Question

If $A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$, then for every positive integer $n,A^n$ is equal to

• A. $\left[\begin{array}{cc}1+2n& 4n\\ n& 1+2n\end{array}\right]$
• B. $\left[\begin{array}{cc}1+2n& -4n\\ n& 1-2n\end{array}\right]$
• C. $\left[\begin{array}{cc}1-2n& 4n\\ n& 1+2n\end{array}\right]$
• D. $Noneofthese$

Answer 1

The correct option is A $\left[\begin{array}{cc}1+2n& 4n\\ n& 1+2n\end{array}\right]$

$A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

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

$=\left[\begin{array}{cc}5& 8\\ 2& 5\end{array}\right]$

$\therefore A^3=\left[\begin{array}{cc}5& 8\\ 2& 5\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

$\therefore A^3=\left[\begin{array}{cc}7& 12\\ 3& 7\end{array}\right]$

$\therefore A^4=\left[\begin{array}{cc}7& 12\\ 3& 7\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

$\therefore A^4=\left[\begin{array}{cc}9& 16\\ 4& 9\end{array}\right]$

$\therefore A^n=\left[\begin{array}{cc}1+2n& 4n\\ n& 1+2n\end{array}\right]$.