If $A =\begin{bmatrix}1&0\\3&1\end{bmatrix}$ and $A^4 =\begin{bmatrix}1&0\\k&1\end{bmatrix}$, then value of $k$ is

12

The correct answer is Option (3) → 12

Given:

$A=\begin{bmatrix}1 & 0 \\ 3 & 1\end{bmatrix}$

Compute powers:

$A^2=\begin{bmatrix}1 & 0 \\ 6 & 1\end{bmatrix}$

$A^3=\begin{bmatrix}1 & 0 \\ 9 & 1\end{bmatrix}$

$A^4=\begin{bmatrix}1 & 0 \\ 12 & 1\end{bmatrix}$

Given $A^4=\begin{bmatrix}1 & 0 \\ k & 1\end{bmatrix}$

Thus $k=12$.

The value of $k$ is 12.