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

Given $A = \begin{bmatrix}1 & 0\\3 & 1\end{bmatrix}$ and $A^4 = \begin{bmatrix}1 & 0\\k & 1\end{bmatrix}$.

Calculate $A^2$:

$A^2 = A \times A$

$A^2 = \begin{bmatrix}1 & 0\\3 & 1\end{bmatrix} \times \begin{bmatrix}1 & 0\\3 & 1\end{bmatrix}$

Position (1,1): $(1 \times 1) + (0 \times 3) = 1$

Position (1,2): $(1 \times 0) + (0 \times 1) = 0$

Position (2,1): $(3 \times 1) + (1 \times 3) = 6$

Position (2,2): $(3 \times 0) + (1 \times 1) = 1$

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

---

Calculate $A^3$:

$A^3 = A^2 \times A$

$A^3 = \begin{bmatrix}1 & 0\\6 & 1\end{bmatrix} \times \begin{bmatrix}1 & 0\\3 & 1\end{bmatrix}$

Position (1,1): $(1 \times 1) + (0 \times 3) = 1$

Position (1,2): $(1 \times 0) + (0 \times 1) = 0$

Position (2,1): $(6 \times 1) + (1 \times 3) = 9$

Position (2,2): $(6 \times 0) + (1 \times 1) = 1$

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

---

Calculate $A^4$:

$A^4 = A^3 \times A$

$A^4 = \begin{bmatrix}1 & 0\\9 & 1\end{bmatrix} \times \begin{bmatrix}1 & 0\\3 & 1\end{bmatrix}$

Position (1,1): $(1 \times 1) + (0 \times 3) = 1$

Position (1,2): $(1 \times 0) + (0 \times 1) = 0$

Position (2,1): $(9 \times 1) + (1 \times 3) = 12$

Position (2,2): $(9 \times 0) + (1 \times 1) = 1$

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

---

Comparing with $A^4 = \begin{bmatrix}1 & 0\\k & 1\end{bmatrix}$, the value of $k = 12$.