#### Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.3 question 50

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:

$A=\left[\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right] and \ \ I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] and \ \ A^{2}+I=k A$

Consider $\\\\A^{2}=A A$

$\\\\ A^{2}=\left[\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right]\\\\\\ =\left[\begin{array}{cc}(-3)(-3)+2(1) & (-3)(2)+2(-1) \\ 1(-3)+(-1)(1) & 1(2)+(-1)(-1)\end{array}\right]\\\\\\ =\left[\begin{array}{cc}9+2 & -6-2 \\ -3-1 & 2+1\end{array}\right] =\left[\begin{array}{cc}11 & -8 \\ -4 & 3\end{array}\right]$

I is an identity matrix

Consider,

$\\\\ k A=k\left[\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right]$

$\\\\ k A=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right] ... (ii)$

Substitute all values in equation $A^{2}+I=k A$, we get

$\\\\ \Rightarrow\left[\begin{array}{cc}11 & -8 \\ -4 & 3\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{cc}11+1 & -8+0 \\ -4+0 & 3+1\end{array}\right]=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{cc}12 & -8 \\ -4 & 4\end{array}\right]=\left[\begin{array}{cc}-3 k & 2 k \\ k & -k\end{array}\right]$

Since, corresponding entries of equal matrices are equal, So

$\begin{array}{l} \Rightarrow-3 k=12 \\\\ \Rightarrow k=-\frac{12}{3} \\\\ \Rightarrow k=-4 \end{array}$

Hence, value of k is -4