Solve this
Question:

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) $\left[\begin{array}{cc}n & n a \\ 0 & n\end{array}\right]$

Solution:

(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.

$\therefore A^{n}=\left[\begin{array}{cc}1 & n a \\ 0 & 1\end{array}\right]$