Mathematics - Matrices Question with Solution | TestHub

Given $M=\left(\begin{array}{cc}\frac{5}{2}& \frac{3}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)$

$M^2=\left(\begin{array}{cc}\frac{5}{2}& \frac{3}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)\left(\begin{array}{cc}\frac{5}{2}& \frac{3}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)=\left(\begin{array}{cc}4& 3\\ -3& -2\end{array}\right)$

$M^3=\left(\begin{array}{cc}\frac{5}{2}& \frac{3}{2}\\ -\frac{3}{2}& -\frac{1}{2}\end{array}\right)\left(\begin{array}{cc}4& 3\\ -3& -2\end{array}\right)=\left(\begin{array}{cc}\frac{11}{2}& \frac{9}{2}\\ -\frac{9}{2}& -\frac{7}{2}\end{array}\right)$

$M^4=\left(\begin{array}{cc}\frac{11}{2}& \frac{9}{2}\\ -\frac{9}{2}& -\frac{7}{2}\end{array}\right)\left(\begin{array}{cc}\frac{11}{2}& \frac{9}{2}\\ -\frac{9}{2}& -\frac{7}{2}\end{array}\right)=\left(\begin{array}{cc}7& 6\\ -6& -5\end{array}\right)$

and so on

i.e. $M^n=\left(\begin{array}{cc}\frac{3n}{2}+1& \frac{3n}{2}\\ -\frac{3n}{2}& -\frac{3n}{2}+1\end{array}\right)$

Hence,

$M^{2022}=\left(\begin{array}{cc}\frac{3\times 2022}{2}+1& \frac{3\times 2022}{2}\\ -\frac{3\times 2022}{2}& -\frac{3\times 2022}{2}+1\end{array}\right)$

$=\left(\begin{array}{cc}3034& 3033\\ -3033& -3032\end{array}\right)$