Question

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

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

Answer 1

Answer: (C)

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

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

$A^2=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 4\\ 0& 1\end{array}\right]$

$A^3=\left[\begin{array}{cc}1& 4\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 6\\ 0& 1\end{array}\right]$

$A^4=\left[\begin{array}{cc}1& 6\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 8\\ 0& 1\end{array}\right]$

Continuing in the same manner we have,

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

Hence, option C is correct.