Question

If $A=\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right]$, then Aⁿ (where n ∈ N) equals

(a) $\left[\begin{array}{cc}1& na\\ 0& 1\end{array}\right]$

(b) $\left[\begin{array}{cc}1& n^{2}a\\ 0& 1\end{array}\right]$

(c) $\left[\begin{array}{cc}1& na\\ 0& 0\end{array}\right]$

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

Answer 1

(a) $\left[\begin{array}{cc}1& na\\ 0& 1\end{array}\right]$

$$\begin{gathered} \mathrm{Here}, \\ A=\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right] \\ \Rightarrow A^2=\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1+0& a+a\\ 0+0& 0+1\end{array}\right]=\left[\begin{array}{cc}1& 2a\\ 0& 1\end{array}\right] \\ {A}^3=A^2\times A=\left[\begin{array}{cc}1& 2a\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& a\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1+0& a+2a\\ 0+0& 0+1\end{array}\right]=\left[\begin{array}{cc}1& 3a\\ 0& 1\end{array}\right] \end{gathered}$$


This pattern is applicable for all natural numbers.

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