Question

If $A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$, then $A^n$ is equal to $(n\in N)$-

• A. $\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}n& n\\ 0& n\end{array}\right]$
• C. $\left[\begin{array}{cc}n& 1\\ 0& 1\end{array}\right]$
• D. none of these

Answer 1

The correct option is A $\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$

$A=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$
$A^2=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$
$A^3=A^2.A=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$
Similarly $A^n=\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$
Ans: A