Mathematics - Matrices Question with Solution | TestHub

$\begin{array}{rl}& |\mathrm{A}|=0\\ & \alpha \beta +6=0\\ & \alpha \beta =-6\\ & \alpha +\beta =1\\ & \Rightarrow \alpha =3,\beta =-2\\ & \mathrm{A}=\left[\begin{array}{ll}3& -1\\ 6& -2\end{array}\right]\\ & {\mathrm{A}}^2=\left[\begin{array}{ll}3& -1\\ 6& -2\end{array}\right]\left[\begin{array}{ll}3& -1\\ 6& -2\end{array}\right]=\left[\begin{array}{ll}3& -1\\ 6& -2\end{array}\right]\\ & \therefore {\mathrm{A}}^2=\mathrm{A}\\ & \mathrm{A}={\mathrm{A}}^2={\mathrm{A}}^3={\mathrm{A}}^4={\mathrm{A}}^5\\ & (\mathrm{I}+\mathrm{A}{)}^8\\ & =\mathrm{I}+{}^8{\mathrm{C}}_1{\mathrm{A}}^7+{}^8{\mathrm{C}}_2{\mathrm{A}}^6+\dots ..+{}^8{\mathrm{C}}_8{\mathrm{A}}^8\\ & =\mathrm{I}+\mathrm{A}\left({}^8{\mathrm{C}}_1+{}^8{\mathrm{C}}_2+\dots ..+{}^8{\mathrm{C}}_8\right)\\ & =\mathrm{I}+\mathrm{A}\left(2^8-1\right)\\ & =\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}765& -255\\ 1530& -510\end{array}\right]\end{array}$
$=\left[\begin{array}{cc}766& -255\\ 1530& -509\end{array}\right]$