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
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 12
Since \(A^4 = A^2\cdot A^2\), it is enough to find \(A^2\) by matrix multiplication:
\( \displaystyle A^2 = \begin{bmatrix}1&0\\3&1\end{bmatrix}\begin{bmatrix}1&0\\3&1\end{bmatrix} = \begin{bmatrix}1&0\\6&1\end{bmatrix} \)
\( \displaystyle A^4 = \begin{bmatrix}1&0\\6&1\end{bmatrix}\begin{bmatrix}1&0\\6&1\end{bmatrix} = \begin{bmatrix}1&0\\12&1\end{bmatrix} \)
Comparing with the given \(A^4\), k = 12.
Tip: \(\begin{bmatrix}1&0\\a&1\end{bmatrix}^n = \begin{bmatrix}1&0\\na&1\end{bmatrix}\): the corner entry is multiplied by n, not raised to the power n. So k = 4 × 3 = 12, not \(3^4 = 81\).