If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ and $A^2 + 7I = kA$, then the value of $k$ is:

5

The correct answer is Option (3) → 5 ##

Given: $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$ = \begin{bmatrix} 3 \times 3 + 1 \times (-1) & 3 \times 1 + 1 \times 2 \\ (-1) \times 3 + 2 \times (-1) & (-1) \times 1 + 2 \times 2 \end{bmatrix}$

$ = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

Now, $A^2 + 7I = kA$

$⇒\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = k \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$⇒\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} = k \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$⇒5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = k \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$⇒k = 5$