Question

If $A=\begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$ 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)

$A^{2}-4A+10I=A$
$\Rightarrow \begin{bmatrix} 1 \, -3 \\ 2 \, \, \, k \end{bmatrix}\begin{bmatrix} 1 \, -3 \\ 2 \, \, \, k \end{bmatrix}-4\begin{bmatrix} 1 \, \, -3 \\ 2 \, \, \, k \end{bmatrix}+10\begin{bmatrix} 1 \, \, \, 0 \\ 0 \, \, \, 1 \end{bmatrix}=\begin{bmatrix} 1 \, \, -3 \\ 2 \, \, k \end{bmatrix}$
$\Rightarrow \begin{bmatrix} -5 & -3-3k \\ 2+2k & -6+k^{2} \end{bmatrix}-\begin{bmatrix} 4 & -12 \\ 8 & 4k \end{bmatrix}+\begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}=\begin{bmatrix} 1 & -3 \\ 2 & k \end{bmatrix}$
$\Rightarrow \begin{bmatrix} 1 \, \, \, 9-3k \\ -6+2k \, 4+k^{2}-4k \end{bmatrix}=\begin{bmatrix} 1 \, \, -3 \\ 2 \, k \end{bmatrix}$
$\Rightarrow 9-3k=-3,-6+2k=2...\left(1\right)$
and $4+k^{2}-4k=k$
$\Rightarrow k^{2} - 5 k + 4 = 0 \Rightarrow k = 4 , \, 1$
But, $k=1$ does not satisfy the Eq $\left(1\right)$ .