Question

If $A=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]$ and $A^2=8A+K{I}_2$, then $K$ is equal to

• A. $-1$
• B. $1$
• C. $-7$
• D. $7$

Answer 1

Answer: (C)

$-7$

$A^2=\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]=\left[\begin{array}{cc}1& 0\\ -8& 49\end{array}\right]$

and $8A+K{I}_2=8\left[\begin{array}{cc}1& 0\\ -1& 7\end{array}\right]+K\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$=\left[\begin{array}{cc}8+K& 0\\ -8& 56+K\end{array}\right]$

$\because A^2=8A+K{I}_2$

$\Rightarrow \left[\begin{array}{cc}1& 0\\ -8& 49\end{array}\right]=\left[\begin{array}{cc}8+K& 0\\ -8& 56+K\end{array}\right]$

On comparing, we get $8+K=1$

$\therefore K=-7$