$A = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}$ and $A^2 - kA - 5I = 0$, then the value of $k$ is:

$5$

The correct answer is Option (2) → $5$ ##

$A = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}$

$A^2=A.A$

$A^2 = \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix}$

$ = \begin{bmatrix} 1+9 & 3+12 \\ 3+12 & 9+16 \end{bmatrix}$

$ = \begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix}$

Now, $A^2 - kA - 5I = 0$ [Given]

$\begin{bmatrix} 10 & 15 \\ 15 & 25 \end{bmatrix} - k \begin{bmatrix} 1 & 3 \\ 3 & 4 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\begin{bmatrix} 10-k-5 & 15-3k-0 \\ 15-3k-0 & 25-4k-5 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\begin{bmatrix} 5-k & 15-3k \\ 15-3k & 20-4k \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

On comparing:

$5 - k = 0$

$ ⇒k = 5$

or, $15 - 3k = 0$

$ ⇒k = 5$

or, $20 - 4k = 0$

$ ⇒k = 5$

Hence, $k = 5$.