Question

If $A=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]$, find $k$ so that $A^2=8A+kI$.

Answer 1

Given that,

$A=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]$

Now,

$\begin{array}{c}\left(A-\lambda I\right)=0\\ \left|\begin{array}{cc}1-\lambda & 0\\ -1& 7-\lambda \end{array}\right|=0\\ \left(\lambda -7\right)\left(\lambda -1\right)=0\\ {\lambda }^2-8\lambda +7=0\\ A^2-8A+7I=0\\ A^2=8A-7I\\ k=-7\end{array}$

On comparing, we get

$k=-7$

Hence, this is the answer.