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#### Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 48 sub question (v) maths textbook solution

$A=\left[\begin{array}{ccc}1 & -2 & -5 \\ 3 & 4 & 0\end{array}\right]$

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

$\left[\begin{array}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right] A=\left[\begin{array}{ccc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{array}\right]$

The matrix given on the LHS of the equation is a $3 \times 2$ matrix and the one given on the RHS of the equation is $3 \times 3$ matrix. So,  A has to be $2 \times 3$ matrixNow, let

$\begin{array}{l} \text { Now, let } A=\left[\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right] \\\\ \Rightarrow\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right]\left[\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \\ \end{array}$

$\begin{array}{l} \Rightarrow\left[\begin{array}{ccc} 2(a)+(-1) d & 2(b)+(-1)(e) & 2(c)+(-1) f \\ 1(a)+0(d) & 1(b)+0(e) & 1(c)+0(f) \\ -3(a)+4(d) & -3(b)+4(e) & -3(c)+4(f) \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \\\\ \Rightarrow\left[\begin{array}{ccc} 2 a-d & 2 b-e & 2 c-f \\ a+0 & b+0 & c+0 \\ -3 a+4 d & -3 b+4 e & -3 c+4 f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \\\\ \Rightarrow\left[\begin{array}{ccc} 2 a-d & 2 b-e & 2 c-f \\ a & b & c \\ -3 a+4 d & -3 b+4 e & -3 c+4 f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \end{array}$

Equating the corresponding elements of the two matrices, we have

$\\\\ a = 1, b = -2 , c = -5 \\\\ then \ \ \, 2a - d = -1$

$\begin{array}{l} \Rightarrow 2(1)-d=-1 \\\\ \Rightarrow 2-d=-1 \\\\ \Rightarrow d=2+1=3 \\\\ \Rightarrow 2 b-e=-8 \\\\ \Rightarrow 2(-2)-e=-8 \\\\ \Rightarrow-4-e=-8 \\\\ \Rightarrow e=-4+8=4 \\\\ \Rightarrow e=4 \end{array}$

$\begin{array}{l} \Rightarrow 2 c-f=-10 \\\\ \Rightarrow 2(-5)-f=-10 \\\\ \Rightarrow-10-f=-10 \\\\ \Rightarrow f=-10+10 \\\\ \Rightarrow f=0 \\ \end{array}$

Hence, $a=1, b=-2, c=-5, d=3, e=4 \text { and } f=0 \\$

Therefore, matrix

$A=\left[\begin{array}{ccc} 1 & -2 & -5 \\ 3 & 4 & 0 \end{array}\right]$