Mathematics - Matrices Question with Solution | TestHub

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

$$\begin{array}{rl}{A}^2& =\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]=\left[\begin{array}{cc}5& -8\\ 2& -3\end{array}\right]\\ A^3& =A^{2}A=\left[\begin{array}{cc}5& -8\\ 2& -3\end{array}\right]\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]=\left[\begin{array}{cc}7& -12\\ 3& -5\end{array}\right]\end{array}$$

Observing $A,A^2,A^3$, we can conclude that $A^n=\left[\begin{array}{cc}1+2n& -4n\\ n& 1-2n\end{array}\right]$.

$$\begin{array}{rl}\det (A^n)& =\left|\begin{array}{cc}1+2n& -4n\\ n& 1-2n\end{array}\right|=(1+2n)(1-2n)-(-4n)(n)\\ & =1-4n^2+4n^2=1\\ \therefore \det (A^{2005})& =1\end{array}$$