#### Provide solution for rd sharma math class class 12 chapter Algebra of matrices exercise 4.3 question 36

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: $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]$ and $A^{2}-8 A+k l=0$

I is an identity matrix. So, $k I=k\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]$

Now, we have to find $A^2$, we get

$A^{2}=A A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right] \\\\ \\ A^{2}=\left[\begin{array}{cc}1+0 & 0+0 \\ -1-7 & 0+49\end{array}\right] \\\\ \\ A^{2}=\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right] ...(i)$

Now, we will find the matrix 8A, we get

$8 A=8\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]=\left[\begin{array}{cc}8 \times 1 & 8 \times 0 \\ 8 \times(-1) & 8 \times 7\end{array}\right] \\\\ 8 \boldsymbol{A}=\left[\begin{array}{cc}8 & 0 \\ -8 & 56\end{array}\right] ... (ii)$

So, substituting corresponding values from equation i & ii in equation

$A^{2}-8 A+k I=0 \\\\ \Rightarrow\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]-\left[\begin{array}{cc}8 & 0 \\ -8 & 56\end{array}\right]+\left[\begin{array}{cc}k & 0 \\ 0 & k\end{array}\right]=0 \\\\\\ \Rightarrow\left[\begin{array}{cc}1-8+k & 0-0+0 \\ -8+8+0 & 49-56+k\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

And to satisfy the above conditions of equality, the corresponding entries of the matrices should be equal

Hence,

1-8+k-0

k=8-1=7

Therefore, the value of k=7