If $A=\begin{bmatrix} 1& 0 \\\frac{1}{3} & 1\end{bmatrix}$ then $A^{48}$ is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\begin{bmatrix} 1& 0 \\16 & 1\end{bmatrix}$
$A=\begin{bmatrix} 1& 0 \\\frac{1}{3} & 1\end{bmatrix}$
$A^2=\begin{bmatrix} 1& 0 \\\frac{2}{3} & 1\end{bmatrix}$
$A^3=\begin{bmatrix} 1& 0 \\\frac{3}{3} & 1\end{bmatrix}$
$⇒A^n=\begin{bmatrix} 1& 0 \\\frac{n}{3} & 1\end{bmatrix}⇒A^{48}=\begin{bmatrix} 1& 0 \\\frac{48}{3} & 1\end{bmatrix}$
$=\begin{bmatrix} 1& 0 \\16 & 1\end{bmatrix}$