Question

If A =$\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$ then $A^n$ =

• A. $\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}2& n\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}1& n\\ 0& 2\end{array}\right]$
• D. $\left[\begin{array}{cc}1& 2n\\ 0& 1\end{array}\right]$

Answer 1

Answer: (D)

$\left[\begin{array}{cc}1& 2n\\ 0& 1\end{array}\right]$

We have, $A=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$
$\Rightarrow A^2=AA=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1\times 1+2\times 0& 1\times 2+2\times 1\\ 0\times 1+1\times 0& 0\times 2+1\times 1\end{array}\right]=\left[\begin{array}{cc}1& 2\times 2\\ 0& 1\end{array}\right]$
$\Rightarrow A^3=A^{2}A=\left[\begin{array}{cc}1& 4\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1\times 1+4\times 0& 1\times 2+4\times 1\\ 0\times 1+1\times 0& 0\times 2+1\times 1\end{array}\right]=\left[\begin{array}{cc}1& 2\times 3\\ 0& 1\end{array}\right]$
on doing this $n$ times we get,
$A^n=\left[\begin{array}{cc}1& 2n\\ 0& 1\end{array}\right]$