Provide solution for rd sharma math class 12 chapter Algebra of matrices exercise 4.3 question 28

Answer: $\lambda=-7$

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:

$A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] and I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \\\\\ A^{2}=5 A+\lambda I \\\\ \Rightarrow\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]=5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+\lambda\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \\\\\\ \Rightarrow\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{cc}\lambda & 0 \\ 0 & \lambda\end{array}\right] \\\\\\\ \Rightarrow\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]=\left[\begin{array}{cc}15+\lambda & 5 \\ -5 & 10+\lambda\end{array}\right]$

Since, corresponding entries of equal matrices are equal, so

$\Rightarrow 8=15+\lambda \\ \Rightarrow \lambda=8-15\\ \Rightarrow \lambda=-7$