#### Provide solution for rd sharma math class 12 chapter Algebra of matrices exercise 4.3 question 45

Answer:  Hence, proved $A^{2}-4 A-5 I=0$

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

Given:

$A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right] \\\\$

Prove:  $A^{2}-4 A-5 I=0$

I is identity matrix

$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$

$\begin {array}{ll}5 I=5\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right] \\\\\\ \mathrm{LHS}=A^{2}-4 A-5 I \end{}$

$\begin {array}{ll}=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]-4\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]-\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right] \\\\\\ =\left[\begin{array}{lll}1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1\end{array}\right]-\left[\begin{array}{lll}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{array}\right]-\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]\\\\\\\ =\left[\begin{array}{lll}9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9\end{array}\right]-\left[\begin{array}{lll}4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4\end{array}\right]-\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right] \\\\\\ =\left[\begin{array}{llll}9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5\end{array}\right] \end{}$

$\begin{array}{l} =\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \\\\ =0 \\\\ \text { LHS=RHS } \\\\ \end{array}$

Hence, $A^{2}-4 A-5 I=0$