Question

If $A=\left[\begin{array}{cc}1& -3\\ 2& k\end{array}\right]$ and $A^2-4A+10I=A$ , then $k$ is equal to

• A. $0$
• B. $-4$
• C. $4$
• D. $1$ or $4$

Answer 1

Answer: (C)

$A^2-4A+10I=A$
$\Rightarrow \left[\begin{array}{c}1-3\\ 2k\end{array}\right]\left[\begin{array}{c}1-3\\ 2k\end{array}\right]-4\left[\begin{array}{c}1-3\\ 2k\end{array}\right]+10\left[\begin{array}{c}10\\ 01\end{array}\right]=\left[\begin{array}{c}1-3\\ 2k\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}-5& -3-3k\\ 2+2k& -6+k^2\end{array}\right]-\left[\begin{array}{cc}4& -12\\ 8& 4k\end{array}\right]+\left[\begin{array}{cc}10& 0\\ 0& 10\end{array}\right]=\left[\begin{array}{cc}1& -3\\ 2& k\end{array}\right]$
$\Rightarrow \left[\begin{array}{c}19-3k\\ -6+2k4+k^2-4k\end{array}\right]=\left[\begin{array}{c}1-3\\ 2k\end{array}\right]$
$\Rightarrow 9-3k=-3,-6+2k=\mathrm{2...}\left(1\right)$
and $4+k^2-4k=k$
$\Rightarrow k^2-5k+4=0\Rightarrow k=4,1$
But, $k=1$ does not satisfy the Eq $\left(1\right)$ .