Mathematics - Matrices Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{BCB}}^{-1}=\mathrm{A}\\ & \Rightarrow \left({\mathrm{BCB}}^{-1}\right)\left({\mathrm{BCB}}^{-1}\right)=\mathrm{A}\cdot \mathrm{A}\\ & \Rightarrow \mathrm{BCI}{\mathrm{CB}}^{-1}={\mathrm{A}}^2\\ & \Rightarrow {\mathrm{BC}}^2{\mathrm{B}}^{-1}={\mathrm{A}}^2\\ & \Rightarrow {\mathrm{B}}^{-1}\left({\mathrm{BC}}^2{\mathrm{B}}^{-1}\right)\mathrm{B}={\mathrm{B}}^{-1}\text{ (A.A)B }\end{array}$
From equation (1) $\begin{array}{rl}& {\mathrm{C}}^2={\mathrm{A}}^{-1}\cdot \mathrm{A}\cdot \mathrm{B}\\ & {\mathrm{C}}^2=\mathrm{B}\end{array}$ $\begin{array}{rl}& \text{ Also }{\mathrm{AB}}^{-1}={\mathrm{A}}^{-1}\\ & \Rightarrow {\mathrm{AB}}^{-1}\cdot \mathrm{A}={\mathrm{A}}^{-1}\mathrm{A}=\mathrm{I}\\ & \Rightarrow {\mathrm{A}}^{-1}\left({\mathrm{AB}}^{-1}\mathrm{A}\right)={\mathrm{A}}^{-1}\mathrm{I}\\ & {\mathrm{B}}^{-1}\mathrm{A}={\mathrm{A}}^{-1}\end{array}$
Now characteristics equation of ${\mathrm{C}}^2$ is $\begin{array}{rl}& \left|C_2-\lambda I\right|=0\\ & |B-\lambda I|=0\end{array}$ $\begin{array}{rl}& \Rightarrow \left|\begin{array}{cc}1-\lambda & 3\\ 1& 5-\lambda \end{array}\right|=0\\ & \Rightarrow (1-\lambda )(5-1)-3=0\Rightarrow \left({\lambda }^2-6\lambda +5\right)-3=0\\ & \Rightarrow {\lambda }^2-6\lambda +2=0\\ & \Rightarrow {\beta }^2-6\mathrm{B}+2\mathrm{I}=0\\ & \Rightarrow {\mathrm{C}}^4-6{\mathrm{C}}^2+2\mathrm{I}=0\\ & \alpha =-6\\ & \beta =2\\ & \therefore 2\beta -\alpha =4+6=10\end{array}$