RD Sharma Class 12 Exercise 4.1 Algebra of Matrices Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 4.1 Algebra of Matrices Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 20, 2022 03:39 PM IST

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

## RD Sharma Class 12 Solutions Chapter 4 Algebra of Matrices - Other Exercise

Algebra of Matrices Exercise 4.1

Algebra of Matrices Exercise 4.1 Question 1

Answer: 1st part: $\left ( 8\times 1 \right ),\left ( 1\times 8 \right ),\left ( 4\times 2 \right ),\left ( 2\times 4 \right )$ and for the 2nd part: $\left ( 5\times 1 \right ),\left ( 1\times 5 \right )$
Given: matrix has 8 elements
Here we have to find the possible order of the given matrix
Hint: If the matrix is of the order $m\times n$ elements, it has $m n$ elements.
Solution: If the matrix has 8 elements, we will find the ordered pairs $m$and $n$.
$m\times n = 8$
Then ordered pairs $m$ and $n$ will be $m\times n$ be $\left ( 8\times 1 \right ),\left ( 1\times 8 \right ),\left ( 4\times 2 \right ),\left ( 2\times 4 \right )$
Now, if it has 5 elements then possible orders are .$\left ( 5\times 1 \right ),\left ( 1\times 5 \right )$

Algebra of Matrices Exercise 4.1 Question 2 (i)

Answer:$a_{22}+b_{21}= 1$
Given:
$A= \left [ a_{ij} \right ]= \begin{bmatrix} 2 & 3 &-5 \\ 1& 4 & 9 \\ 0& 7 &-2 \end{bmatrix}$ and $B= \left [ b_{ij} \right ]= \begin{bmatrix} 2 &-1 \\ -3& 4\\ 1& 2 \end{bmatrix}$
Here we have to find out the values of $a_{22}+b_{21}$
Hint:
Simply we select the elements in the matrix which elements required and simplify
Solution: We know that
$A= \left [ a_{ij} \right ]= \begin{bmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23}\\ a_{31}& a_{32} & a_{33} \end{bmatrix}$ $\cdot \cdot \cdot \left ( i \right )$
$B= \left [ b_{ij} \right ]= \begin{bmatrix} b_{11} &b_{12} \\ a_{21} & a_{22} \\ a_{31}& a_{32} \end{bmatrix}$ $\cdot \cdot \cdot \left ( ii \right )$
Also given that
$A= \left [ a_{ij} \right ]= \begin{bmatrix} 2 & 3 &-5 \\ 1& 4 & 9 \\ 0& 7 &-2 \end{bmatrix}$ and $B= \left [ b_{ij} \right ]= \begin{bmatrix} 2 &-1 \\ -3& 4\\ 1& 2 \end{bmatrix}$
Now comparing with eqn (i) and (ii) we have,
$a_{22}= 4$, $b_{21}= -3$
Hence $a_{22}+b_{21}= 4+\left ( -3 \right )= 1$

Algebra of Matrices Exercise 4.1 Question 2 (i)

Answer:$20$
Given: Here given that
$A= \left [ a_{ij} \right ]= \begin{bmatrix} 2 & 3 &-5 \\ 1& 4 & 9 \\ 0& 7 &-2 \end{bmatrix}$ and $B= \left [ b_{ij} \right ]= \begin{bmatrix} 2 &-1 \\ -3& 4\\ 1& 2 \end{bmatrix}$
Here we have to find out the values of $a_{11}b_{11}+a_{22}b_{22}= 1$
Hint: Simply we select the elements in the matrix which elements required and simplify
Solution: We know that
$A= \left [ a_{ij} \right ]= \begin{bmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23}\\ a_{31}& a_{32} & a_{33} \end{bmatrix}$ $\cdot \cdot \cdot \left ( i \right )$
$B= \left [ b_{ij} \right ]= \begin{bmatrix} b_{11} &b_{12} \\ a_{21} & a_{22} \\ a_{31}& a_{32} \end{bmatrix}$ $\cdot \cdot \cdot \left ( ii \right )$
Also given that
$A= \left [ a_{ij} \right ]= \begin{bmatrix} 2 & 3 &-5 \\ 1& 4 & 9 \\ 0& 7 &-2 \end{bmatrix}$ and $B= \left [ b_{ij} \right ]= \begin{bmatrix} 2 &-1 \\ -3& 4\\ 1& 2 \end{bmatrix}$
Now comparing with eqn(i) and (ii) we have,
$a_{11}= 2$ $a_{22}= 4$
$b_{11}= 2$ $b_{22}= 4$
Hence, $a_{11}b_{11}+a_{22}b_{22}= 2\times 2+4\times 4$
$= 4+16$
$a_{11}b_{11}+a_{22}b_{22}= 20$

Algebra of Matrices Exercise 4.1 Question 3

Answer: The order of matrix $R_{1}= 1\times 4$
And the order of matrix $C_{2}= 3\times 1$
Given: $A$ be a matrix of order $3\times 4$
Here we have to determine the order of matrices $R_{1}$ and $C_{2}$
Hint: We have to write the order along row and again along column
Solution: Here $A$ be a matrix of order $3\times 4$
So, $A= \left [ a_{ij} \right ]_{3\times 4}$
$R_{1}$= first row of $A= \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \end{bmatrix}$
So, order of Matrix $R_{1}= 1\times 4$
Again, $C_{2}$ = second column of
$A= \begin{bmatrix} a_{12}\\ a_{22}\\ a_{32} \end{bmatrix}$
Therefore, order of $C_{2}= 3\times 1$ .

Algebra of Matrices Exercise 4.1 Question 4 (i)
Answer: $a_{ij}$$= i\times j$,$A= \begin{bmatrix} 1 &2 &3 \\ 2& 4& 6 \end{bmatrix}$

Given: Here given that matrix of order $2\times 3$
$A= \left [ a_{ij} \right ]_{2\times 3}$
Here we have to construct$2\times 3$ matrix as $a_{ij}$$= i\times j$

Hint: We have to construct the matrix according to the question
Solution: Given $a_{ij}$$= i\times j$
Let $A= \left [ a_{ij} \right ]_{2\times 3}$
So, the elements in a $2\times 3$ matrix are $a_{11},a_{12},a_{13},a_{21},a_{22},a_{23}$
$A= \begin{bmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= 1\times 1= 1\\a_{12}= 1\times 2= 2\\a_{13}= 1\times 3= 3$ $\! \! \! \! \! \! \! \! \! a_{21}= 2\times 1= 2\\a_{22}= 2\times 2= 4\\a_{23}= 2\times 3= 6$
Substituting these values in Matrix $A$, we get
$A= \begin{bmatrix} 1 &2 &3 \\ 2& 4& 6 \end{bmatrix}$
Hence this is the required answer.

Algebra of Matrices Exercise 4.1 Question 4 (ii)

Answer: $A= \begin{bmatrix} 1 &0 &-1 \\ 3& 2& 1 \end{bmatrix}$
Given: Here given that matrix of order $2\times 3$
$A= \left [ a_{ij} \right ]_{2\times 3}$
Here we have to construct $2\times 3$ matrix as $a_{ij}= 2i-j$
Hint: We have to construct the matrix according to the question
Solution: Let $A= \left [ a_{ij} \right ]_{2\times 3}$
So, the elements in a $2\times 3$ matrix are $a_{11},a_{12},a_{13},a_{21},a_{22},a_{23}$
$A= \begin{bmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= 2\times 1-1= 2-1= 1\\a_{12}= 2\times 1-2= 2-2= 0\\a_{13}= 2\times 1- 3= 2-3= -1$ $\! \! \! \! \! \! \! \! \! a_{21}= 2\times 2-1= 4-1= 3\\a_{22}= 2\times 2-2= 4-2= 2\\a_{23}= 2\times 2- 3= 4-3= 1$
Substituting these values in Matrix $A$, we get
$A= \begin{bmatrix} 1 &0 &-1 \\ 3& 2& 1 \end{bmatrix}$
Hence this is the required answer.

Algebra of Matrices Exercise 4.1 Question 4 (iii)

Answer: $A= \begin{bmatrix} 2 &3 &4 \\ 3& 4& 5 \end{bmatrix}$
Given: Here given that matrix of order $2\times 3$
$A= \left [ a_{ij} \right ]_{2\times 3}$
Here we have to construct the matrix according to $a_{ij}= 2i+j$
Hint: First we have to simply adding the row elements with column element as given
question and then construct matrix.
Solution: Let $A= \left [ a_{ij} \right ]_{2\times 3}$
So, the elements in a $2\times 3$ matrix are $a_{11},a_{12},a_{13},a_{21},a_{22},a_{23}$
$A= \begin{bmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= 1+ 1= 2\\a_{12}= 1+2= 3\\a_{13}= 1+3= 4$ $\! \! \! \! \! \! \! \! \! a_{21}= 2+ 1= 3\\a_{22}= 2+ 2= 4\\a_{23}= 2+ 3= 5$
Substituting these values in Matrix $A$, we get
$A= \begin{bmatrix} 2 &3 &4 \\ 3& 4& 5 \end{bmatrix}$
Hence this is the required answer.

Algebra of Matrices Exercise 4.1 Question 4 (iv)

Answer:$A= \begin{bmatrix} 2 &4.5 &8 \\ 4.5& 8& 12.5 \end{bmatrix}$
Given: Here given that matrix of order $2\times 3$
$A= \left [ a_{ij} \right ]_{2\times 3}$
Here we have to construct the matrix according to $a_{ij}= \frac{\left ( i+j \right )^{2}}{2}$
Hint: Adding row and column element and squaring then divide by 2
Solution: Let $A= \left [ a_{ij} \right ]_{2\times 3}$ and given that $a_{ij}= \frac{\left ( i+j \right )^{2}}{2}$
So, the elements in a $2\times 3$ matrix are $a_{11},a_{12},a_{13},a_{21},a_{22},a_{23}$
$A= \begin{bmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} & a_{22} & a_{23} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= \frac{\left ( 1+1 \right )^{2}}{2}=\frac{4}{2} = 2\\a_{12}=\frac{\left ( 1+2 \right )^{2}}{2}=\frac{\left ( 3 \right )^{2}}{2} = \frac{9}{2}= 4.5\\a_{13}=\frac{\left ( 1+3 \right )^{2}}{2}= \frac{\left ( 4 \right )^{2}}{2}= \frac{16}{2}= 8$ $\! \! \! \! \! \! \! \! \! a_{21}=\frac{\left ( 2+ 1 \right )^{2}}{2}= \frac{\left ( 3 \right )^{2}}{2}= \frac{9}{2}= 4.5\\a_{22}=\frac{\left ( 2+ 2 \right )^{2}}{2} = \frac{\left ( 4 \right )^{2}}{2}= \frac{16}{2}= 8\\a_{23}=\frac{\left ( 2+ 3 \right )^{2}}{2} = \frac{\left ( 5 \right )^{2}}{2}= \frac{25}{2}= 12.5$
Substituting these values in Matrix $A$, we get
$A= \begin{bmatrix} 2 &4.5 &8 \\ 4.5& 8& 12.5 \end{bmatrix}$
Hence this is the required answer.

Algebra of Matrices Exercise 4.1 Question 5 (i)

Answer: $A= \begin{bmatrix} 2 &\frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}$
Given: $\frac{\left ( i+j \right )^{2}}{2}$
Here we have to construct $2\times 2$ matrix according to $\frac{\left ( i+j \right )^{2}}{2}$
Hint: Substitute required values in the $2\times 2$ matrix
Solution: Let $A= \left [ a_{ij} \right ]_{2\times 2}$
So, the elements in a $2\times 2$ are $a_{11},a_{12},a_{21},a_{22}$
$A= \begin{bmatrix} a_{11} &a_{12} \\ a_{21} & a_{22} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= \frac{\left ( 1+1 \right )^{2}}{2}= \frac{4}{2}= 2\\a_{12}= \frac{\left ( 1+2 \right )^{2}}{2}= \frac{\left ( 3 \right )^{2}}{2}= \frac{9}{2}= 4.5$ $\! \! \! \! \! \! \! \! \! a_{21}= \frac{\left ( 2+1 \right )^{2}}{2}=\frac{\left ( 3 \right )^{2}}{2} = \frac{9}{2}= 4.5\\a_{22}= \frac{\left ( 2+2 \right )^{2}}{2}= \frac{\left ( 4 \right )^{2}}{2}= \frac{16}{2}= 8$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} 2 &\frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 5 (ii)

Answer:$A= \begin{bmatrix} 0 &\frac{1}{2} \\ \frac{1}{2} & 0 \end{bmatrix}$
Given: $\frac{\left ( i-j \right )^{2}}{2}$
Here we have to construct $2\times 2$ matrix according to $\frac{\left ( i-j \right )^{2}}{2}$
Hint: Substitute required values in the $2\times 2$ matrix
Solution: Let $A= \left [ a_{ij} \right ]_{2\times 2}$
So, the elements in a $2\times 2$ are $a_{11},a_{12},a_{21},a_{22}$
$A= \begin{bmatrix} a_{11} &a_{12} \\ a_{21} & a_{22} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= \frac{\left ( 1-1 \right )^{2}}{2}= 0\\a_{12}= \frac{\left ( 1-2 \right )^{2}}{2}= \frac{\left ( 1 \right )^{2}}{2}= \frac{1}{2}$ $\! \! \! \! \! \! \! \! \! a_{21}= \frac{\left ( 2-1 \right )^{2}}{2}=\frac{1 }{2}\\a_{22}= \frac{\left ( 2-2 \right )^{2}}{2}=0$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} 0 &\frac{1}{2} \\ \frac{1}{2} & 0 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 5 (iii)

Answer: $A= \begin{bmatrix} \frac{1}{2} &\frac{9}{2} \\ \0 & 2 \end{bmatrix}$
Given:$a_{ij}= \frac{\left ( i-2j \right )^{2}}{2}$
Here we have to construct $2\times 2$ matrix according to $\frac{\left ( i-2j \right )^{2}}{2}$
Hint: Substitute required values in the $2\times 2$ matrix
Solution: Let $A= \left [ a_{ij} \right ]_{2\times 2}$
So, the elements in a $2\times 2$ are $a_{11},a_{12},a_{21},a_{22}$
$A= \begin{bmatrix} a_{11} &a_{12} \\ a_{21} & a_{22} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= \frac{\left ( 1-2\times 1 \right )^{2}}{2}= \frac{-1^{2}}{2}= \frac{1}{2}\\a_{12}= \frac{\left ( 1-2\times 2 \right )^{2}}{2}= \frac{\left ( -3 \right )^{2}}{2}= \frac{9}{2}$ $\! \! \! \! \! \! \! \! \! a_{21}= \frac{\left ( 2-1 \times 1\right )^{2}}{2}=\frac{\left ( 0 \right )^{2}}{2} = 0\\a_{22}= \frac{\left ( 2-2 \times 2\right )^{2}}{2}= \frac{ -2 ^{2}}{2}= \frac{4}{2}= 2$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} \frac{1}{2} &\frac{9}{2} \\ \0 & 2 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 5 (iv)

Answer: $A= \begin{bmatrix} \frac{9}{2} &8 \\ \\ \frac{25}{2} & 18 \end{bmatrix}$
Given: $a_{ij}= \frac{\left ( 2i+j \right )^{2}}{2}$
Here we have to construct $2\times 2$ matrix according to $\frac{\left ( 2i+j \right )^{2}}{2}$
Hint: Substitute required values in the $2\times 2$ matrix
Solution: Let $A= \left [ a_{ij} \right ]_{2\times 2}$
So, the elements in a $2\times 2$ are $a_{11},a_{12},a_{21},a_{22}$
$A= \begin{bmatrix} a_{11} &a_{12} \\ a_{21} & a_{22} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= \frac{\left ( 2\times 1+1 \right )^{2}}{2}= \frac{3^{2}}{2}= \frac{9}{2}\\a_{12}= \frac{\left ( 2\times 1+2 \right )^{2}}{2}= \frac{\left ( 4 \right )^{2}}{2}= \frac{16}{2}= 8$ $\! \! \! \! \! \! \! \! \! a_{21}= \frac{\left ( 2 \times 2+1\right )^{2}}{2}=\frac{\left ( 5 \right )^{2}}{2} = \frac{25}{2}\\a_{22}= \frac{\left ( 2 \times 2+2\right )^{2}}{2}= \frac{ 6 ^{2}}{2}= \frac{36}{2}= 18$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} \frac{9}{2} &8 \\ \\ \frac{25}{2} & 18 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 5 (v)

Answer:$A= \begin{bmatrix} \frac{1}{2} &2 \\ \\ \frac{1}{2} & 1 \end{bmatrix}$
Given: $a_{ij}= \frac{\left | 2 i-3j \right |}{2}$
Here we have to construct $2\times 2$ matrix according to $\frac{\left | 2 i-3j \right |}{2}$
Hint: Substitute required values in the $2\times 2$ matrix according $\frac{\left | 2 i-3j \right |}{2}$
Solution: Let $A= \left [ a_{ij} \right ]_{2\times 2}$
So, the elements in a $2\times 2$ are $a_{11},a_{12},a_{21},a_{22}$
$A= \begin{bmatrix} a_{11} &a_{12} \\ a_{21} & a_{22} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= \frac{\left | 2\times 1-3\times 1 \right |}{2}= \frac{1}{2}\\ \\a_{12}= \frac{ \left |2\times 1-3\times 2 \right |}{2}= \frac{ 4 }{2}= 2$ $\! \! \! \! \! \! \! \! \! a_{21}= \frac{ \left | 2\times 2-3\times 1 \right |}{2}=\frac{4-3 }{2} = \frac{1}{2}\\ \\a_{22}= \frac{\left | 2\times 2-3\times 2 \right | }{2}= \frac{ 2}{2}= 1$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} \frac{1}{2} &2 \\ \\ \frac{1}{2} & 1 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 5 (vi)

Answer: $A= \begin{bmatrix} 1 &\frac{1}{2} \\ \\ \frac{5}{2} & 2 \end{bmatrix}$
Given: $a_{ij}= \frac{\left |-3i+j \right |}{2}$
Here we have to construct $2\times 2$ matrix according to $a_{ij}= \frac{\left |-3i+j \right |}{2}$
Hint: Substitute required values in the $2\times 2$ matrix
Solution: Let $A= \left [ a_{ij} \right ]_{2\times 2}$
So, the elements in a $2\times 2$ are $a_{11},a_{12},a_{21},a_{22}$
$A= \begin{bmatrix} a_{11} &a_{12} \\ a_{21} & a_{22} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= \frac{\left | -3\times 1+1 \right |}{2}= \frac{2}{2}= 1\\ \\a_{12}= \frac{ \left |-3\times 1+ 2 \right |}{2}= \frac{ 1 }{2}$ $\! \! \! \! \! \! \! \! \! a_{21}= \frac{ \left | -3\times 2+1 \right |}{2}=\frac{5 }{2}\\ \\a_{22}= \frac{\left | -3\times 2+2 \right | }{2}= \frac{ 4}{2}= 2$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} 1 &\frac{1}{2} \\ \\ \frac{5}{2} & 2 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 5 (vii)

Answer:$A= \begin{bmatrix} e^{2x}\sin x &e^{2x}\sin 2x \\ e^{4x} \sin x& e^{4x}\sin 2x \end{bmatrix}$
Given: $a_{ij}= e^{2x}\sin xj$
Here we have to construct $2\times 2$ matrix according to $e^{2x}\sin xj$
Hint: Putting the value of each row and column element according to the question in matrix
Solution: Let $A= \left [ a_{ij} \right ]_{2\times 2}$ $= e^{2x}\sin xj$
So, the elements in a $2\times 2$ are $a_{11},a_{12},a_{21},a_{22}$
$A= \begin{bmatrix} a_{11} &a_{12} \\ a_{21} & a_{22} \end{bmatrix}$
$\! \! \! \! \! \! \! \! \! a_{11}= e^{2\times 1x}\sin x\times 1= e^{2x}\sin x\\a_{12}= e^{2\times 1x}\sin x\times 2= e^{2x}\sin 2x$ $\! \! \! \! \! \! \! \! \! a_{21}= e^{2\times 2x}\sin x\times 1= e^{4x}\sin x\\a_{22}= e^{2\times 2x}\sin x\times 2= e^{4x}\sin 2x$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} e^{2x}\sin x &e^{2x}\sin 2x \\ e^{4x} \sin x& e^{4x}\sin 2x \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 6 (i)

Answer: $A= \begin{bmatrix} 2 &3 &4 &5 \\ 3 &4 &5 &6 \\ 4 &5 &6 &7 \end{bmatrix}$
Given:$a_{ij}= \left ( i+j \right )$
Here we have to construct $3\times 4$ matrix according to $\left ( i+j \right )$
Hint: Find the sum of $i$ and $j$ for each element.
Solution: Here $a_{ij}= \left ( i+j \right )$
Let $A= \left [ a_{ij} \right ]_{3\times 4}$
So, $A= \begin{bmatrix} a_{11} &a_{12} &a_{13} & a_{14}\\ a_{21} &a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} &a_{34} \end{bmatrix}_{3\times 4}$
$\! \! \! \! \! \! \! \! \! a_{11}= 1+1= 2\\a_{12}= 1+2= 3\\a_{13}= 1+4=4\\a_{14}= 1+4= 5$ $\! \! \! \! \! \! \! \! \! a_{21}= 2+1= 3\\a_{22}= 2+2= 4\\a_{23}= 2+3=5\\a_{24}= 2+4= 6$ $\! \! \! \! \! \! \! \! \! a_{31}= 3+1= 4\\a_{32}= 3+2= 5\\a_{33}= 3+3=6\\a_{34}= 3+4= 7$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} 2 &3 &4 &5 \\ 3 &4 &5 &6 \\ 4 &5 &6 &7 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 6 (ii)

Answer: $A= \begin{bmatrix} 0 &-1 &-2 &-3\\ 1 &0 &-1 &-2 \\ 2 &1 &0 &-1 \end{bmatrix}$
Given: $a_{ij}= \left ( i-j \right )$
Here we have to construct $3\times 4$ matrix according to $\left ( i-j \right )$
Hint: Find the sum of $i$ and $j$ for each element.
Solution: Here $a_{ij}= \left ( i-j \right )$
Let $A= \left [ a_{ij} \right ]_{3\times 4}$
So, $A= \begin{bmatrix} a_{11} &a_{12} &a_{13} & a_{14}\\ a_{21} &a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} &a_{34} \end{bmatrix}_{3\times 4}$
$\! \! \! \! \! \! \! \! \! a_{11}= 1-1= 0\\a_{12}= 1-2= -1\\a_{13}= 1-3=-2\\a_{14}= 1-4= -3$ $\! \! \! \! \! \! \! \! \! a_{21}= 2-1= 1\\a_{22}= 2-2= 0\\a_{23}= 2-3=-1\\a_{24}= 2-4= -2$ $\! \! \! \! \! \! \! \! \! a_{31}= 3-1= 2\\a_{32}= 3-2= 1\\a_{33}= 3-3=0\\a_{34}= 3-4= -1$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} 0 &-1 &-2 &-3\\ 1 &0 &-1 &-2 \\ 2 &1 &0 &-1 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 6 (iii)

Answer: $A= \begin{bmatrix} 2 &2 &2 &2\\ 4 &4 &4 &4 \\ 6 &6 &6 &6 \end{bmatrix}$
Given:$a_{ij}= \left (2i \right )$
Here we have to construct $3\times 4$ matrix according to $\left (2i \right )$
Hint: Substitute the required value according $\left (2i \right )$
Solution: Here $a_{ij}= \left (2i \right )$
Let $A= \left [ a_{ij} \right ]_{3\times 4}$
So, $A= \begin{bmatrix} a_{11} &a_{12} &a_{13} & a_{14}\\ a_{21} &a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} &a_{34} \end{bmatrix}_{3\times 4}$
$\! \! \! \! \! \! \! \! \! a_{11}= 2\times 1= 2\\a_{12}= 2\times 1= 2\\a_{13}= 2\times 1=2\\a_{14}= 2\times 1= 2$ $\! \! \! \! \! \! \! \! \! a_{21}= 2\times 2= 4\\a_{22}= 2\times 2= 4\\a_{23}= 2\times 2=4\\a_{24}= 2\times 2= 4$ $\! \! \! \! \! \! \! \! \! a_{31}= 2\times 3= 6\\a_{32}= 2\times 3= 6\\a_{33}= 2\times 3=6\\a_{34}= 2\times 3= 6$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} 2 &2 &2 &2\\ 4 &4 &4 &4 \\ 6 &6 &6 &6 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 6 (v)

Answer: $A= \begin{bmatrix} 1 &\frac{1}{2} &0 &\frac{1}{2}\\ \\\frac{5}{2} &2 &\frac{3}{2} &1 \\\\ 4 &\frac{7}{2} &3 &\frac{5}{2} \end{bmatrix}$
Given:$a_{ij}= \frac{1}{2}\left | -3i+j \right |$
Here we have to construct $3\times 4$ matrix according to $a_{ij}= \frac{1}{2}\left | -3i+j \right |$
Hint: We have to find all the elements of matrix according to $\frac{1}{2}\left | -3i+j \right |$
Solution: Here $a_{ij}= \frac{1}{2}\left | -3i+j \right |$
Let $A= \left [ a_{ij} \right ]_{3\times 4}$
So, $A= \begin{bmatrix} a_{11} &a_{12} &a_{13} & a_{14}\\ a_{21} &a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} &a_{34} \end{bmatrix}_{3\times 4}$
$\! \! \! \! \! \! \! \! \! a_{11}= \frac{1}{2} \left | -3\times 1 +1\right |= \frac{1}{2}\times 2= 1\\\\a_{12}=\frac{1}{2}\left | -3\times 1+2 \right |= \frac{1}{2}\times 1= \frac{1}{2}\\\\a_{13}=\frac{1}{2}\left | -3\times 1+3 \right |= \frac{1}{2} \times 0= 0\\\\a_{14}=\frac{1}{2}\left | -3\times 1+4 \right | = \frac{1}{2}\times 1= \frac{1}{2}$ $\! \! \! \! \! \! \! \! \! a_{21}\frac{1}{2}\left | -3\times 2+1 \right |= \frac{1}{2}\times 5= \frac{5}{2}\\\\a_{22}\frac{1}{2}\left | -3\times 2+2 \right |= \frac{1}{2}\times 4= 2\\\\a_{23}\frac{1}{2}\left | -3\times 2+3 \right |= \frac{1}{2}\times 3= \frac{3}{2}\\\\a_{24}\frac{1}{2}\left | -3\times 2+4 \right |= \frac{1}{2}\times 2= 1$ $\! \! \! \! \! \! \! \! \! a_{31}= \frac{1}{2}\left | -3\times 3+1 \right |= \frac{1}{2}\times 8= 4\\\\a_{32}= \frac{1}{2}\left | -3\times 3+2 \right |= \frac{1}{2}\times7 = \frac{7}{2}\\\\a_{33}= \frac{1}{2}\left | -3\times 3+3 \right |= \frac{1}{2}\times 6= 3\\\\a_{34}= \frac{1}{2}\left | -3\times 3+4 \right |= \frac{1}{2}\times 5= \frac{5}{2}$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} 1 &\frac{1}{2} &0 &\frac{1}{2}\\ \\\frac{5}{2} &2 &\frac{3}{2} &1 \\\\ 4 &\frac{7}{2} &3 &\frac{5}{2} \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 7 (i)

$A= \begin{bmatrix} 3 &\frac{5}{2} &\frac{7}{3} \\ \\ 6 &5 & \frac{14}{3}\\ \\ 9 & \frac{15}{2} &7 \\ \\ 12 & 10 & \frac{28}{3} \end{bmatrix}$
Given:$a_{ij}= 2i+\frac{i}{j}$
Here we have to construct $4\times 3$ matrix according to $a_{ij}= 2i+\frac{i}{j}$
Hint: First we will find all the elements of matrix according to $2i+\frac{i}{j}$
Solution: Here $a_{ij}= 2i+\frac{i}{j}$
Let $A= \left [ a_{ij} \right ]_{4\times 3}$
So, The elements in a $4\times 3$ matrix are $a_{11},a_{21},a_{31},a_{41},a_{12},a_{22},a_{32},a_{42},a_{13},a_{23},a_{33},a_{43}$
$A= \begin{bmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31} &a_{32} &a_{33} \\ a_{41} & a_{42} & a_{43} \end{bmatrix}_{4\times 3}$
$\! \! \! \! \! \! \! \! \! a_{11}= 2\times 1+\frac{1}{1}= 2+1= 3\\\\a_{12}= 2\times 1+\frac{1}{2}= 2+\frac{1}{2}= \frac{5}{2}\\\\a_{13}= 2\times 1+\frac{1}{3}= 2+\frac{1}{3}= \frac{7}{2}$ $\! \! \! \! \! \! \! \! \! a_{21}= 2\times 2+\frac{2}{1}= 4+2= 6\\\\a_{22}= 2\times 2+\frac{2}{2}= 4+1= 5\\\\a_{23}= 2\times 2+\frac{2}{3}= 4+\frac{2}{3}= \frac{14}{3}$

$\! \! \! \! \! \! \! \! \! a_{31}= 2\times 3+\frac{3}{1}= 6+3= 9\\\\a_{32}= 2\times 3+\frac{3}{2}= 6+\frac{3}{2}= \frac{15}{2}\\\\a_{33}= 2\times 3+\frac{3}{3}= 6+1= 7$ $\! \! \! \! \! \! \! \! \! a_{41}= 2\times 4+\frac{4}{1}= 8+4= 12\\\\a_{42}= 2\times 4+\frac{4}{2}= 8+2= 10\\\\a_{43}= 2\times 4+\frac{4}{3}= 8+\frac{4}{3}= \frac{28}{3}$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} 3 &\frac{5}{2} &\frac{7}{3} \\ \\ 6 &5 & \frac{14}{3}\\ \\ 9 & \frac{15}{2} &7 \\ \\ 12 & 10 & \frac{28}{3} \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 7 (ii)

$A= \begin{bmatrix} 0 &-\frac{1}{3} &-\frac{1}{2} \\ \\ \frac{1}{3} &0 & -\frac{1}{5}\\ \\ \frac{1}{2} & \frac{1}{5} &0 \\ \\ \frac{3}{5} & \frac{1}{3} & \frac{1}{7} \end{bmatrix}$
Given: Here $a_{ij}= \frac{i-j}{i+j}$
Here we have to construct $4\times 3$ matrix according to $\frac{i-j}{i+j}$
Hint: First we will find all the elements of matrix according to $\frac{i-j}{i+j}$
Solution: Here $a_{ij}= \frac{i-j}{i+j}$
Let $A= \left [ a_{ij} \right ]_{4\times 3}$
So, the elements in a $4\times 3$ matrix are $a_{11},a_{21},a_{31},a_{41},a_{12},a_{22},a_{32},a_{42},a_{13},a_{23},a_{33},a_{43}$
$A= \begin{bmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31} &a_{32} &a_{33} \\ a_{41} & a_{42} & a_{43} \end{bmatrix}_{4\times 3}$
$\! \! \! \! \! \! \! \! \! a_{11}=\frac{1-1}{1+1}= \frac{0}{2}= 0 \\\\a_{12}=\frac{1-2}{1+2}=- \frac{1}{3} \\\\a_{13}= \frac{1-3}{1+3}= -\frac{2}{4}= \frac{1}{2}$ $\! \! \! \! \! \! \! \! \! a_{21}=\frac{2-1}{2+1}= \frac{1}{3} \\\\a_{22}=\frac{2-2}{2+2}= \frac{0}{4}= 0 \\\\a_{23}= \frac{2-3}{2+3}= -\frac{1}{5}$

$\! \! \! \! \! \! \! \! \! a_{31}=\frac{3-1}{3+1}= \frac{2}{4} = \frac{1}{2}\\\\a_{32}=\frac{3-2}{3+2}= \frac{1}{5} \\\\a_{33}= \frac{3-3}{3+3}= \frac{0}{6}= 0$ $\! \! \! \! \! \! \! \! \! a_{41}=\frac{4-1}{4+1}= \frac{3}{5} \\\\a_{42}=\frac{4-2}{4+2}= \frac{2}{6}= \frac{1}{3} \\\\a_{43}= \frac{4-3}{4+3}= \frac{1}{7}$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} 0 &-\frac{1}{3} &-\frac{1}{2} \\ \\ \frac{1}{3} &0 & -\frac{1}{5}\\ \\ \frac{1}{2} & \frac{1}{5} &0 \\ \\ \frac{3}{5} & \frac{1}{3} & \frac{1}{7} \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 7 (iii)

$A= \begin{bmatrix} 1 &1 &1\\ 2 &2 & 2\\ 3 & 3 &3\\ 4& 4 & 4 \end{bmatrix}$
Given:$a_{ij}= i$
Here we have to construct $4\times 3$ matrix according to $a_{ij}= i$
Hint: First we will find all the elements of matrix according to $a_{ij}= i$
Solution: Here $a_{ij}= i$
Let $A= \left [ a_{ij} \right ]_{4\times 3}$
So, The elements in a $4\times 3$ matrix are $a_{11},a_{21},a_{31},a_{41},a_{12},a_{22},a_{32},a_{42},a_{13},a_{23},a_{33},a_{43}$
$A= \begin{bmatrix} a_{11} & a_{12} &a_{13} \\ a_{21} &a_{22} &a_{23} \\ a_{31} &a_{32} &a_{33} \\ a_{41} & a_{42} & a_{43} \end{bmatrix}_{4\times 3}$
$\! \! \! \! \! \! \! \! \! a_{11}= 1 \\a_{12}=1 \\a_{13}=1$ $\! \! \! \! \! \! \! \! \! a_{21}=2 \\a_{22}=2\\a_{23}=2$ $\! \! \! \! \! \! \! \! \! a_{31}=3\\a_{32}=3\\a_{33}=3$ $\! \! \! \! \! \! \! \! \! a_{41}=4\\a_{42}=4\\a_{43}=4$
Substituting these values in Matrix $A$ , we get
$A= \begin{bmatrix} 1 &1 &1\\ 2 &2 & 2\\ 3 & 3 &3\\ 4& 4 & 4 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 8

Answer:$a= 0,b=5,x=2$ and $y= -1$
Given: Here given that

$\begin{bmatrix} 3x+4y & 2 &x-2y \\ a+b &2a-b & -1 \end{bmatrix}$ $= \begin{bmatrix} 2 & 2 &4\\ 5&-5 & -1 \end{bmatrix}$

We have to find the value of $a,b,x$ and $y$
Hint: If two matrices are equal then the elements of each matrix are also equal.
Solution: Given that two matrices are equal
$\therefore$ By equating them, we get
$3x+4y$ ……(i)
$x-2y$ ……(ii)
$a+b= 5$ ……. (iii)
$2a-b= -5$ …….. (iv)
Multiplying equation (ii) by 2 and adding to equation (i), we get
$\! \! \! \! \! \! \! \! 3x+4y+2x-4y= 2+8\\\Rightarrow 5x= 10\\\Rightarrow x= 2$
Now substituting the value of $x$ in eqn (i), we get
$3\times 2+4y= 2\\\Rightarrow 6 +4y= 2\\\Rightarrow 4y= -4\\\Rightarrow y= -1$
Now by adding eqn(iii) and eqn (iv)
$a+b+2a-b=5+\left ( -5 \right )$
$\Rightarrow 3a= 5-5$
$\Rightarrow a= 0$

Now, again substituting the value of $a$ in eqn(iii), we get
$a+b= 5$
$\Rightarrow b=5$

Hence, $a=0,b=5,x=2$ and $y=-1$

Algebra of Matrices Exercise 4.1 Question 10

Answer: $a= 1,b=2,c=3,d=4$
Given:$\begin{bmatrix} 2a+b &a-2b\\ 5c-d &4c+3d \end{bmatrix}$$= \begin{bmatrix} 4 &-3\\ 11&24 \end{bmatrix}$
Here we have to find out the values of $a,b,c$ and $d$.
Hint: If two matrices are equal then the elements of each matrix are also equal.
Solution:
Given that two matrices are equal
$\therefore$ By equating them, we get
$2a+b=4$ ….. (i)
$a-2b=-3$ ……(ii)
$5c-d=11$ ……(iii)
and $4c+3d=24$ ……(iv)
Multiplying eqn(i) by 2 and adding to eqn(ii) we get
$4a+2b+a-2b=8-3$
$\Rightarrow 5a=5$
$\Rightarrow a=1$
Now, substituting the value of a in eqn(i)
$2\times 1+b=4$
$\Rightarrow b=4-2$
$\Rightarrow b=2$
Multiplying eqn(iii) by 3 and adding to eqn(iv) we get
$15c-3d+4c+3d=33+24\\\\\Rightarrow 19c=57\\\\\Rightarrow c=3$
Now, substituting the value of c in eqn(iv) we get
$4\times 3+3d=24\\\\\Rightarrow 12+3d=24\\\\\Rightarrow 3d=24-12\\\\\Rightarrow d=\frac{12}{3}\\\\\Rightarrow d=4$
Hence $a= 1,b=2,c=3,d=4$

Algebra of Matrices Exercise 4.1 Question 11

Answer: $x=11,y=9,z=3$
Given: $A=B$
$\begin{bmatrix} x-2 & 3 &2z \\ 18z &y+2 & 6z \end{bmatrix}$ $= \begin{bmatrix} y & z &6\\ 6y&x & 2y \end{bmatrix}$
Hint: If $A=B\Rightarrow$ If two matrices are equal then the elements of each matrix are also equal.
Solution: Here $A=B$
$\begin{bmatrix} x-2 & 3 &2z \\ 18z &y+2 & 6z \end{bmatrix}$ $= \begin{bmatrix} y & z &6\\ 6y&x & 2y \end{bmatrix}$
Since corresponding entries of equal matrices are equal, So
$x-2=y$ ….. (i)
$3=z$ ….. (ii)
$2z=6$ ….. (iii)
$18z=6y$ ….. (iv)
$y+2=x$ ….. (v)
$6z=2y$ ….. (vi)
Equation (ii) gives $z=3$
Putting the value of z in eqn(iv) we get
$18\times 3=6y\\\Rightarrow y=9$
Putting the value of $y$ in eqn(v) we get
$y+2=x\\\Rightarrow 9+2=x\\\Rightarrow x=11$
Hence $x=11,y=9,z=3$

Algebra of Matrices Exercise 4.1 Question 12

Answer:$x=3,y=7,z=-2,w=14$
Given:$\begin{bmatrix} x &3x-y \\ 2x+z & 3y-w \end{bmatrix}$$= \begin{bmatrix} 3 &2\\ 4 & 7 \end{bmatrix}$
We have to find the value of $x,y,z$ and $w$
Hint: If two matrices are equal then the elements of each matrix are also equal.
Solution: Let $\begin{bmatrix} x &3x-y \\ 2x+z & 3y-w \end{bmatrix}$$= \begin{bmatrix} 3 &2\\ 4 & 7 \end{bmatrix}$
Since corresponding entries of equal matrices are equal, So
$x=3$ ….. (i)
$3x-y=2$ ….. (ii)
$2x+z=4$ ….. (iii)
$3y-w=7$ ….. (iv)
Put the value of $x=3$ in eqn(ii) we get
$3\times 3-y=2\\\Rightarrow y=9-2\\\Rightarrow y=7$
Put the value of $y=7$ in eqn(iv) we get
$3y-w=7\\\Rightarrow 3\times 7-w=7\\\Rightarrow w=21-7\\\Rightarrow w=14$
Put the value of $x=3$ in eqn(iii) we get
$2x+z=4\\\Rightarrow 2\times 3+z=4\\\Rightarrow z=4-6\\\Rightarrow z=-2$
Hence, $x=3,y=7,z=-2,w=14$

Algebra of Matrices Exercise 4.1 Question 13

Answer:$x=1,y=2,z=4$ and $w=5$
Given: $\begin{bmatrix} x-y & z \\ 2x-y & w \end{bmatrix}$$= \begin{bmatrix} -1 &4\\ 0 & 5 \end{bmatrix}$
We have to find the value of $x,y,z$ and $w$
Hint: If two matrices are equal then the elements of each matrix are also equal.
Solution: Here $\begin{bmatrix} x-y & z \\ 2x-y & w \end{bmatrix}$$= \begin{bmatrix} -1 &4\\ 0 & 5 \end{bmatrix}$
Since corresponding entries of equal matrices are equal, So
$x-y=-1$ ….. (i)
$z=4$ ….. (ii)
$2x-y=0$ ….. (iii)
$w=5$ ….. (iv)
Solving eqn(i) and (iii), we get
$\! \! \! \! \! \! x-y=-1\\2x-y=0\\-x=-1$
Putting the value of $x=1$ in eqn (i), we get
$1-y=-1\\\Rightarrow y=1+1\\\Rightarrow y=2$
Equation (ii) and (iv) gives the values of $z$ and $w$ respectively. So, $z=4,w=5$
Hence, $x=1,y=2,z=4$ and $w=5$

Algebra of Matrices Exercise 4.1 Question 14

Answer: $x=-3,y=-5,z=2,a=-2,b=-7,c=-1$
Given:$\begin{bmatrix} x+3 &z+4 & 2y-7\\ 4x+6 &a-1 &0 \\ b-3& 3b &z+2c \end{bmatrix}$ $= \begin{bmatrix} 0 &6 &3y-2 \\ 2x & -3 &2c+2 \\ 2b+4 & -21 & 0 \end{bmatrix}$
Here we have to find out all the values of $x,y,z,a,b,c$
Hint: By definition of equal matrices is $A=\left [ a_{ij} \right ]_{m\times n}$ and $B=\left [ b_{ij} \right ]_{m\times n}$ are equal then
$a_{ij}=b_{ij}$ for $i=1,2,3....m$ and $j=1,2,3....n$
Solution: Given that $\begin{bmatrix} x+3 &z+4 & 2y-7\\ 4x+6 &a-1 &0 \\ b-3& 3b &z+2c \end{bmatrix}$ $= \begin{bmatrix} 0 &6 &3y-2 \\ 2x & -3 &2c+2 \\ 2b+4 & -21 & 0 \end{bmatrix}$
Equating the entries, we get
$x+3=0\\\Rightarrow x=-3$ ; $z+4=6\\\Rightarrow z=2$ ; $2y-7=3y-2\\\Rightarrow 2y-3y=-2+7\\\Rightarrow -y=+5\\\Rightarrow y=-5$
Similarly,
$a-1=-3$ and $z+2c=0$
$\Rightarrow a=-3+1$ and $2+2c=0$
$\Rightarrow a=-2$ and $c=-1$
Lastly,
$b-3=2b+4\\\Rightarrow b-2b=4+3\\\Rightarrow -b=7\\\Rightarrow b=-7$
Hence, $x=-3,y=-5,z=2,a=-2,b=-7,c=-1$

Algebra of Matrices Exercise 4.1 Question 15

Answer:$x+y=7$ or $-3$
Given: Given that
$\begin{bmatrix} 2x+1 &5x \\ 0 &y^{2}+1 \end{bmatrix}$$= \begin{bmatrix} x+3 &10\\ 0 & 26 \end{bmatrix}$
Here we have to calculate the value of $x+y$
Hint: Here we will use equality of matrices.
Solution: Here $\begin{bmatrix} 2x+1 &5x \\ 0 &y^{2}+1 \end{bmatrix}$$= \begin{bmatrix} x+3 &10\\ 0 & 26 \end{bmatrix}$
The corresponding entries of equal matrices are equal, So
$2x+1=x+3\\\Rightarrow 2x-x= 3-1\\\Rightarrow x=2$
$y^{2}+1=26\\\Rightarrow y^{2}=26-1\\\Rightarrow y=\sqrt{25}\\\Rightarrow y=\pm 5$
Case 1: If $x=2$ and $y=+5$
$\Rightarrow x+y=2+5=7$
Case 2: If $x=2$ and $y=-5$
$\Rightarrow x+y=2+\left ( -5 \right )=-3$
Hence $x+y=7$ or $-3$

Algebra of Matrices Exercise 4.1 Question 16

$\! \! \! \! \! \! \! \! \! x=2\: or\: 4\\y=4\: or\: 2\\z=-6\\w=4$
Given: $\begin{bmatrix} xy &4 \\ z+6&x+y \end{bmatrix}=\begin{bmatrix} 8 &w \\ 0& 6 \end{bmatrix}$
Here we have to find the value of $x,y,z,w$
Hint: Here we will use equality of matrices.
Solution:
$\begin{bmatrix} xy &4 \\ z+6&x+y \end{bmatrix}=\begin{bmatrix} 8 &w \\ 0& 6 \end{bmatrix}$
The corresponding entries of equal matrices are equal, So
$xy=8$ ….. (i)
$w=4$ ….. (ii)
$z+6=0$ ….. (iii)
$x+y=6$ ….. (iv)
From equation (ii) and (iii) we get
$z=-6$ and $w=4$
From eqn (iv) we have,
$x+y=6\Rightarrow x=6-y$
Substituting the value of $x$ in eqn (i), we get
$\! \! \! \! \! \! \! \! \left ( 6-y \right )y=8\\\Rightarrow y^{2}-6y+8=0\\\Rightarrow \left ( y-2 \right )\left ( y-4 \right )=0\\\Rightarrow y=2 \: or\: 4$
Substituting the value of $y$ in eqn (i), we get
$x=4,2$
Hence value of $x,y,z,w$ are $4,2;2,4;-6$ and $4$ respectively

Algebra of Matrices Exercise 4.1 Question 17 (i)

Answer:$\left [ a \right ]_{1\times 1}$
Given: Here we have to give an example of row matrix which is also a column matrix
Hint:
Solution: We know that
Order of row matrix = $1\times n$
Order of column matrix = $m\times 1$
So, order of a row as column matrix = $1\times 1$
Hence required matrix =$\left [ a \right ]_{1\times 1}$

Algebra of Matrices Exercise 4.1 Question 17 (ii)

Answer: $\begin{bmatrix} 4 & 0 & 0\\ 0 &-3 &0 \\ 0& 0 &2 \end{bmatrix}$
Here we have to give an example of diagonal matrix which is not scalar
Given:
Hint: A diagonal matrix has only $a_{11}\: a_{22}\: a_{33}$ for a $3\times 3$ matrix such that $a_{11}\: a_{22}\: a_{33}$ are equal or different and all other entries zero
Solution: We know that a diagonal matrix has only $a_{11}\: a_{22}\: a_{33}$ for a $3\times 3$ matrix such that $a_{11}\: a_{22}\: a_{33}$ are equal or different and all other entries zero. While scalar matrix has $a_{11}= a_{22}= a_{33}= m$ (say)
So, a diagonal matrix which is not scalar must have $a_{11}\neq a_{22}\neq a_{33}$ and $a_{ij}=0$ for $i\neq j$ .
Hence Required matrix = $\begin{bmatrix} 4 & 0 & 0\\ 0 &-3 &0 \\ 0& 0 &2 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 17 (iii)

Answer:$\begin{bmatrix} 3 & 2 &-1 \\ 0 & 4 &3 \\ 0 & 0 & -6 \end{bmatrix}$
Here we have to create an example of triangular matrix.
Given:
Hint: A triangular matrix is a square matrix
Solution: We know that a triangular matrix is a square matrix
$A=\left [ a_{ij} \right ]$ such that$a_{ij}=0$ for $i> j$

Hence required matrix

Algebra of Matrices Exercise 4.1 Question 18

Answer: $A=\begin{bmatrix} 5 &3 & 4\\ 7 &2 &3 \end{bmatrix}$
and $B=\begin{bmatrix} 8 &7 & 6\\ 10 &5 &7 \end{bmatrix}$
Given: Here the sales figure of two car dealers during January 2013
Here we have to write $2\times 3$ summarizing sales data.
Hint: Simply summarize for dealer A and dealer B
Solution: Given data is
For January 2013:
$\begin{matrix} &Delux & Premium & Stadard\\ dealer A & 5 & 3& 4\\ dealerB &7 & 2 & 3 \end{matrix}$
Hence, $A=\begin{bmatrix} Delux &Premium &Standard \\ 5& 3 & 4\\ 7 &2 &3 \end{bmatrix}$
And For January – February:
$\begin{matrix} & Deluxe &Premium & Stadard\\ dealerA & 8 & 7& 6\\ dealerB &10 & 5 & 7 \end{matrix}$
Hence, $B=\begin{bmatrix} &Deluxe &Premium &Stadard \\ dealerA & 8 &7 &6 \\ dealerB & 10 &5 & 7 \end{bmatrix}$

Algebra of Matrices Exercise 4.1 Question 19

Answer: $A$ and $B$ are not equal for any value of$y$

Given: $A=\begin{bmatrix} 2x+1 &2y \\ 0 & y^{2}+2 \end{bmatrix}$ and

$B=\begin{bmatrix} x+3 &y^{2}+2 \\ 0 & -6 \end{bmatrix}$

We have to find out the value of $x$ and $y$
Hint: We will use equality of matrices.
Solution: Here $A=B$
Since equal matrix have all corresponding entries equal. So,

$2x+1=x+3$ ….. (i)

$2y=y^{2}+2$ ….. (ii)

$y2-5y=-6$ ….. (iii)

Solving equation (i), We get

$2x+1=x+3\\\Rightarrow 2x-x=3-1\\\Rightarrow x=2$

Solving equation (ii), We get

$2y=y^{2}+2\\\Rightarrow y^{2}-2y+2=0\\\Rightarrow D=b^{2}-4ac$

$\! \! \! \! \! \! \! \! \! =\left ( -2 \right )^{2}-4\left ( 1 \right )\left ( 2 \right )\\=4-8\\=-4$

Here $D< 0$
So, there is no real value of $y$ from equation (ii)
Solving equation (iii), We get

$y^{2}5y=-6\\\Rightarrow y^{2}-5y+6=0\\\left ( y-3 \right )\left ( y-2 \right )=0\\y=3\: or\: 2$

From solution of equation (i) (ii) and (iii)
we can say that $A$ and $B$ cannot equal for any value of$y$

Algebra of Matrices Exercise 4.1 Question 20

Answer:$x= 3,y= 1$
Given:$\begin{bmatrix} x+10 & y^{2}+2y\\ 0& -4 \end{bmatrix}$$= \begin{bmatrix} 3x+4 & 3\\ 0& y^{2}-5y \end{bmatrix}$
We have to find out the value of $x$ and $y$
Hint: We will use equality of matrices.
Solution: Here $\begin{bmatrix} x+10 & y^{2}+2y\\ 0& -4 \end{bmatrix}$$= \begin{bmatrix} 3x+4 & 3\\ 0& y^{2}-5y \end{bmatrix}$
Since, corresponding entries of equal matrices are equal. So,
$x+10=3x+4$ ….. (i)
$y^{2}+2y= 3$ ….. (ii)
$-4= y^{2}-5y$ ….. (iii)
Solving equation (i), We get
$x+10= 3x+4\\\Rightarrow 2x=6\\\Rightarrow x=3$
Solving equation (ii), We get
$y^{2}+2y-3=0\\\Rightarrow y+3y-y-3=0\\\Rightarrow y\left ( y+3 \right )-1\left ( y+3 \right )=0\\\Rightarrow y=1\: or\: -3$
Solving equation (iii), We get
$-4=y^{2}-5y\\\Rightarrow y^{2}-5y+4=0\\\Rightarrow y^{2}-4y-y+4=0\\\Rightarrow y\left ( y-1 \right )-1( \left ( y-4 \right ) =0\\\Rightarrow \left ( y-1 \right )\left ( y-4 \right )=0\\\Rightarrow y=1\: or\: 4$
From equation (ii) and (iii)
We have common value of $y= 1$
So, $x= 3,y= 1$

Algebra of Matrices Exercise 4.1 Question 21

Answer:$A= B$
Given:
$A= \begin{bmatrix} a+4 &3b \\ 8 &-6 \end{bmatrix},B\begin{bmatrix} 2a+2 &b^{2}+2 \\ 8 & b^{2}-10 \end{bmatrix}$
We have to find out the value of $a$ and $b$
Hint: We will use equality of matrices.
Solution:
$A= \begin{bmatrix} a+4 &3b \\ 8 &-6 \end{bmatrix},B\begin{bmatrix} 2a+2 &b^{2}+2 \\ 8 & b^{2}-10 \end{bmatrix}$
$\because A= B$
Corresponding elements of two equal matrix are equal.
$a+4= 2a+2$ ….. (i)
$3b=b^{2}+2$ ….. (ii)
$b^{2}-10= -6$ ….. (iii)
Solving equation (i), We get
$a+4=2a+2\\\Rightarrow 2a-a=4-2\\\Rightarrow a=2$
Solving equation (ii), We get
$b^{2}-3b+2=0\\\Rightarrow b^{2}-2b-b+2= 0\\\Rightarrow b\left ( b-2 \right )-1\left ( b-2 \right )= 0\\\Rightarrow \left ( b-1 \right )\left ( b-2 \right )= 0\\\Rightarrow = 1\: or\: 2$
Solving equation (iii), We get
$b^{2}-10= -6\\\Rightarrow b^{2}= -6+10\\\Rightarrow b^{2}= 4\\\Rightarrow b= \pm 2$
Here, we have common value of $b= 2$ from equation (ii) and (iii)
Hence, $a=2,b= 2$

The class 12 RD Sharma chapter 4 exercise 4.1 solution contains the chapter Algebra of Matrices which explores Order of the matrix, Matrix formation, addition, subtraction, multiplication, etc., of matrices along with types of Matrices like null matrices, diagonal Matrices, and triangular matrices. Exercise 4.1 contains 39 questions including subparts, on two levels based on these concepts.

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Chapter-wise RD Sharma Class 12 Solutions

1. What are the benefits of using RD Sharma class 12th exercise 4.1 Solutions?

RD Sharma class 12th exercise 4.1 Solutions can be extremely helpful to students who will be appearing for their board exams. In addition, students can use these answers to test their knowledge at home and develop their math skills.

2. What does the 4th chapter of the Class 12 Mathematics book contain?

The 4th chapter of the book contains advanced concepts and problems on matrices. For example, there will be sums on addition, subtraction, division, and multiplication of matrices. You will also explore types of Matrices like skew-symmetric matrix, null matrix, symmetric matrix, etc.

3. Who can use class 12 RD Sharma chapter 4 exercise 4.1 solution?

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