If \( A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \) then \( A^{2}-5 A \) is equal to

Given that, \( A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \)
So, \( A^{2}=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] \)
\( =\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] \)
And
\( 5 A=5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]=\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right] \)
Therefore,
\[
\begin{array}{l}
A^{2}-5 A=\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]-\left[\begin{array}{cc}
15 & 5 \\
-5 & 10
\end{array}\right] \\
=\left[\begin{array}{cc}
-7 & 0 \\
0 & -7
\end{array}\right]=-7 I
\end{array}
\]