2}=\left[\begin{array}{cc} \alpha^{2}+\beta \gamma & \alpha \beta-\alpha \beta \\ \alpha \gamma-\alpha \gamma & \gamma \beta+\alpha^{2} \end{array}\right] \\ &I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}$
$\left[\begin{array}{cc} \alpha^{2}+\beta \gamma & \alpha \beta-\alpha \beta \\ \alpha \gamma-\alpha \gamma & \gamma \beta+\alpha^{2} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
So,$\begin{aligned} &\alpha^{2}+\beta \gamma=1 \\ &\alpha^{2}+\beta \gamma-1=0 \\ &1-\alpha^{2}-\beta \gamma=0 \end{aligned}$
Hence, the correct option is (C).
Given:$A=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \text { and } \mathrm{A}^{2}=\mathrm{I}$
Solution: $\begin{aligned} &A^{2}=\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \times\left[\begin{array}{cc} \alpha & \beta \\ \gamma & -\alpha \end{array}\right] \\ &A^{2}=\left[\begin{array}{cc} \alpha^{2}+\beta \gamma & \alpha \beta-\alpha \beta \\ \alpha \gamma-\alpha \gamma & \gamma \beta+\alpha^{2} \end{array}\right] \\ &I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \end{aligned}$
$\left[\begin{array}{cc} \alpha^{2}+\beta \gamma & \alpha \beta-\alpha \beta \\ \alpha \gamma-\alpha \gamma & \gamma \beta+\alpha^{2} \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
So,$\begin{aligned} &\alpha^{2}+\beta \gamma=1 \\ &\alpha^{2}+\beta \gamma-1=0 \\ &1-\alpha^{2}-\beta \gamma=0 \end{aligned}$
Hence, the correct option is (C).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 15
Answer: The correct option is (C).
Given:$S=\left[S_{i j}\right] \text { and } S_{\ddot{j}}=\mathrm{K}$
Solution: $\begin{aligned} &S=\left[S_{i j}\right] \\ &S=\left[\begin{array}{cc} K & 0 \\ 0 & K \end{array}\right] \quad\left(\mathrm{As}_{3} S_{i j}=\mathrm{K}\right) \end{aligned}$
$\text { Let } A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]$ (Square Matrix)
$\begin{aligned} A S &=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \times\left[\begin{array}{cc} K & 0 \\ 0 & K \end{array}\right] \\ &=\left[\begin{array}{ll} K a_{11} & K a_{12} \\ K a_{21} & K a_{22} \end{array}\right] \\ &=K\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] \\ &=K A \end{aligned}$
Hence,
$A S=S A=K A$
So, the correct option is (C).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 16
Answer: The correct option is (C).Given: A is a Square Matrix and $A^{2}=A$
Solution:
$(I+A)^{3}=I^{3}+A^{3}+3 A^{2} I+3 A I^{2}$ $\text { (Using the Identity, } \left.(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\right)$
$(I+A)^{3}=I+A^{2}(A)+3 A I+3 A$ (‘I’ stands for Identity Matrix)
$\begin{aligned} &(I+A)^{3}=I+A+3 A+3 A \\ &(I+A)^{3}=I+7 A \\ &(I+A)^{3}-7 A=I \end{aligned}$
So, the correct option is (C).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 17
Answer: The correct option is (B).Given: A is both Symmetric and Skew-Symmetric.
Solution:
If a matrix A is both Symmetric and Skew-Symmetric,
$A=A$ and $A=-A$
Comparing both the equations,
$\begin{aligned} &A=-A \\ &A+A=0 \\ &2 A=0 \\ &A=0 \end{aligned}$
Then A is a Zero Matrix.
So, the correct option is (B).
4 Algebra of Matrices Exercise Multiple Choice Questions 18
Answer: The correct option is (A).
Given: Matrix = $\left[\begin{array}{ccc} 0 & 5 & -7 \\ -5 & 0 & 11 \\ 7 & -11 & 0 \end{array}\right]$
Solution:
$\text { Let } \mathrm{A}=\left[\begin{array}{ccc} 0 & 5 & -7 \\ -5 & 0 & 11 \\ 7 & -11 & 0 \end{array}\right]$
$\begin{aligned} &A^{T}=\left[\begin{array}{ccc} 0 & -5 & 7 \\ 5 & 0 & -11 \\ -7 & 11 & 0 \end{array}\right] \\ &-A=\left[\begin{array}{ccc} 0 & -5 & 7 \\ 5 & 0 & -11 \\ -7 & 11 & 0 \end{array}\right] \end{aligned}$
$\because A^{T}=-A$
Therefore, the given matrix is a Skew-Symmetric matrix.
So, the correct option is (A).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 19
Answer: The correct option is (D).Given: A is a Square Matrix.
Solution:
If A is a Square matrix,
$\begin{aligned} &\text { Let } A=\left[\begin{array}{ll} 1 & 2 \\ 1 & 0 \end{array}\right] \\ &\begin{aligned} A A &=\left[\begin{array}{ll} 1 & 2 \\ 1 & 0 \end{array}\right] \times\left[\begin{array}{ll} 1 & 2 \\ 1 & 0 \end{array}\right] \\ &=\left[\begin{array}{ll} 3 & 2 \\ 1 & 2 \end{array}\right] \end{aligned} \end{aligned}$
Then, AA is neither of the matrices given in the options of the question.
So, the correct option is (D).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 21
Answer: The correct option is (C), x = yGiven: $A=\left[\begin{array}{ll} 5 & x \\ y & 0 \end{array}\right]$
Solution:
Then$\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ll} 5 & y \\ x & 0 \end{array}\right]$
So,$\mathrm{A}=\mathrm{A}^{\mathrm{T}}$
$\begin{aligned} &\Rightarrow\left[\begin{array}{ll} 5 & x \\ y & 0 \end{array}\right]=\left[\begin{array}{ll} 5 & y \\ x & 0 \end{array}\right] \\ &\Rightarrow \mathrm{x}=\mathrm{y} \end{aligned}$
So, the correct option is (C).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 21
Answer:
Answer: The correct option is (C), x = yGiven: $A=\left[\begin{array}{ll} 5 & x \\ y & 0 \end{array}\right]$
Solution:
Then$\mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ll} 5 & y \\ x & 0 \end{array}\right]$
So,$\mathrm{A}=\mathrm{A}^{\mathrm{T}}$
$\begin{aligned} &\Rightarrow\left[\begin{array}{ll} 5 & x \\ y & 0 \end{array}\right]=\left[\begin{array}{ll} 5 & y \\ x & 0 \end{array}\right] \\ &\Rightarrow \mathrm{x}=\mathrm{y} \end{aligned}$
So, the correct option is (C).
Answer: The correct option is (A), $3\times 4$
Given:$\mathrm{A}^{\mathrm{T}} \mathrm{B}$ and $\mathrm{BA}^{\mathrm{T}}$ are both defined.
Solution:
As $\mathrm{A}^{\mathrm{T}} \mathrm{B}$ and $\mathrm{BA}^{\mathrm{T}}$ are both defined, the number of columns in B should be equal to the number of rows in A for BA and also, the number of columns in A should be equal to the number of rows in B for BA.
So, the order of the matrix is $B=3\times 4$
Thus, the correct option is (A).
Given:$\mathrm{A}^{\mathrm{T}} \mathrm{B}$ and $\mathrm{BA}^{\mathrm{T}}$ are both defined.
Solution:
As $\mathrm{A}^{\mathrm{T}} \mathrm{B}$ and $\mathrm{BA}^{\mathrm{T}}$ are both defined, the number of columns in B should be equal to the number of rows in A for BA and also, the number of columns in A should be equal to the number of rows in B for BA.
So, the order of the matrix is $B=3\times 4$
Thus, the correct option is (A).
4 Algebra of Matrices Exercise Multiple choice Questions Question 23:
Answer: The correct option is (D), none of these.Given:$\mathrm{A}=\left[a_{i j}\right]$ is a Square matrix of even order such that $a_{i j}=i^{2}-j^{2}$
Solution:
From $a_{i j}=i^{2}-j^{2}$
$\begin{gathered} \Rightarrow a_{11}=12-12=0 \\ a_{12}=12-22=-10 \\ a_{21}=22-12=10 \\ a_{22}=22-22=0 \end{gathered}$
$\begin{aligned} &\therefore A=\left[\begin{array}{cc} 0 & -10 \\ 10 & 0 \end{array}\right] \\ &A^{T}=\left[\begin{array}{cc} 0 & 10 \\ -10 & 0 \end{array}\right] \\ &-A=\left[\begin{array}{cc} 0 & 10 \\ -10 & 0 \end{array}\right] \end{aligned}$
So,$A^{T}=-A$
$\begin{aligned} |A| &=0(0)-(-10) 10 \\ &=0+100 \\ &=100 \neq 0 \end{aligned}$
So, the correct option is (D).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 24
Answer: The correct option is $\text { (C), } \theta=2 n \pi+\frac{\pi}{3}, n \in Z$Given:$A=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$
Solution:
Now,
$A=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$
$\begin{aligned} &\Rightarrow A^{T}=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \\ &A+A^{T}=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]+\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &\Rightarrow 2 \cos \theta=1 \\ &\cos \theta=\frac{1}{2} \\ &\Rightarrow \theta=2 n \pi+\frac{\pi}{3}\{n \in Z\} \end{aligned}$
So, the correct option is (C).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 25
Answer: The correct option is (A).
$\frac{1}{2}\left(A+A^{\prime}\right)=\left[\begin{array}{ccc} 2 & 2 & -4 \\ 2 & 3 & 4 \\ -4 & 4 & 2 \end{array}\right]$
Given:$A=\left[\begin{array}{ccc} 2 & 0 & -3 \\ 4 & 3 & 1 \\ -5 & 7 & 2 \end{array}\right]$
Solution:
Then $A^{\prime}=\left[\begin{array}{ccc} 2 & 4 & -5 \\ 0 & 3 & 7 \\ -3 & 1 & 2 \end{array}\right]$
As the sum is expressed as,
$\begin{aligned} B &=\frac{1}{2}\left(A+A^{\prime}\right) \\ &=\frac{1}{2}\left[\left[\begin{array}{ccc} 2 & 0 & -3 \\ 4 & 3 & 1 \\ -5 & 7 & 2 \end{array}\right]+\left[\begin{array}{ccc} 2 & 4 & -5 \\ 0 & 3 & 7 \\ -3 & 1 & 2 \end{array}\right]\right] \\ &=\frac{1}{2}\left[\begin{array}{ccc} 4 & 4 & -8 \\ 4 & 6 & 8 \\ -8 & 8 & 4 \end{array}\right] \\ &=\left[\begin{array}{ccc} 2 & 2 & -4 \\ 2 & 3 & 4 \\ -4 & 4 & 2 \end{array}\right] \end{aligned}$
So, the correct option is (A).
Answer: The correct option is (A), $\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$
Given: Matrix
Solution: Since Scalar matrix is a matrix whose all off-diagonal elements are zero and all on-diagonal elements are equal, so the Scalar matrix is $\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$
Thus, the correct option is (A).
Given: Matrix
Solution: Since Scalar matrix is a matrix whose all off-diagonal elements are zero and all on-diagonal elements are equal, so the Scalar matrix is $\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]$
Thus, the correct option is (A).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 27
Answer: The correct option is (D), 512.Given: The Matrix is of order $3\times 3$
Solution:
$\text { Let } A=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]$
No. of elements = 9
Order = $3\times 3$
Every item in this matrix can be filled in two ways either by 0 or by 1.
So, the number of possible matrices is = $2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=512$
Thus, the correct option is (D).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 28
Answer: The correct option is (D), Not possible to find.
Given:
Solution: $\left[\begin{array}{cc} 3 x+7 & 5 \\ y+1 & 2-3 x \end{array}\right] \text { and }\left[\begin{array}{cc} 0 & y-2 \\ 8 & 4 \end{array}\right]$are equal
Then from the matrices,
$\begin{aligned} &3 x+7=0 \\ &x=\frac{-7}{3} \\ &y-2=5 \\ &y=7 \\ &2-3 x=4 \\ &3 x=-2 \\ &x=\frac{-2}{3} \end{aligned}$
These values of $x$ are not equal to each other, so it is not possible to find.
Thus, the correct option is (D).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 29
Answer: The correct option is $(C) ;-6,-4,-9$Given:$A=\left[\begin{array}{cc} 0 & 2 \\ 3 & -4 \end{array}\right] \text { and } k A=\left[\begin{array}{cc} 0 & 3 a \\ 2 b & 24 \end{array}\right]$
Solution:
From Matrix,
$\begin{aligned} &k A=\left[\begin{array}{cc} 0 & 3 a \\ 2 b & 24 \end{array}\right] \\ &\Rightarrow k A=\left[\begin{array}{cc} 0 & 2 k \\ 3 k & -4 k \end{array}\right] \\ &\Rightarrow\left[\begin{array}{cc} 0 & 2 k \\ 3 k & -4 k \end{array}\right]=\left[\begin{array}{cc} 0 & 3 a \\ 2 b & 24 \end{array}\right] \end{aligned}$
Now comparing the Equations,
$\begin{aligned} &-4 k=24 \Rightarrow k=-6 \\ &3 k=2 b \Rightarrow 3(-6)=2 b \\ &\Rightarrow 2 b=-18 \\ &\Rightarrow b=-9 \\ &3 a=2 k \Rightarrow 3 a=2(-6) \\ &\Rightarrow 3 a=-12 \\ &\Rightarrow a=-4 \end{aligned}$
So the values are k = -6, a= -4 and b = -9
Thus, the correct option is (C).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 30
Answer: The correct option is $\text { (A); } I \cos \theta+J \sin \theta$Given: $\begin{aligned} &I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &J=\left[\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right] \\ &B=\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \end{aligned}$
Solution:
From the Matrices,
$\begin{aligned} &I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &\Rightarrow I \cos \theta=\left[\begin{array}{cc} \cos \theta & 0 \\ 0 & \cos \theta \end{array}\right] \\ &\begin{aligned} J \sin \theta=\left[\begin{array}{cc} 0 & \sin \theta \\ -\sin \theta & 0 \end{array}\right] \mathbf{I} \end{aligned} \\ &\begin{array}{l} I \cos \theta+J \sin \theta=\left[\begin{array}{cc} \cos \theta & 0 \\ 0 & \cos \theta \end{array}\right]+\left[\begin{array}{cc} 0 & \sin \theta \\ -\sin \theta & 0 \end{array}\right] \\ =\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right] \end{array} \\ &\Rightarrow B=I \cos \theta+J \sin \theta \end{aligned}$
Thus, the correct option is (C).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 31
Answer: The correct option is (A); 17Given:$A=\left[\begin{array}{ccc} 1 & -5 & 7 \\ 0 & 7 & 9 \\ 11 & 8 & 9 \end{array}\right]$
Solution: As the trace of a matrix is the sum of diagonal elements,
1 + 7 + 9 = 17
Trace = 17
Thus, the correct option is (A).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 32
Answer: The correct option is (A).Given:$A=\left[a_{i j}\right]_{m \times n} \text { and } a_{i j}=K$
Solution:
$\because A=\left[a_{\ddot{j}}\right]_{m \times n}$
Trace of A, i.e,
$\begin{aligned} \operatorname{tr}(A)=\sum a_{i j}^{n i}=1 &=a_{11}+a_{22}+a_{33}+\ldots .+a_{n n} \\ &=K+K+K+K \ldots . \quad \text { (n times) } \\ &=K(n) \\ &=n K \end{aligned}$
Thus, the correct option is (A).
4 Algebra of Matrices Exercise Multiple Choice questionsQuestion 33
Answer: The correct option is (D).Given:$A=\left[\begin{array}{lll} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{array}\right]$
Solution:
$A=\left[\begin{array}{lll} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{array}\right]=4\left[\begin{array}{lll} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right]=4 \mathrm{I}$
Therefore, it is not matching with any of the given matrix options.
Thus, the correct option is (D).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 34
Answer: The correct option is (D).Given: The order of the matrix = $3\times 3$
Solution:
$\text { Let } A=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]$
No. of elements = 9
Order = $3\times 3$
Every item in this matrix can be filled in two ways either by 0 or by 1.
So, the number of possible matrices is $=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2=512$
Thus, the correct option is (D).
4 Algebra of Matrices Exercise Multiple Choice Questions Question 35
Answer: The correct option is (B).Given: $\left[\begin{array}{cc} 2 x+y & 4 x \\ 5 x-7 & 4 x \end{array}\right]=\left[\begin{array}{cc} 7 & 7 y-13 \\ y & x+6 \end{array}\right]$
Solution:
Comparing the Equations,$\begin{aligned} &2 x+y=7 \text { and } 4 x=x+6 \\ &3 x=6 \Rightarrow x=2 \\ &2(2)+y=7 \\ &y=7-4 \\ &y=3 \\ &x=2 \text { and } y=3 \end{aligned}$
Thus, the correct option is (B).
4 Algebra of Matrices Exercise MCQs Question 36Answer: The correct option is (A).
Given: $A^{2}=I$is a square matrix
Hint: Using $(a-b)^{3} \text { and }(a+b)^{3}$
Solution:
$\begin{aligned} &(A-I)^{3}+(A+I)^{3}-7 A=A^{3}-I^{3}-3 A^{2} I+3 A I^{2}+A^{3}+I^{3}+3 A^{2} I+3 A I^{2}-7 A \\ &{\left[\begin{array}{l} (a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2} \\ (a+b)^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2} \end{array}\right]} \\ &=A^{2} \cdot A-I-3 A^{2} \cdot I+3 A I+A^{2} \cdot A+I+3 A^{2} \cdot I+3 A I-7 A \quad\left[\because I^{2}=I\right. \\ &=I \cdot A-I-3 I \cdot I+3 A+I A+I+3 I \cdot I+3 A-7 A \\ &=A-I-3 I+3 A+A+I+3 I+3 A-7 A \\ &=8 A-7 A=A \end{aligned}$
4 Algebra of Matrices Exercise Multiple Choice Questions Question 37\
Answer:$B\times n$
Given: if A and B are two matrices of order $3\times m$and $3\times n$ respectively and $m=n$, then the order of $5A-2B$is
Hint: you must know the subtraction of the matrices
Solution:
Order of the matrix $A=3\times m$
Order of the matrix $B=3\times n$
Now we know in matrices;
Addition or subtraction is possible only if their orders are same
i.e. order of A must be equal to the order or B
$\begin{aligned} &\Rightarrow 3 \times 3=3 \times n \\ &\Rightarrow 3 \times n=3 \times n \end{aligned}$ [Also $m=n$,is given]
Order of $5A-2B$ is also equal to $3\times n$
4 Algebra of Matrices Exercise Multiple Choice Questions Question 38
Answer:$m\times n$Given: If A is a matrix of order $m\times n$ and B is the matrix such that $AB^{T}$and $B^{T}A$ are both defined. Then the order of the matrix B is
Hint: compare order of matrices
Solution:
Order of the matrix $A=m \times n$
Order of the matrix $B=p \times q$
Order of the matrix $B^{T}=q \times p$
We are given $AB^{T}$ exists
Number of columns in A= Number of rows in $B^{T}$
$n=q$
Also we are given $B^{T}A$ exists
Number of columns in $B^{T}$= Number of rows in A
$p=m$
$\therefore$Order of B becomes $m\times n$
Answer: Skew symmetric matrix
Given: If A and B are matrices of same order then $AB^{T}-BA^{T}$is
Hint: You must know about properties of transpose
Solution:
Given that order of matrices A and B are same
Then we check it by taking transpose of $AB^{T}-BA^{T}$
i.e.$\begin{array}{ll} \left(A B^{T}-B A^{T}\right)^{T}=\left(A B^{T}\right)^{T}-\left(B A^{T}\right)^{T} & {\left[(A-B)^{T}=A^{T}-B^{T}\right]} \\ =\left(B^{T}\right)^{T}(A)^{T}-\left(A^{T}\right)^{T} B^{T} & {\left[(A B)^{T}=B^{T} A^{T}\right]} \\ =B A^{T}-A B^{T} & {\left[\left(A^{T}\right)^{T}=A\right]} \\ =\left(A B^{T}-B A^{T}\right) \end{array}$
$AB^{T}-BA^{T}$ Is a skew symmetric matrix
4 Algebra of Matrices Exercise Multiple Choice Questions Question 40
Answer:$I$Given: If matrix $A=\left[a_{i j}\right]_{2 \times 2} \text { where } a_{i j}=\left\{\begin{array}{ll} 1 & \text { if } i \neq j \\ 0 & \text { if } i=j \end{array} \text { then } A^{2}\right.$
Hint: find $A^{2}$
Solution:
We have$A=\left[a_{i j}\right]_{2 \times 2} \text { where } \begin{cases}1 & \text { if } i \neq j \\ 0 & \text { if } i=j\end{cases}$
Let A be a square matrix
$\text { i.e. } A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]$
Now for $i \neq j \text { i.e. } a_{12}=1 \text { and } a_{21}=1$
And for $i=j \text { i.e. } a_{11}=0 \text { and } a_{22}=0$
$\begin{aligned} &\therefore A=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \\ &\therefore A^{2}=\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=I \end{aligned}$
4 Algebra of Matrices Exercise Multiple Choice Questions Question 41
Answer:$\frac{1}{2} I$Given: $A=\frac{1}{\pi}\left[\begin{array}{ll} \sin ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x) \end{array}\right] ; B=\frac{1}{\pi}\left[\begin{array}{ll} -\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}(\pi x) \end{array}\right]$
Hint:$\text { Use } \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \text { and } \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$
Solution:
$\begin{aligned} &A-B=\frac{1}{\pi}\left[\left[\begin{array}{cc} \sin (\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x) \end{array}\right]-\left[\begin{array}{cc} -\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right) & -\tan ^{-1}\left(\frac{\pi}{x}\right) \end{array}\right]\right]\\ &=\frac{1}{\pi}\left[\begin{array}{ll} \sin (\pi x)+\cos ^{-1}(\pi x) & \tan ^{-1}\left(\frac{x}{\pi}\right)-\tan ^{-1}\left(\frac{x}{\pi}\right) \\ \sin ^{-1}\left(\frac{x}{\pi}\right)-\sin ^{-1}\left(\frac{x}{\pi}\right) & \cot ^{-1}(\pi x)+\tan ^{-1}\left(\frac{\pi}{x}\right) \end{array}\right]\\ &=\frac{1}{\pi}\left[\begin{array}{cc} \frac{\pi}{2} & 0 \\ 0 & \frac{\pi}{2} \end{array}\right] \quad\left[\begin{array}{l} \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\ \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2} \end{array}\right]\\ &=\frac{1}{\pi} \times \frac{\pi}{2} I=\frac{1}{2} I \end{aligned}$
4 Algebra of Matrices Exercise Multiple Choice Questions Question 42
Answer: $A^{2}-B^{2}+B A-A B$Given: If A and B are square matrices of the same order then $(A+B)(A-B)=$
Hint: You must know the multiplying the matrices
Solution:
$\begin{aligned} &(A+B)(A-B) \\ &=A(A-B)+B(A-B) \\ &=A^{2}-A B+B A-B^{2} \\ &=A^{2}-B^{2}+B A-A B \end{aligned}$
4 Algebra of Matrices Exercise Multiple Choice Questions Question 43
Answer: AB and BA are both definedGiven: $A=\left[\begin{array}{ccc} 2 & -1 & 3 \\ -4 & 5 & 1 \end{array}\right], B=\left[\begin{array}{cc} 2 & 3 \\ 4 & -2 \\ 1 & 5 \end{array}\right]$
Hint: Check order of both matrices
Solution:
Hence order of the matrix $A=2\times 3$
And order of the matrix $B=3\times 2$
$\therefore$ Columns of matrix A= Rows of matrix B
$\Rightarrow$AB is defined
Also here column of matrix B= Rows of matrix A
$\Rightarrow BA$ is defined
4 Algebra of Matrices Exercise Multiple Choice Questions Question 44
Answer: Skew symmetricGiven: $A=\left[\begin{array}{ccc} 0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0 \end{array}\right]$
Hint: matrix is skew symmetric if $A^{T}=-A$
Solution:
$\begin{aligned} &A=\left[\begin{array}{ccc} 0 & -5 & 8 \\ 5 & 0 & 12 \\ -8 & -12 & 0 \end{array}\right] \\ &A^{T}=\left[\begin{array}{ccc} 0 & 5 & -8 \\ -5 & 0 & -12 \\ 8 & 12 & 0 \end{array}\right]=-\left[\begin{array}{ccc} 0 & -5 & 8 \\ 5 & 0 & -12 \\ -8 & -12 & 0 \end{array}\right]=-A \\ &\Rightarrow A^{T}=-A \end{aligned}$
$\therefore$ A is a skew symmetric matrix
4 Algebra of Matrices Exercise Multiple Choice Questions Question 45
Answer: Diagonal matrix
Given: $A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]$
Hint: You must know about diagonal matrix
Solution:
$A=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{array}\right]$
Here all the elements of the matrix except the diagonal elements are 0
$\therefore$ A is a diagonal matrix
4 Algebra of Matrices Exercise Multiple Choice Questions Question 46
Answer: Skew symmetric matrix
Given: A and B are symmetric matrices of same order.
Hint: You must know about properties of transpose of a matrix
Solution:
Given that A and B are symmetric matrices
$\begin{aligned} &\Rightarrow A^{T}=A \text { and } B^{T}=B\\ &\text { Now }\\ &\left(A B^{T}-B A^{T}\right)^{T}=\left(A B^{T}\right)^{T}-\left(B A^{T}\right)^{T} \quad\left[\because(A-B)^{T}=A^{T}-B^{T}\right]\\ &=\left(B^{T}\right)^{T} A^{T}-\left(A^{T}\right)^{T} B^{T} \quad\left[\because(A B)^{T}=B^{T} A^{T}\right]\\ &=B A^{T}-A B^{T} \quad\left[\because\left(A^{T}\right)^{T}=A\right] \end{aligned}$
$\begin{aligned} &=-\left(A B^{T}-B A^{T}\right)\\ &\Rightarrow A B^{T}-B A^{T} \text { Is a skew symmetric matrix } \end{aligned}$
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Chapter-wise RD Sharma Class 12 Solutions
- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable
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