#### Provide solution for rd sharma math class class 12 chapter Algebra of matrices exercise 4.3 question 32

Answer: Hence, proved $A^{2}-12 A-I=0$

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: $A=\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]$

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

To show that $A^{2}-12 A-I=0$

Now, we will find the matrix for $A^2$, we get

$A^{2}=A A=\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right]\\\\ A^{2}=\left[\begin{array}{ll}25+36 & 15+21 \\ 60+84 & 36+49\end{array}\right]\\\\A^{2}=\left[\begin{array}{cc}61 & 36 \\ 144 & 85\end{array}\right] .... (i)$

Now, we will find the matrix for 12A, we get

$12 A=12\left[\begin{array}{cc}5 & 3 \\ 12 & 7\end{array}\right] \\\\\ 12 A=\left[\begin{array}{cc}60 & 36 \\ 144 & 84\end{array}\right] \quad \ldots(i i)$

So, substituting corresponding values from equation i & ii in $A^{2}-12 A-I$

we get

$=\left[\begin{array}{cc} 61 & 36 \\ 144 & 85 \end{array}\right]-\left[\begin{array}{cc} 60 & 36 \\ 144 & 84 \end{array}\right]-\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\\\$

$\left[\begin{array}{cc} 61-60-1 & 36-36-0 \\ 144-144-0 & 85-84-1 \end{array}\right]$

$\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0 \\ A^{2}-12 A-I=0$

Hence, matrix A is the root of the given equation.