If \(A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right], I=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\) and \(A^{2}=8 A+k I\), then the value of \(K\) is

\(A^2=A \times A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]\)
Given \(A^2=8 A+K I\)
\(=8\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]+ k \left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}8 & 0 \\ -8 & 56\end{array}\right]+\left[\begin{array}{cc} k & 0 \\ 0 & k \end{array}\right]\)
\(\therefore\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]=\left[\begin{array}{cc}8+ k & 0 \\ -8 & 56+ k \end{array}\right] \Rightarrow 8+ k =1 \Rightarrow k =-7\)