If $A=\left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right]$, then $A^{n}($ where $n \in N)$ equals
(a) $\left[\begin{array}{cc}1 & n a \\ 0 & 1\end{array}\right]$
(b) $\left[\begin{array}{cc}1 & n^{2} a \\ 0 & 1\end{array}\right]$
(c) $\left[\begin{array}{cc}1 & n a \\ 0 & 0\end{array}\right]$
(d) $\left[\begin{array}{cc}n & n a \\ 0 & n\end{array}\right]$
(a) $\left[\begin{array}{cc}1 & n a \\ 0 & 1\end{array}\right]$
Here,
$A=\left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1+0 & a+a \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{cc}1 & 2 a \\ 0 & 1\end{array}\right]$
$A^{3}=A^{2} \times A=\left[\begin{array}{cc}1 & 2 a \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & a \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1+0 & a+2 a \\ 0+0 & 0+1\end{array}\right]=\left[\begin{array}{cc}1 & 3 a \\ 0 & 1\end{array}\right]$
This pattern is applicable for all natural numbers.
$\therefore A^{n}=\left[\begin{array}{cc}1 & n a \\ 0 & 1\end{array}\right]$
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