RD Sharma Solutions Class 12 Mathematics Chapter 4 VSA

# RD Sharma Solutions Class 12 Mathematics Chapter 4 VSA

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 01:22 PM IST

The RD Sharma books are the most recommended solutions for the class 12 students. The majority of the students find it hard to maintain a good consistency in scoring marks in mathematics. Even though a few chapters like the 4th Chapter Algebra of matrices, the twists and other complexities present make the students perplexed. It is highly recommended for them to use the RD Sharma Class 12th VSA book.

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

## Algebra of Matrices Excercise: VSA

Algebra of Matrices Exercise Very Short Asnwer Question 1

Answer: $AB$ exist yes. Its order is $m \times p$
Hint: You must be aware with the order of matrix
Given:$A$ is a matrix $m\times n$
$B$ is a matrix $n\times p$
Solution: Matrix multiplication can be possible when $1^{st}$ matrix’s column and $2^{nd}$matrix’s rows are similar.
$\rightarrow$So, here, $A$’s column (i=n) is equal to $B$’s row(j=n) So, $AB$ can be exist.
$\rightarrow$Its order is $m \times p$ order.

Algebra of Matrices Exercise Very Short Asnwer Question 2

Answer:$AB$ is a $2\times 2$ matrix
$BA$ is a $3\times 3$ matrix
Hint: To solve this type of question, we must know the basics of matrix multiplication.
Given:$A=\left [ 2 1 4 4 1 5 \right ] B=\begin{bmatrix} 3 &-1 \\ 2 &2 \\ 1 & 3 \end{bmatrix}$
Solution: Here, $A$ is $2\times 3$ matrix $B$ is $3\times 2$ matrix
$\rightarrow$So, $AB$ should have $2\times 2$ order
and $BA$ should have $3\times 3$ order

Answer:$AB=\begin{bmatrix} -7\\+2 \end{bmatrix}$$=\begin{bmatrix} -7\\2 \end{bmatrix}$
Hint: Here, we use the basic multiplication rule, rule is one’s column and second’s row should be match, then multiplication is possible.
Given:$A=\begin{bmatrix} 4 & 3\\ 1 & 2 \end{bmatrix}$,$B=\begin{bmatrix} -4\\3 \end{bmatrix}$
Solution:
Here A has 2 number of column and B has 2 number of row
$AB = \begin{bmatrix} 4 &3 \\ 1& 2 \end{bmatrix}\begin{bmatrix} -4\\3 \end{bmatrix}=\begin{bmatrix} -16+9 \\ -4+6 \end{bmatrix}=\begin{bmatrix} -7\\+2 \end{bmatrix}$
$AB=\begin{bmatrix} -7\\2 \end{bmatrix}$

Algebra of Matrices Exercise Very Short Asnwer Question 4

Answer:$AA^{T}=\begin{bmatrix} 1& 2 & 3\\ 2 & 4 &6 \\ 3& 6 &9 \end{bmatrix}$
Hint: Here , we use basic concept of matrix transpose.
Given:$A=\begin{bmatrix} 1\\ 2\\ 3\\ \end{bmatrix}$
Solution:
$A\times A^{T}=\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}$$A\times A^{T}=\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3 \end{bmatrix}$
Here multiplication is possible.
$AA^{T}=\left [1\times 1 \: 1\times2\: 1\times3 \: 2\times1\: 2\times2 \: 2\times3\: 3\times1\: 3\times2 \: 3\times 3 \right ]=\left [ 1 \: 2\: 3\: 2\: 4\: 6 \: 3\: 6\: 9 \right ]$

Answer:$A=\begin{bmatrix} 0 &0 \\ 1 & 0 \end{bmatrix}$,$B=\begin{bmatrix} 0 &0 \\ 1 & 2 \end{bmatrix}$
Hint: Here, we use basics of matrices.
Given: $AB=0$ and both has $2\times2$ order
Solution:$AB=0$
So, $AB=\begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}$
Example
$A=\begin{bmatrix} 0 &0 \\ 1 & 0 \end{bmatrix}$,$B=\begin{bmatrix} 0 &0 \\ 1 & 2 \end{bmatrix}$
So, now AB must be null

Algebra of Matrices Exercise Very Short Asnwer Question 6

Answer:$A+A^{T}=\begin{bmatrix} 4 & 8\\ 8 & 14 \end{bmatrix}$
Hint: Here we use basic of transpose matrix
Given:$A=\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}$
Solution:
$A+A^{T}$
$A=\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix},A^{T}=\begin{bmatrix} 2 &5 \\ 3 & 7 \end{bmatrix}$
$A+A^{T}=\begin{bmatrix} 2 & 3\\ 5 & 7 \end{bmatrix}+\begin{bmatrix} 2 & 5\\ 3 & 7 \end{bmatrix}=\begin{bmatrix} 4 &8 \\ 8 &14 \end{bmatrix}$

### Algebra of Matrices Exercise Very Short Asnwer Question 7

Answer:$A^{2}=\begin{bmatrix} i^{2} &0 \\ 0 & i^{2} \end{bmatrix}$
Hint: Here we use concept of matrix multiplication
Given:$A=\begin{bmatrix} i &0 \\ 0 & i \end{bmatrix}$
Solution:
$A=\begin{bmatrix} i &0 \\ 0 & i \end{bmatrix}$
$A^{2}=\begin{bmatrix} i &0 \\ 0 & i \end{bmatrix}\begin{bmatrix} i &0 \\ 0 & i \end{bmatrix}$
$=\begin{bmatrix} i^{2}+0 &0+0 \\ 0+0 &0+ i^{2} \end{bmatrix}=\begin{bmatrix} i^{2} &0 \\ 0 & i^{2} \end{bmatrix}$

Algebra of Matrices Exercise Very Short Asnwer Question 8

Answer:$x=\frac{\pi}{3}$
Hint: Here we use basic of transpose matrix
Given:$A=\begin{bmatrix} \cos x & \sin x\\ -\sin x & \cos x \end{bmatrix}A+A^{T}=I$
Solution:
$A+A^{T}=I$
$\begin{bmatrix} \cos x & \sin x\\ -\sin x & \cos x \end{bmatrix}+\begin{bmatrix} \cos x & -sin x\\ \sin x & \cos x \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix} \cos x+\cos x & \sin x-\sin x\\ -\sin x+\sin x & \cos x+\cos x \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 2\cos x & 0\\ 0&2 \cos x \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$
So,$\cos x=\frac{1}{2}$
$x=\frac{\pi}{3}$

Answer:$A= \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}$
Hint: Here we find that if we use the concept of matrix transpose, we can use also matrix algebra
Given:$A_{T}=\begin{bmatrix} \cos x &\sin x \\ -\sin x & \cos x \end{bmatrix}$
Solution:
$AA^{T}=\begin{bmatrix} \cos x &-\sin x \\ \sin x & \cos x \end{bmatrix}\begin{bmatrix} \cos x &\sin x \\ -\sin x & \cos x \end{bmatrix}$
$=\begin{bmatrix} \cos^{2} x+\sin^{2} x &\cos x\sin x-\sin x\cos x \\ \sin x \cos x-\cos x\sin x& \sin^{2}x\cos^{2} x \end{bmatrix}$
$= \begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}$ because $\left [ \cos^{2}x+\sin^{2}x=1 \right ]$

Algebra of Matrices Exercise Very Short Asnwer Question 10

Answer:$y=\left ( -2 \right )\: and\: x=0$
Hint: Here we use the concept of identity matrix
Given:$\begin{bmatrix} 1 & 0\\ y & 5 \end{bmatrix}+2\begin{bmatrix} x &0 \\ 1& -2 \end{bmatrix}=I$
Solution:
\begin{aligned} &{\left[\begin{array}{ll} 1 & 0 \\ y & 5 \end{array}\right]+2\left[\begin{array}{cc} x & 0 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\ &{\left[\begin{array}{ll} 1 & 0 \\ y & 5 \end{array}\right]+\left[\begin{array}{cc} 2 x & 0 \\ 2 & -4 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\ &{\left[\begin{array}{ll} 2 x+1 & 0 \\ y+2 & 1 \end{array}\right]=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]} \\ &\text { So } 2 x+1=1 \\ &2 x=0 \\ &\begin{array}{l} x=0 \\ y+2=0 \end{array} \\ &y=(-2) \end{aligned}

#### Algebra of Matrices Exercise Very Short Asnwer Question 11

Answer:$k=2$
Hint: Here we use the concept of matrix scalar
Given:
$A=\begin{bmatrix} 1 & -1\\ -1 &1 \end{bmatrix}$
$A^{2}=kA$
Solution:
$A=\begin{bmatrix} 1 & -1\\ -1 &1 \end{bmatrix}$
$A=\begin{bmatrix} 1 & -1\\ -1 &1 \end{bmatrix}\times \begin{bmatrix} 1 & -1\\ -1 &1 \end{bmatrix}$
$A=\begin{bmatrix} 1+1 & -1-1\\ -1-1 &1 +1\end{bmatrix}=\begin{bmatrix} 2 & -2\\ -2 &2 \end{bmatrix}$
$A^{2}=kA$
$\begin{bmatrix} 2 & -2\\ -2 &2 \end{bmatrix}=k\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$
$2\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}=k\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}$
$k=2$

Answer:$\lambda=8$
Hint: Here, we use the basic concept of Algebra
Given:
$A=\begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}$
$A^{^{4}}=\lambda A$
Solution:
$A^{^{4}}= A^{2}\times A^{2}$
$A^{2}=\begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}\times \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}$
$A^{2}=\begin{bmatrix} 1 +1& 1+1\\ 1+1 & 1+1 \end{bmatrix}=\begin{bmatrix} 2 & 2\\ 2 & 2 \end{bmatrix}$
$A^{2}\times A^{2}=\begin{bmatrix} 2 & 2\\ 2 & 2 \end{bmatrix}\begin{bmatrix} 2 & 2\\ 2 & 2 \end{bmatrix}$
$=\begin{bmatrix} 4+4 & 4+4\\ 4+4 & 4+4 \end{bmatrix}=\begin{bmatrix} 8 & 8\\ 8 & 8 \end{bmatrix}$
$A^{4}=\begin{bmatrix} 8 & 8\\ 8 & 8 \end{bmatrix}$
$A^{^{4}}=\lambda A$
$\begin{bmatrix} 8 & 8\\ 8 & 8 \end{bmatrix}=\lambda \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}$
$\begin{bmatrix} 8 & 8\\ 8 & 8 \end{bmatrix}= \begin{bmatrix} \lambda & \lambda\\ \lambda & \lambda \end{bmatrix}$
So,$\lambda=8$
Answer:$A^{2}=I=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0& 0 &1 \end{bmatrix}$
Hint: Here, we use the concept of matrix multiplication
Given:$A=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0& 0 &1 \end{bmatrix} \: A^{2}=?$
Solution:
$A^{2}=\begin{bmatrix} -1 & 0 & 0\\ 0 & -1 & 0\\ 0& 0 &-1 \end{bmatrix} \times \begin{bmatrix} -1 & 0 & 0\\ 0 & -1 & 0\\ 0& 0 &-1 \end{bmatrix}$
$=\begin{bmatrix} 1+0+0 & 0+0+0 & 0+0+0\\ 0+0+0 & 1+0+0 & 0+0+0\\ 0+0+0& 0+0+0 &1+0+0 \end{bmatrix}$
$=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0& 0&1 \end{bmatrix}$
$A^{2}=I$
Note: To solve this type of question, must care about matrix multiplication rule.
Answer:$A^{3}=A=\begin{bmatrix} -1 & 0 &0 \\ 0& -1 & 0\\ 0 & 0&-1 \end{bmatrix}$
Hint: Here, we use the concept of matrix multiplication
Given:$A=\begin{bmatrix} -1 & 0 &0 \\ 0& -1 & 0\\ 0 & 0&-1 \end{bmatrix}A^{3}=?$
Solution:
$A^{3}=\left [-1\: 0 \: 0 \: 0 \: -1\: 0\: 0 \: 0 \: -1 \right ] \times A^{2}A^{2}=\left [ -1 \:0\: 0\: 0 \:-1\: 0\: 0\: 0\: -1 \right ] \times\left [-1\: 0\: 0\: 0\: -1\: 0 \:0 \:0\: -1 \right ] =\left [ 1 \:0\: 0\: 0\: 1 \:0 \:0 \:0 \:1 \right ] A^{3}=\left [ 1\: 0\: 0\: 0\: 1\: 0\: 0\: 0\: 1 \right ] \times\left [ -1 \:0 \:0 \:0 \:-1\: 0\: 0\: 0\: -1 \right ]$
$A^{3}=\begin{bmatrix} -1+0+0 &0+0+0 &0+0+0 \\ 0+0+0 &0+0-1 &0+0+0 \\ 0+0+0 &0+0+0 &0+0-1 \end{bmatrix}$
$A^{3}=\begin{bmatrix} -1 & 0 &0 \\ 0& -1 & 0\\ 0 & 0&-1 \end{bmatrix}$
$A^{3}=A$

Answer: $A^{4}=\begin{bmatrix} 81 & 0\\ 0 & 81 \end{bmatrix}$
Hint: Here, we use the basic concept of matrix multiplication
Given:$A=\begin{bmatrix}-3 & 0\\ 0 & -3 \end{bmatrix} find A^{4}$
Solution:
$A^{4}=A^{2}\times A^{2}$
$A^{2}= \begin{bmatrix} -3 & 0\\ 0 & -3 \end{bmatrix}\times\begin{bmatrix} -3 & 0\\ 0 & -3 \end{bmatrix}$
$A^{2}= \begin{bmatrix} 9+0 & 0+0\\ 0+0 & 0+9 \end{bmatrix}=\begin{bmatrix} 9& 0\\ 0 & 9 \end{bmatrix}$
$A^{2}\times A^{2}= \begin{bmatrix} 9 & 0\\ 0& 9 \end{bmatrix}\times\begin{bmatrix} 9& 0\\ 0 & 9 \end{bmatrix}$
$A^{4}= \begin{bmatrix} 81+0 & 0+0\\ 0+0& 81+0 \end{bmatrix}=\begin{bmatrix} 81& 0\\ 0 & 81 \end{bmatrix}$
Note: Whenever you solve the matrix multiplication’s question must follow their rules.

Algebra of Matrices Exercise Very Short Asnwer Question 16

Answer:$x=-2$
Hint: To solve this, we must follow the basic rules of matrix multiplication
Example: One’s number of column is equal to second’s number of rows
Given:$\begin{bmatrix} x &2 \end{bmatrix}\begin{bmatrix} 3\\4 \end{bmatrix}=2$
Solution:
$\begin{bmatrix} x &2 \end{bmatrix}\begin{bmatrix} 3\\4 \end{bmatrix}=2$
$3x+8=2$
$3x=-6$
$x=-2$
Here, multiplication rules are followed

Answer:$A=\begin{bmatrix} 3 &5 \\ 4& 6 \end{bmatrix}$
Hint: To solve this, we should follow the basic concepts of matrix order
Given:$a_{ij}=i+2j \: and\: a\: is\: 2\times2\: matrix$
Solution:
Let $A=\begin{bmatrix} a_{11} &a_{12}\\ a_{21} &a_{22} \end{bmatrix}$
$A=\begin{bmatrix} 1+2\left ( 1 \right ) &1+2\left ( 2 \right )\\ 2+2\left ( 1 \right ) &2+2\left ( 2 \right ) \end{bmatrix}$
$A=\begin{bmatrix} 3 &5\\ 4 &6 \end{bmatrix}$
Answer:$A=\begin{bmatrix} {1} & {-9}\\ {-2}& {4} \end{bmatrix}$
Hint: Here we use the basic algebra of matrix

Given:
$A+\begin{bmatrix} 23 & -14 \end{bmatrix}=\begin{bmatrix} 3 &-6 &-38 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} a_{11} & a_{12}\\ a_{21}& a_{22} \end{bmatrix}$
So,
$\begin{bmatrix} a_{11} &a_{12} &a_{21} &a_{22} \end{bmatrix}+\begin{bmatrix} 23 & -14 \end{bmatrix}=\begin{bmatrix} 3 & -6 & 38 \end{bmatrix}a_{11}+2=3=1a_{12}+3=-6=-9a_{21}+\left ( -1 \right )$
So, $A=\begin{bmatrix} {1} & {-9}\\ {-2}& {4} \end{bmatrix}$

Answer:$A$is skew symmetric matrix
Hint: Here we should know the basics of matrix, skew and symmetric matrix
Given:$A$ is square matrix and $a_{ij}=i^{2}-j^{2}$
Solution: Let $A$ is square $2\times2$ matrix
So, $A=\begin{bmatrix} 1^{2}-1^{2} &1^{2}-2^{2} \\ 2^{2}-1^{2}& 2^{2} -2^{2}\end{bmatrix}$
$A=\begin{bmatrix} 0 &-3 \\ 3 & 0 \end{bmatrix}$
So, Let $A^{T}$
$A^{T}=\begin{bmatrix} 0 &3 \\ -3& 0 \end{bmatrix}$
So, Here $A^{T}=-A$
So, $A$is skew symmetric

Algebra of Matrices Exercise Very Short Asnwer Question 20

Answer:$AA^{T}$ is symmetric matrix
Hint: Here we should know the concept of matrix, skew and skew-symmetric matrix
Given:$AA^{T}$is symmetric or skew symmetric if $A$is square
Solution: $A$is square matrix
Let $A$is $2\times 2$ matrix
So, $A=\begin{bmatrix} 1 &3 \\ 2& 4 \end{bmatrix}$
So, $A\times A^{T}=\begin{bmatrix} 1 &3 \\ 2& 4 \end{bmatrix}\begin{bmatrix} 1 &2 \\ 3 &4 \end{bmatrix}$
$=\begin{bmatrix} 1+9 &2+12 \\ 2+12 & 4+16 \end{bmatrix}$
$=\begin{bmatrix} 10 &14 \\ 14 & 20 \end{bmatrix}$
$\left ( AA^{T} \right )^{T}=\begin{bmatrix} 10 &14 \\ 14 & 20 \end{bmatrix}$
So, here we can say that$AA^{T}$is symmetric matrix

Answer:$\sum _{i}a_{ii}=0$
Hint: Here to solve this we use the basic concept of skew symmetric matrix
Given: $A=\left [ a_{ij} \right ]$is skew symmetric
Write value of $\sum _{i}a_{ii}=?$
Solution: Let $A$is a $m\times n$matrix
Where $a_{ij}=i$denotes row
J denotes column
$I=\sum _{i=1}^{m}a_{ii}$
$=a_{11}+a_{22}+a\cdot \cdot \cdot \cdot \cdot a_{mn}$
$A=\left [ a_{ij} \right ]_{m\times n} where\: i\leq m,j\leq n$
Here$A$ is skew symmetric
So, $-A=A^{T}$
$-a_{ij}=a_{ji}$
$2a_{ij}=0$
$a_{ij}=0$
So, if $j=i$
$a_{ii}=0$
So,$\sum _{i}a_{ii}=0$

Algebra of Matrices Exercise Very Short Asnwer Question 22

Answer: $\sum _{i}\sum _{j}a_{ij}=0$
Hint: Here to solve this we use the basic concept of skew symmetric matrix
Given:$A=\left [ a_{ij} \right ]$is a skew symmetric matrix
Solution:
$a_{ij}=-a_{ji}$ $\left [ For \: all \: values \: of \: i,j\right ]a_{ii} =-a_{ii} \left [ For\: all\: values \: of \: i \right ]a_{ii}=0$
Now,
$\sum _{i}\sum_{j}a_{ij}=a_{11}+a_{12}+a_{13}+\cdot \cdot \cdot \cdot \cdot +a_{21}$
$a_{ij}=a_{11}+a_{12}+a_{13}+\cdot \cdot \cdot \cdot \cdot -a_{12}$
$\sum _{i}\sum_{j}a_{ij}=0$
$a_{ij}=0$

Algebra of Matrices Exercise Very Short Asnwer Question 23

Answer: $AB=BA$
Both must be commutative to each other
Hint: Here to solve this we use the basic concept of skew symmetric matrix
Given: $A \: and \: B$ is symmetric matrix
Solution:
$A \: and \: B$ is symmetric then
$A=A^{T}\cdot \cdot \cdot \cdot \cdot \left ( i \right )$
$B=B^{T}\cdot \cdot \cdot \cdot \cdot \left ( ii \right )$
→So, let AB can be symmetric$AB=\left ( AB^{T} \right ) =B^{T}A^{T} =BA$
From equation (i) and (ii)$AB=BA$
→So, condition is $A \: and \: B$ matrix must be commutative and $AB=BA$

Algebra of Matrices Exercise Very Short Asnwer Question 24

Answer:$ABA^{T}$is skew symmetric matrix
Hint: Here to solve this we use the basic concept of skew symmetric matrix
Given:$B$ is skew symmetric matrix
$ABA^{T}$is skew symmetric or symmetric matrix
Solution:
$B$is Skew symmetric
So, $B^{T}=-B$
So, $ABA^{T}$
$\rightarrow$Let take transpose
$=ABA^{T}=\left ( ABA^{T} \right )^{T}=AB^{T}A^{T}=A\left ( -B \right )A^{T}$
$ABA^{T}=-\left ( BA \right )A^{T}$
$\rightarrow$So, we clearly see that it satisfy the condition of skew symmetric
So, $ABA^{T}$is skew symmetric matrix

4 Algebra of Matrices Exercise Very Short Asnwer Question 25

Answer:$ABA^{T}$is symmetric matrix
Hint: Here to solve this we use the basic concept of skew symmetric matrix
Given: $B$is symmetric matrix
So, $ABA^{T}=?$
Solution:$B$is symmetric matrix
So, $B^{T}=B$ ……(i)
$\rightarrow$Let $ABA^{T}$and take transpose of it
Then
$=ABA^{T}=\left ( ABA^{T} \right )^{T}=AB^{T}A^{T}=ABA^{T}=ABA^{T}$
So, $ABA^{T}$is symmetric matrix

4 Algebra of Matrices Exercise Very Short Asnwer Question 26

Answer:$\lambda=\left ( -1 \right )^{n}$
Hint: Here to solve this we use the basic concept of skew symmetric matrix
Given: $\left ( A^{n} \right )^{T}=\lambda A'' \: \lambda=?$
Solution: A is skew symmetric
\begin{aligned} &A^{T}=-A \\ &\qquad \begin{aligned} \left(A^{n}\right)^{T}=\left(A^{n-1} \times A^{1}\right)^{T} &=A^{T} \times\left(A^{n-1}\right)^{T} \quad\left[(A B)^{T}=B^{T} A^{T}\right] \\ =(-A) \times\left(A^{n-2} \times A\right)^{T} &=(-A) A^{T}\left(A^{n-2}\right)^{T} \\ =(-A)(-A)\left(A^{n-3} \times A\right)^{T} &=(-A)(-A)\left(A^{1} A^{1}\right)^{T} \end{aligned} \\ &\qquad(-1)^{n-2} \times A^{n-2}\left(A^{T} \times A^{T}\right) =(-1)^{n-2} \times A^{n-2}(A \times A) \quad= \\ &(-1)^{n}(-1)^{-2} A^{n} \quad=(-1)^{n} \frac{1}{(-1)^{2}} \times A^{n}=(-1)^{n} A^{n} \end{aligned}
So, $\lambda=\left ( -1 \right )^{n}$

4 Algebra of Matrices Exercise Very Short Asnwer Question 27

Answer:$\left ( A^{n} \right )^{T}=A^{n}$
Hint: We must know the basic of symmetric
Given:$A$is symmetric
Solution: $A=A^{T}$ as $A$is symmetric
$\left ( A^{n} \right )^{T}=\left ( A\times A\times A\times A\times\cdot \cdot \cdot \cdot \cdot \cdot \times A^{n}\right ) ^{T} =A^{T}\times A^{T}\times A^{T}\times A^{T}\cdot \cdot \cdot \cdot \left ( A^{T} \right )^{n} =(A^{T})^{n}=A^{n}$
So, $\left ( A^{n} \right )^{T}=A^{n}$

Algebra of Matrices Exercise Very Short Asnwer Question 28

Answer:$n$is even So$A^{n}$ is symmetric
Hint: We must know the basic algebra
Given:$A$is a skew symmetric matrix
Solution:$A$where $n\epsilon A$
$A^{T}=-A$ ($A$is a skew symmetric matrix)
\begin{aligned} \left(A^{n}\right)^{T}=\left(A^{n-1} \times A^{1}\right)^{T} &=A^{T} \times\left(A^{n-1}\right)^{T} \quad\left[(A B)^{T}=B^{T} A^{T}\right] & \\ =(-A) \times\left(A^{n-2} \times A\right)^{T} &=(-A) A^{T}\left(A^{n-2}\right)^{T} &=(-A)(-A)\left(A^{n-3} \times A\right)^{T} \end{aligned}
As the same sequence will be continue ,so it can be written in general term as $=\left ( -1 \right )^{n}A^{n}=1A^{n}$
So, n is even.
Answer:$A^{n}$ is skew symmetric matrix
Hint: To solve this, we must knew the basic of matrix
Given:$A$ is skew symmetric and is n odd number
Solution:
$A^{T}=-A$
$\left ( A^{n} \right )^{T}=\left ( A^{T} \right )^{n}$
$\left ( A^{n} \right )^{T}=-A^{n}$
$\left ( A^{T} \right )^{n}=-A^{n}$
So here $A^{n}$ is skew symmetric matrix

Algebra of Matrices Exercise Very Short Asnwer Question 30

Answer:$AB-BA$is skew symmetric matrix
Hint: We must know the basic of matrix multiplication
Given:$A$and $B$are symmetric matrix
Solution:
$A^{T}=A$ ……(i)
$B^{T}=B$ ……(ii)
Let $AB-BA$ and take transpose
$\left ( AB-BA \right )^{T}=\left ( AB \right )^{T}-\left ( BA \right )^{T}$
$=B^{T}A^{T}-A^{T}B^{T}$
$= BA-AB$ [From eqn(i) and (ii)]
$AB-BA=-\left ( AB-BA \right )$
So, $AB-BA$is skew symmetric matrix

Algebra of Matrices Exercise Very Short Asnwer Question 31

Answer:$\begin{bmatrix} 0 &0 &0 &0 \end{bmatrix}$ or Null matrix
Hint: we should use the basic concept of symmetric as well as skew symmetric matrix,
For example,
$A^{T}=A$ ($A$is symmetric matrix)
$A^{T}=-A$ ($A$is skew symmetric matrix)
Given: Write a square matrix which is both symmetric as well as skew symmetric
Solution: Let’s take basic properties of symmetric and skew symmetric
$\begin{gathered} \quad A^{T}=A \\ -A^{T}=-A \\ \hline A^{T}-A^{T}=A-(-A) \\ 0=2 A \\ A=0 \end{gathered}$
So, a skew symmetric or symmetric matrix can be a square matrix’s.
Example is,
$O_{2\times2}=\begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}$

Answer:$x=3 \: and\: y=3$
Hint: We can use the basic union concept of matrices.
Given:$\begin{bmatrix} 2 & 6\\ 0& 2x \end{bmatrix}+\begin{bmatrix} y &0 \\ 1& 2 \end{bmatrix}=\begin{bmatrix} 5 &6 \\ 1 & 8 \end{bmatrix}$
Solution:
$2\begin{bmatrix} 1 & 3\\ 0& x \end{bmatrix}+\begin{bmatrix} y &0 \\ 1 &2 \end{bmatrix}= \begin{bmatrix} 5 &6 \\ 1 & 8 \end{bmatrix}$
$\begin{bmatrix} 2 & 6\\ 0& 2x \end{bmatrix}+\begin{bmatrix} y &0 \\ 1 &2 \end{bmatrix}= \begin{bmatrix} 5 &6 \\ 1 & 8 \end{bmatrix}$
$\begin{bmatrix} 2+y &6+0 \\ 0+1 & 2x+2 \end{bmatrix}=\begin{bmatrix} 5 & 6\\ 1 & 8 \end{bmatrix}$
$\begin{bmatrix} 2+y=5 &6+0=6 \\ 0+1=1 & 2x+2=8 \end{bmatrix}$
Here, $2+y=5$ and $2x+2=8$
$y=3$ $2x=6$
$x=3$
Here we find $y=3$ and $x=3$
Note: We must use the union algebra concept of matrix.

Algebra of Matrices Exercise Very Short Asnwer Question 33

Answer:$x=2\: and\: y=7$
Hint: We can use the basic union concept of matrices.
Given: if $\begin{bmatrix} x+3 &4 \\ y-4 & x+y \end{bmatrix}=\begin{bmatrix} 5 &4 \\ 3 & 9 \end{bmatrix}$ Find x and y
Solution:
Here,
$x+3=y$
$x=2$ .......(i)
$\therefore y-4=3$
$y=7$ .........(ii)
$\therefore x+y=9$
$=2+7=9$ [From (i) and (ii)]
RHS=LHS
So,$x=2\: and\: y=7$
Answer: $x=2$
Hint: Here we use the basic concept of matrix.
Given: If $\begin{bmatrix} 2x-y & 5\\ 3 & y \end{bmatrix}=\begin{bmatrix} 6 &5 \\ 3& -2 \end{bmatrix}$
Solution:
Here,Both side has $2\times 2$ matrix
So we can write
$y=\left ( -2 \right )$ .....(i)
$2x-y=6$
$2x-\left ( -2 \right )=6$ [From (i)]
$\! \! \! \! \! \! \! \! \! 2x+2=6\\\\2x=6-2\\\\2x=4\\\\x=\frac{4}{2}=2\\\\x=2$

Algebra of Matrices Exercise Very Short Asnwer Question 35

Answer:$y=1$
Hint: Here we use the basic concept of matrix.
Given:$\begin{bmatrix} x-y & 2\\ x & 5 \end{bmatrix}\begin{bmatrix} 2 & 2\\ 3 & 5 \end{bmatrix}y=?$
Solution:
Here,Both side has $2\times2$ matrix
So we can write
$x=3$ ......(i)
$x-y=2$
$3-y=2$ [From (i)]
$3-2=y$
$y=1$

Algebra of Matrices Exercise Very Short Asnwer Question 36

Answer:$x=1 \: and\: y=\left ( -2 \right )$
Hint: Here we use the basic concept of matrix.
Given: $[3 x+y-y 2 y-x 3]=[12-53]$
Solution:
Here, Both side has $2\times2$ matrix
So we can write
$-y=2\\y=\left ( -2 \right )$ ...(i)
$\therefore 3x+y=1$
$3x+\left ( -2 \right )=1$
$\therefore 3x=3\\$
$x=1\\$ .....(i)
$\! \! \! \! \! \! \! \! LHS=2y-x=-5\\\\=2\left ( -2 \right )-1\\\\=-4-1\\\\=-5$
=RHS

Algebra of Matrices Exercise Very Short Asnwer Question 37

Answer:$AA^{T}=\left [ 14 \right ]$
Hint: Here, we use the basic transpose matrix concept
Given:$A=\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$ with $AA^{T}$
Solution:
$A=\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$
$A^{T}=\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\left [ For\: transpose \right ]$
$A\times A^{T}=\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$
$=\begin{bmatrix} 1 & 2 & 3 \end{bmatrix}_{1\times3} \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}_{3\times1}$
=Here both sides coloumn and row is similar.So,let's use matrix multiplication
$= \left [ 1\times1+2\times2+3\times3 \right ]$
$= \left [ 1+4+9\right ]$
$A\times A^{T}=\left [ 14 \right ]$

Algebra of Matrices Exercise Very Short Asnwer Question 38

Answer:$x=3$
Hint: Here we use the basic concept of algebra
Given:$\begin{bmatrix} 2x+y & 3y\\ 0 & 4 \end{bmatrix}=\begin{bmatrix} 6 &0 \\ 6& 4 \end{bmatrix}$
Solution:
Here Both side has $2\times 2$ matrix
So we ca write
$3y=0\\$
$y=0\\$ .............(i)
$\therefore 2x+y=6$
$2x+0=6$ [ y=0 from eqn (i)]
$\therefore 2x=6$
$x=3$
Note: You should must match the order of matrix.

Algebra of Matrices Exercise Very Short Asnwer Question 39

Answer:$A+A^{T}=\begin{bmatrix} 2 & 5\\ 5 & 8 \end{bmatrix}$
Hint: Here we use basic transpose of matrix
Given:$A=\begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix} Find\: A+A^{T}$
Solution:
Let's find $A^{T}$ first
$A^{T}=\begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}$ [For tranpose]
$A+A^{T}=\begin{bmatrix} 1 & 3\\ 2 & 4 \end{bmatrix}+\begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}$
$= \begin{bmatrix} 1+1 & 3+2\\ 3+2 & 4+4 \end{bmatrix}=\begin{bmatrix} 2 & 5\\ 5 & 8 \end{bmatrix}$

Algebra of Matrices Exercise Very Short Asnwer Question 40

Answer:$a= 8$
Hint: Here, we use basic union concept of matrix
Given:$\begin{bmatrix} a-b &2 \\ 5 & b \end{bmatrix}=\begin{bmatrix} 6 &5 \\ 2 & 2 \end{bmatrix}$ Find a
Solution:
Here, Both side has $2\times 2$ matrix
So we can write
$b=2$ .....(i)
$a-b=6$
$a-2=6$ [From eqn (i)]
$\therefore a=8$
So, value of a is 8.

Algebra of Matrices Exercise Very Short Asnwer Question 41

Answer:$AB$ is order of $3\times3$ matrix
Hint: Here we should use the concept of multiplication.
Given:
$A$ is order of $3\times4$ matrix
$B$ is order of $4\times3$ matrix
Find order $A\times B$
Solution:$A_{3\times 4}\times B_{4\times3}$
Here, $A_{R_{1}\times C_{1}}\times B_{R_{2}\times C_{2}}$
→So, C1 and R2 is same So, mutiplication is possible.So,A×B matrix's order is 3×3 matrix

Algebra of Matrices Exercise Very Short Asnwer Question 42

Answer:$\alpha=0$
Hint: Here we use basic concept of identity matrix
Given:
$A=\begin{bmatrix} \cos \alpha & -\sin \alpha\\ \sin \alpha & \cos \alpha \end{bmatrix}$ is idetnify matrix
Find $\alpha$
Solution:$A_{2\times2}=\begin{bmatrix} \cos \alpha &-\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}_{2\times2}$
A is identity matrix A=
\begin{aligned} &A \text { is identity matrix } A=[\cos \cos \alpha-\sin \sin \alpha \sin \sin \alpha \cos \cos \alpha]_{2 \times 2}\\ &=\left[\begin{array}{lll} 1001 \end{array}\right]_{2 \times 2} \cos \cos \alpha=1 \sin \sin \alpha=0 \alpha=0^{\circ}\left[\text { For } \alpha \in 0^{\circ} \text { to } 360^{\circ}\right] \alpha\\ &=0 \text { radian } \end{aligned}

Algebra of Matrices Exercise Very Short Asnwer Question 43

Answer:$k=17$
Hint: Here we use the basic concept of multiplication
Given: $\begin{bmatrix} 1 & 2\\ 3& 4 \end{bmatrix}\begin{bmatrix} 3 & 1\\ 2 & 5 \end{bmatrix}=\begin{bmatrix} 7& 11\\ k &23 \end{bmatrix}$
Solution:
$\begin{bmatrix} 1 & 2\\ 3& 4 \end{bmatrix}\begin{bmatrix} 3 & 1\\ 2 & 5 \end{bmatrix}$
$=\begin{bmatrix} 1\times 3+2\times2 & 1\times1+2\times5\\ 3\times3+4\times2& 3\times1+4\times5 \end{bmatrix}$
$=\begin{bmatrix} 3+4 &1+10 \\ 9+8& 3+20 \end{bmatrix}$
$=\begin{bmatrix} 7 &11 \\ 17& 23 \end{bmatrix}$
On Comparing, $k=17$

Algebra of Matrices Exercise Very Short Asnwer Question 44

Answer: $\left ( I+A \right )^{2}-3A=I$
Hint: Here we use the basic knowledge of identity and square matrix
Given:
$I$ is a identity matirx
$A$ is a square matrix
$A^{2}=A$ then find the value of$\left ( I+A \right )^{2}-3A$
Solution:
\begin{aligned} I \text { is a identity } & \text { matrix }=(I+A)^{2}-3 A=(I+A) \times(I+A)-3 A \\ &=[I * I+A * I+I * A+A * A] \\ &-3 A[\text { using matrix distributing law } A(B+C)=A B+A C] \\ &=\left[I+A+A+A^{2}\right]-3 A=I+2 A+A-3 A=I+3 A-3 A=I \end{aligned}
So, $\left ( I+A \right )^{2}-3A=I$

Algebra of Matrices Exercise Very Short Asnwer Question 45

Answer: $B=\begin{bmatrix} 1 & 1\\ 1 & 3 \end{bmatrix}$
Hint: Here we use the basic concept of symmetric and skew symmetric matrix
Given:
$A=\left [ 1\: 2\: 0\: 3 \right ]$ is written as
B+C where B is symmertric and C is skew symmetric matrix Find B
Solution:
$A=B+C$ .....(i) $\left [ B=B^{T}symmetric C^{T}=-C skew symmetric \right ]$
$\rightarrow Let\: us\: take\: the\: transpose\: of\: eqn(i)A^{T}=\left ( B+C \right )^{T}A^{T}=B^{T}+C^{T}$
$\! \! \! \! \! \! \! \because\left((A+B)^{T}=A^{T}+B^{T}\right) A^{T}=B-C \quad \ldots .(i i) \text { From }(i) \text { and }(i i) B=\frac{1}{2}\left(A+A^{T}\right) C \\$
$=\frac{1}{2}\left(A-A^{T}\right) B=\frac{1}{2}\left ( \left [ 1\: 2\: 0\: 3 \right ]+\left [ 1\: 0\: 2\: 3 \right ]\right )B=\frac{1}{2}\left [ 1+1\: 2+0\: 0+2\: 3+3 \right ]$
$=\frac{1}{2}\left [ 2\: 2\: 2\: 6 \right ]=\left [ 1\: 1\: 1\: 3 \right ]$

Algebra of Matrices Exercise Very Short Asnwer Question 46

Answer:$B \: is \: 2\times3$ order matrix
Hint: Here, we use the basic concept of transpose of matrix
Given:$A \: is \: 2\times3$ matrix and $B$ is matrix such that $A^{T}B\: and \: BA^{T}$both are defined then what is order of $B$
Solution:
$A \: is \: 2\times3$ matrix
So,
$A^{T}$ is $3\times 2$ matrix
Here, $A^{T}B$ is exist
$=A^{T} _{3\times2}\: B_{m\times n}$
So,$m=2$ in $B$
Here $BA^{T}$ is exist
$=B_{2\times n}\: A^{T}_{3\times 2}$
So,$n=3$ in $B$
So,$B$ is $2\times3$ matrix

Algebra of Matrices Exercise Very Short Asnwer Question 47

Answer: Number of possible matrices = 16
Hint: Here, we use the basic concept of matrix order
Given:$2\times2$ matrix
Solution: The order of matrix = $2\times2$
Number of elements = 4
Number of entries = $2\left [ 0\: or \: 1 \right ]$
Number of possible matrices = $2^{n}$
$\! \! \! \! \! \! 2^{4}=2\times2\times2\times2\\=4\times4\\=16$

Algebra of Matrices Exercise Very Short Asnwer Question 48

Answer:$y=2$
Hint: Here we use the basic concept of union algebraic matrix.
Given:$\begin{bmatrix} x & x-y\\ 2x+y& 7 \end{bmatrix}=\begin{bmatrix} 3 & 1\\ 8 & 7 \end{bmatrix}$
Solution:
Here both are $2\times2$ matrix
So,
$\! \! \! \! \! \! x=3\; \; \; \; ......(i)\\\\x-y=1\\\\3-y=1$[From eqn (i)]
$3-1=y$
$y=2$ ...(ii)
$\rightarrow$ Let's check both value
$2x+y=8$
LHS = $2x+y$
$=2\left ( 3 \right )+2$
$=6+2$
$=8$
=RHS

4 Algebra of Matrices Exercise Very Short Asnwer Question 49

Answer:$m_{\left ( 1\times5 \right )} \: or \: m_{\left ( 5\times1 \right )}$
Hint: Here we use the basic fundamental concept of matrix order
Given: If a matrix has 5 elements, write all possible orders it can have
Solution:
$A_{m\times n}\\=number\: of\: elements\\=m\times n\\=mn\\=5\times 1\\=1\times5$
So, Possible order is $\left ( 1\times5 \right )$or $\left ( 5\times1 \right )$

Algebra of Matrices Exercise Very Short Asnwer Question 50

Answer:$a_{12}=\frac{1}{2}$
Hint: Here we can use the basic concept of order of matrix
Given:
2×2 matrix, A=an, whose elements are given by $a_{n}=\frac{i}{j}$
Solution:$a_{12}$
Here,$i=1$
$j=2$
So,$a_{12}=\frac{i}{j}$
$a_{12}=\frac{1}{2}$

4 Algebra of Matrices Exercise Very Short Asnwer Question 51

Answer:$x=3\: and\: y=-4$
Hint: Here we use basic concept of scalar matrix
Given:$x\begin{bmatrix} 2\\3 \end{bmatrix}+y\begin{bmatrix} -1\\1 \end{bmatrix}=\begin{bmatrix} 10\\5 \end{bmatrix}x=? y=?$
Solution:
\begin{aligned} &\text { Here, we use scalar matrix }\\ &\left[\begin{array}{c} 2 x \\ 3 x \end{array}\right]+\left[\begin{array}{c} -y \\ y \end{array}\right]=\left[\begin{array}{l} 10 \\ 5 \end{array}\right]\\ &\left[\begin{array}{c} 2 x-y \\ 3 x+y \end{array}\right]=\left[\begin{array}{l} 10 \\ 5 \end{array}\right]\\ &2 x-y=10\\ &\frac{3 x+y=5}{5 x=15} \end{aligned}
\begin{aligned} &5 x=15 \\ &\begin{array}{l} x=3 \\ 2 x-y=10 \\ 2(3)-y=10 \quad \rightarrow[\text { From eqn }(1)] \\ 6-y=10 \\ y=6-10 \\ y=-4 \end{array} \end{aligned}

4 Algebra of Matrices Exercise Very Short Asnwer Question 52

Answer:$A=\begin{bmatrix} 8 &-3 &5 \\ -2 & -3 & -6 \end{bmatrix}$
Hint: Here we use basic concept of subtraction of two matrix
Given: $\begin{bmatrix} 9 &-1 &4 \\ -2 & 1 & 3 \end{bmatrix}=A+\begin{bmatrix} 1 &2 &-1 \\ 0 & 4 & 9 \end{bmatrix}$
Solution:
\begin{aligned} &A=\left[\begin{array}{ccc} 9 & -1 & 4 \\ -2 & 1 & 3 \end{array}\right]-\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 4 & 9 \end{array}\right] \\ &=\left[\begin{array}{ccc} 9-1 & -1-2 & 4-(-1) \\ -2-0 & 1-4 & 3-9 \end{array}\right] \\ &=\left[\begin{array}{ccc} 8 & -3 & 5 \\ -2 & -3 & -6 \end{array}\right] \end{aligned}

Algebra of Matrices Exercise Very Short Asnwer Question 53

Answer:$b=2$
Hint: Here we use basic concept of algebra
Given:$\begin{bmatrix} a-b & 2a+c\\ 2a-b &3c+d \end{bmatrix}=\begin{bmatrix} -1 &5 \\ 0 & 13 \end{bmatrix}$
Solution:
Here both sides has $2\times2$ matrix
So,we can write
$a-b=-1$ ....1i)
$2a-b=0$ .....(2)
$\rightarrow$Let's multiply (-2) with eqn (1)
$-2a+2b=+2$
$\! \! \! \! \! \! \! \! \! \! 2a-b=0\\ \overline{0+b=2}$
$b=2$

Algebra of Matrices Exercise Very Short Asnwer Question 54

Answer: $x=2$
Hint: Here we use basic concept of skew symmetric matrix
Given:$A\left[ 0\: 1 -2-1\: 0-3\: x-3\: 0 \right] \\$ is skew symmetric matrix. Find x
Solution:
Here $A=-A^{T}\left[ 0\: 1 -2-1\: 0-3\: x-3\: 0 \right] \\$
$=-1[0\: -1 \times 10\: -3-2-3\: 0][0\: 1-2-1\: 0-3 x-3\: 0] \\ =[0\: 1-x-1-x\: 3\: 2\: 3\: 0]-2=-x x=2 \: x=2$

Algebra of Matrices Exercise Very Short Asnwer Question 55

Answer: $p=4$
Hint: Here we use basic concept of matrix
Given: $A=\begin{bmatrix} 2 & -2\\ -2 &2 \end{bmatrix}$ and $A^{2}=pA$ Find $p$
Solution:
\begin{aligned} &\text { Here } A^{2}=A \times A \\ &=\left[\begin{array}{cc} 2 & -2 \\ -2 & 2 \end{array}\right] \times\left[\begin{array}{cc} 2 & -2 \\ -2 & 2 \end{array}\right] \\ &=\left[\begin{array}{cc} 4+4 & -4-4 \\ -4-4 & 4+4 \end{array}\right] \\ &A^{2}=\left[\begin{array}{cc} 8 & -8 \\ -8 & 8 \end{array}\right] \\ &A^{2}=p A \\ &{\left[\begin{array}{cc} 8 & -8 \\ -8 & 8 \end{array}\right]=p\left[\begin{array}{cc} 2 & -2 \\ -2 & 2 \end{array}\right]} \\ &\begin{array}{ll} 4\left[\begin{array}{cc} 2 & -2 \\ -2 & 2 \end{array}\right]=p\left[\begin{array}{cc} 2 & -2 \\ -2 & 2 \end{array}\right] \\ p =4 \end{array} \end{aligned}

Algebra of Matrices Exercise Very Short Asnwer Question 56

Answer: $7A\left ( I+A \right )^{3}=-I$
Hint: Here we use basic concept of square and identity matrix
Given:
$A$ is square matrix $A^{2}=A$
$7A\left ( I+A \right )^{3}=?$
Solution:
\begin{aligned} &7 A-(I+A)^{3} \\ &=7 A-\left[I^{3}+A^{3}+3 I A(I+A)\right] \\ &=7 A-\left[I+A^{3}+3 A(I+A)\right] \\ &=7 A-\left[I+A^{3}+3 A I+3 A^{2}\right] \\ &=7 A-\left[I+A^{3}+3 A+3 A^{2}\right] \\ &=7 A-I-A^{3}-3 A-3 A^{2} \\ &=4 A-I-A^{3}-3 A^{2} \\ &=4 A-3 A-\left[A^{2} \times A\right]-I \quad\left[A^{2}=A\right] \\ &=4 A-3 A-[A \times A]-I \\ &=4 A-3 A-A-I \\ &=4 A-4 A-I \\ &7 A-(I+A)^{3}=-I \end{aligned}

Algebra of Matrices Exercise Very Short Asnwer Question 57

Answer:$x = 2 , y = -8$
Hint: Here we use basic concept of sum of matrix
Given:$2\begin{bmatrix} 3 &4 \\ 5 & x \end{bmatrix}+\begin{bmatrix} 1 &y \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 7 & 0\\ 10 &5 \end{bmatrix}$
Solution:
Here both sides has 2×3matrix$2\left [3 \: 4\: 5 \: x \right ]+\left [ 1\: y\: 0\: 1 \right ]=\left [ 7\: 0\: 10\: 5 \right ]\left [ 6\: 8\: 10\: 2x \right ]+\left [ 1\: y\: 0\: 1 \right ]$
$=\left [ 7\: 0\: 10\: 5 \right ] So,y+8=0y=-82x+1=52x=4x=2\: So,x-y$
$= -2-\left ( -8 \right )x-y=10$

Algebra of Matrices Exercise Very Short Asnwer Question 58

Answer: $x=2$
Hint: Here we use basic multiplication algebric matrix’s concept
Given:$\begin{bmatrix} x &1 \end{bmatrix}\begin{bmatrix} 10& -20 \end{bmatrix}=0$ find x
Solution:
Here 1st matrix has 2 columns and 2nd matrix has 2 rowsSo, multiplication is possible
$\begin{bmatrix} x &1 \end{bmatrix}\begin{bmatrix} 10& -20 \end{bmatrix}=0\begin{bmatrix} x\times 1+\left ( -2 \right )0 +0 \end{bmatrix}=0\begin{bmatrix} x -20 \end{bmatrix}_{1\times2}=\begin{bmatrix} 0 &0 \end{bmatrix}_{1\times2}\begin{bmatrix} x & -20 \end{bmatrix}=0x-2=0x=2$

4 Algebra of Matrices Exercise Very Short Asnwer Question 59

Answer:$a-2b=0$
Hint: Here we use basic concept of matrix
Given: $\begin{bmatrix} a+4& 3b\\ 8 & -6 \end{bmatrix}\begin{bmatrix} 2a+2& b+2\\ 8 & a-8b \end{bmatrix}=0$ find $a-2b$
Solution:
$So,\\ a+4=2 a+2\\ 2 a-a=4-2\\ a=2 \quad \ldots \ldots .(1)\\ 3 b=b+2\\ 2 b=2\\ b=1 \quad \ldots \ldots(2)\\a-2b=2-2(1)\; \; \; \; \; \; \; \; \; \; \; \; \; \; From (1)\: and(2)\\=2-2\\a-2b=0$

Algebra of Matrices Exercise Very Short Asnwer Question 60

Answer:$A=\begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$
Hint: Here we use basic concept of symmetric and skew symmetric matrix
Given:$2\times2$matrix is symmetric and skew symmetric matrix
Solution:
$A^{T}=A$ (A is symmetric matrix)
$A^{T}=-A$ (A is skew symmetric matrix)
$\overline{A^{T}-A^{T}=A-\left ( -A \right )}$
$0=2A$
$A=0$
$A=\begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix}$

Algebra of Matrices Exercise Very Short Asnwer Question 61

Answer:$x+y+z=0$
Hint: Here we use basic concept of matrix
Given:$\begin{bmatrix} xy & 4\\ z+6 & x+y \end{bmatrix}=\begin{bmatrix} 8 &w \\ 0& 6 \end{bmatrix}=0$ find $x+y+z$
Solution:
Both sides has $2\times2$ order of matrix
So,
$z+6=0$
$z=-6$ ........(1)
$x+y=6$ ...........(2)
So,
$x+y+z$
$=6+(-6)$ [From eqn (1) and (2)]
$x+y+z=0$

4 Algebra of Matrices Exercise Very Short Asnwer Question 62

Answer:$A=\begin{bmatrix} 4\frac{1}{2}\: \frac{5}{2}16 \\ \end{bmatrix}_{2\times2}$
Hint: Here we use basic concept of matrix
Given:$A=a_{ij} \; \; a_{ij}=\frac{1}{2}\left [ -3i+j \right ] if \: i\neq j; a_{ij}=(i+j)^{2} if\: i=j$
Solution:
$\! \! \! \! \! \! \! \! A=\begin{bmatrix} a_{11} &a_{12} &a_{21} &a_{22} \end{bmatrix}Given \: a_{ij}=\frac{1}{2}\begin{bmatrix} -3i +j \end{bmatrix}if\: i\neq j;a_{ij}=(i+j)^{2}if\: i=ja_{11}=\begin{bmatrix} 1+1 \end{bmatrix}^{2}$
$\! \! \! \! \! =4a_{ij}=\frac{1}{2}\begin{bmatrix} -3(1)+2 \end{bmatrix}=\frac{1}{2}a_{21}=\frac{1}{2}\begin{bmatrix} -3(2) +1\end{bmatrix}=\frac{5}{2}a_{22}=\begin{bmatrix} 2+2 \end{bmatrix}^{2}=16A$
$=\begin{bmatrix} a_{11} &a_{12} &a_{21} &a_{22} \end{bmatrix}=\begin{bmatrix} 4\frac{1}{2}\: \frac{5}{2}16 \\ \end{bmatrix}_{2\times2}$

4 Algebra of Matrices Exercise Very Short Asnwer Question 63

Answer:$\left ( x,y \right ) =\left ( -1,1 \right )$
Hint: Here we use basic multiplication concept
Given:
$\begin{bmatrix} x+y \\ x-y \end{bmatrix}= \begin{bmatrix} 2& 1\\ 4 &3 \end{bmatrix}\begin{bmatrix} 1\\-2 \end{bmatrix}$ find $\left ( x,y \right )$
Solution:
Here in RHS In both matrix 1st has 2 coloumn and 2nd has row
So,multiplication is possible
$\rightarrow$So, Let's do it
\begin{aligned} &\left[\begin{array}{l} x+y \\ x-y \end{array}\right]=\left[\begin{array}{l} 2-2 \\ 4-6 \end{array}\right]\\ &\left[\begin{array}{c} x+y \\ x-y \end{array}\right]_{2 \times 1}=\left[\begin{array}{c} 0 \\ -2 \end{array}\right]_{1 \times 2}\\ &x+y=0\\ &\frac{x-y=-2}{2 x=-2}\\ &\begin{array}{|l|l|} \hline x=-1 \\ \hline \end{array}\\ &x+y=0\\ &-1+y=0 \end{aligned}

4 Algebra of Matrices Exercise Very Short Asnwer Question 64

Answer:$a=\frac{-2}{3} \: and \: b=\frac{3}{2}$
Hint: Here we use basic concept of matrix
Given:$A= \left [ 0 \: 2b -2\: 3\: 1 \: 3 \: 3a\: 3 -1 \right ]$ is symmetric
Solution:
$A^{T}=A \: as \: A$ is symmetric$\left [ 0 \: 3 \: 3a\: 2b \: 1\: 3 \: -2\: 3\: -1 \right ]$$= \left [ 0 \: 2b\: -2\: 3 \: 1\: 3\: 3a\: 3 -1 \right ]$
So,
$3a=-2\: and\: 2b=3$
$a=\frac{-2}{3}\; \; \; \; \; b=\frac{3}{2}$

4 Algebra of Matrices Exercise Very Short Asnwer Question 65

Answer: Possible matrix is 81 ways
Hint: Here we use basic concept of matrix algebra
Given: All possible matrix $2\times2$with entry 1,2,3
Solution:
$Let \begin{bmatrix} a & b\\ c & d \end{bmatrix}_{2\times2}$
$\! \! \! \! \! \! \! \! So, a=1,2\: or\: 3=3 ways\\ b=1,2\: or\: 3=3 ways\\ c=1,2\: or\: 3=3 ways\\ d=1,2\: or \: 3=3 ways\\ Total \: possible \: ways =3^{4}\\ =81\: ways$

4 Algebra of Matrices Exercise Very Short Asnwer Question 66

Answer:$A=[2],\left ( 1\times1 \right )$ matrix
Hint: Here we use basic of algebra
Given: $[2 1 3]$ $[ -1\: 0 -1 -1 \: 1\: 0\: 0 \: 1\: 1]$ $[1 \: 0 -1 ]=A$ find $A$
Solution:
Here 1st has 3 columns and 2nd has 3 rows , 2nd has 3 columns and 3rd has 3 rowsSo, multiplication is possible
$A=[2 1 3]$ $[ -1\: 0 -1 -1 \: 1\: 0\: 0 \: 1\: 1]$ $[1 \: 0 -1 ]A$
$=[-2-1+0\: 0+1+3-2+0-3][1\: 0-1]$
$A=[-3\: 4-5] [1\: 0-1]A=[-3+0+5]$
$A=[2]$

4 Algebra of Matrices Exercise Very Short Asnwer Question 67

Answer:$P=\begin{bmatrix} 3 &6 \\ 6& 9 \end{bmatrix}$
Hint: Here we use basic of matrix
Given:$A=\begin{bmatrix} 3 &5 \\ 7& 9 \end{bmatrix}A=P+Q$
Solution:
$A=P+Q \: \: \: \: \: \cdot \cdot \cdot \cdot (1)$
Let's take tranpose of equation (1)
\begin{aligned} A^{T} &=(P+Q)^{T} \\ A^{T} &=P^{T}+Q^{T} \\ A^{T} &=P-Q \quad \text { ......(2) } \quad\left(P^{T}=\mathrm{P} \text { and } Q^{T}=-\mathrm{Q}\right) \end{aligned}
From eqn (1) and (2)
\begin{aligned} P &=\frac{1}{2}\left(A+A^{T}\right) \\ P &=\frac{1}{2}\left(\left[\begin{array}{ll} 3 & 5 \\ 7 & 9 \end{array}\right]+\left[\begin{array}{ll} 3 & 7 \\ 5 & 9 \end{array}\right]\right)=\frac{1}{2}\left[\begin{array}{cc} 6 & 12 \\ 12 & 18 \end{array}\right]=\left[\begin{array}{ll} 3 & 6 \\ 6 & 9 \end{array}\right] \end{aligned}

4 Algebra of Matrices Exercise Very Short Asnwer Question 68

Answer:$AB$is order of $3\times4$

Hint: Here we use basic of order of matrix
Given: Find order of $AB$
Solution:
$A=A_{[3\times2]}$
$B=B_{[2\times4]}$
Here in both A's column is same as B's rows
So multiplication is possible
$A\times B=A_{[3\times 2]}\times B_{[2\times4]}=AB_{[3\times4]}$

Answer: $a=-2\: and\: b=3$
Hint: Here we use basic of matrix
Given:$A=\begin{bmatrix} 0 & a &-3 \\ 2& 0 &-1 \\ b& 1 & 0 \end{bmatrix}$ is skew symmetric find a and b
Solution:
$A^{T}=-AA=-1[0 \: 2\: b \: a\: 0\: 1 -3 -1\: 0] A=[0 -2 -b -a\: 0 -1\: 3 \: 1\: 0 ]$
$[ 0\: a -3 \: 2 \: 0 -1\: b \: 1 \: 0]= [0 -2 -b -a\: 0 -1 \: 3 \: 1 \: 0]a=-2b=3$

The Chapter Algebra of matrices is not challenging for most students, yet the absence of an excellent resource to verify the answers is entirely missing. In RD Sharma Class 12th VSA 4th chapter in mathematics, the concept of matrices is dealt with in a detailed manner. These concepts include Addition, Subtraction, Multiplication of matrices, Inverse of Matrices, Transpose Matrices, Directions in Matrices using vectors, Skew matrices, Matrix Determinants, and Rank of Matrices. All these topics are asked in the RD Sharma Class 12th VSA section. There are sixty-nine questions in this section. Hence, the RD Sharma Class 12 Chapter 4 VSA solutions can be referred to.

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Chapter-wise RD Sharma Class 12 Solutions

### Frequently Asked Questions (FAQs)

1. Where can I verify my VSA of Chapter 4 class 12 mathematics?

You can very well refer to the RD Sharma Class 12th VSA solutions book to cross-check or find answers for it. The answers are provided with many tricks that save your time in finding the solutions quickly.

2. Where can I get a free copy of the class 12 RD Sharma chapter 4 exercise FBQ Solutions book?

Anyone can download the RD Sharma Solution books from the Career360 website for free. All you need to do is, visit the website and search for the solutions book according to the subject and grade.

3. Is it enough to write the VSA at my public exams as given in the RD Sharma solutions book?

The experts provide the answers in the RD Sharma Class 12 Chapter 4 VSA. Hence, you can proceed to write the answers in the same way as given in the book.

4. Are the solutions given in the RD Sharma books easier for every student to understand?

The book consists of many methods of finding the solution for a mathematical question. Therefore students who are toppers as well as the average learners can explore and find the methods that they feel are easy to adapt.

5. Do the RD Sharma books follow the NCERT pattern?

Yes, the RD Sharma books follow the NCERT pattern. Hence, it becomes useful for CBSE students to use it for their exam preparations.

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