Provide solution for rd sharma math class class 12 chapter Algebra of matrices exercise 4.3 question 19

Answer: $a_{43} = 8 \ \ and \ \ a _{ 22} =0$

Hint:

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

$a_{43}$ means element from 4th row and 3rd column and $\ \ a _{ 22} =0$ means element from 2nd row and 2nd column.

Given:

$\left[\begin{array}{ccc}0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 3 & 2 \\ 4 & 0 & 4\end{array}\right]\left[\begin{array}{cc}2 & -1 \\ -3 & 2 \\ 4 & 3\end{array}\right]\left[\begin{array}{ccccc}0 & 1 & -1 & 2 & -2 \\ 3 & -3 & 4 & -4 & 0\end{array}\right]$

Firstly, we will multiply first two matrices

Then,

$A=\left[\begin{array}{cc}0-3+0 & 0+2+0 \\ 4+0+8 & -2+0+6 \\ 0-9+8 & 0+6+6 \\ 8+0+16 & -4+0+12\end{array}\right]\left[\begin{array}{ccccc}0 & 1 & -1 & 2 & -2 \\ 3 & -3 & 4 & -4 & 0\end{array}\right] \\\\\\ A=\left[\begin{array}{cc}-3 & 2 \\ 12 & 4 \\ -1 & 12 \\ 24 & 8\end{array}\right]\left[\begin{array}{ccccc}0 & 1 & -1 & 2 & -2 \\ 3 & -3 & 4 & -4 & 0\end{array}\right] \\\\\\ A=\left[\begin{array}{ccccc}0+6 & -3-6 & 3+8 & -6-8 & 6+0 \\ 0+12 & 12-12 & -12+16 & 24-16 & -24+0 \\ 0+36 & -1-36 & 1+48 & -2-48 & 2+0 \\ 0+24 & 24-24 & -24+32 & 48-32 & -48+0\end{array}\right] \\\\ \\ A=\left[\begin{array}{ccccc}6 & -9 & 11 & -14 & 6 \\ 12 & 0 & 4 & 8 & -24 \\ 36 & -37 & 49 & -50 & 2 \\ 24 & 0 & 8 & 16 & -48\end{array}\right]$

Then,$a_{43} = 8 \ \ and \ \ a _{ 22} =0$