Question

If $A^2=8A+kI$ where $A=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]$ then $k$ is

• A. $7$
• B. $-7$
• C. $1$
• D. $-1$

Answer 1

Answer: (B)

$-7$

Given $A=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]$
$A^2=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -8& 49\end{array}\right]$
$8A+kI=\left[\begin{array}{cc}8& 0\\ -8& 56\end{array}\right]+\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]=\left[\begin{array}{cc}8+k& 0\\ -8& 56+k\end{array}\right]$
$\Rightarrow 8+k=1$
$\Rightarrow k=-7$