Question

Prove that : $A^2-6A-17I_2=0$, where $A=\left[\begin{array}{cc}2& -3\\ 3& 4\end{array}\right]$ and hence if det$\left(A^{-1}\right)=\frac{1}{m}$.Find $m$

Answer 1

$A=\left[\begin{array}{cc}2& -3\\ 3& 4\end{array}\right]A^2=\left[\begin{array}{cc}2& -3\\ 3& 4\end{array}\right]\times \left[\begin{array}{cc}2& -3\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}4-9& -6-12\\ 6+12& -9+16\end{array}\right]=\left[\begin{array}{cc}-5& -18\\ 18& 7\end{array}\right]6A=\left[\begin{array}{cc}12& -18\\ 18& 24\end{array}\right]A^2-6A+17I_2=\left[\begin{array}{cc}-5-12+17& -18+18\\ 18-18& 7-24+17\end{array}\right]$

$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0$ [Proved]

$A^{-1}=\frac{1}{17}{\left[\begin{array}{cc}4& -3\\ 3& 2\end{array}\right]}^T=\frac{1}{17}\left[\begin{array}{cc}4& 3\\ -3& 2\end{array}\right]det\left(A^{-1}\right)=\frac{8+9}{17\times 17}=\frac{1}{17}$