#### Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 33

Answer: $A^{2}-5 A-14 I=0$

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

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

Given: $A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$

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

$A^{2}=A A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]=\left[\begin{array}{cc}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right] \\\\ A^{2}=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right] ... (i)$

Now, we will find the matrix for 5A, we get

$5 A=5\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right] \\\\ 5 A=\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right] ... ( ii)$

So, substitute corresponding values from equation i & ii in eqn $A^{2}-5 A-14 I$

we get

$=\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-14\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right] \\ \\\\ =\left[\begin{array}{cc}29 & -25 \\ -20 & 24\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right] \\\\\\ =\left[\begin{array}{cc}29-15-14 & -25+25-0 \\ -20+20-0 & 24-10-14\end{array}\right] \\\\\\ \left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right] = 0$