If $A =\begin{bmatrix}1&a\\0&1\end{bmatrix}$, then $A^n$ (where $n ∈ N$) equals
Answer & explanation
Correct answer: option 1
We have,
$A =\begin{bmatrix}1&a\\0&1\end{bmatrix}$
$∴A^2 =\begin{bmatrix}1&a\\0&1\end{bmatrix}\begin{bmatrix}1&a\\0&1\end{bmatrix}=\begin{bmatrix}1&2a\\0&1\end{bmatrix}$
It can proved by the principle of mathematical induction that $A^n=\begin{bmatrix}1&na\\0&1\end{bmatrix}$ for all $n ∈ N$.