Mathematics - Matrices Question with Solution | TestHub

$\begin{array}{rl}& \mathrm{P}=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\\ & \because {\mathrm{P}}^{\mathrm{T}}\mathrm{P}=\mathrm{I}\\ & \mathrm{B}=\mathrm{PAPT}\end{array}$ Pre multiply by ${\mathrm{P}}^{\mathrm{T}}$ ( Given) ${\mathrm{P}}^{\mathrm{T}}\mathrm{B}={\mathrm{P}}^{\mathrm{T}}\mathrm{P}{\mathrm{AP}}^{\mathrm{T}}={\mathrm{AP}}^{\mathrm{T}}$ Now post multiply by P ${\mathrm{P}}^{\mathrm{T}}\mathrm{BP}={\mathrm{AP}}^{\mathrm{T}}\mathrm{P}=\mathrm{A}$
${\mathrm{A}}^2={\mathrm{P}}^{\mathrm{T}}{\mathrm{B}}^2\mathrm{P}$ Similarly $A^{10}=P^T{B}^{10}P=C$ $\begin{array}{rl}& A=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -2\\ 0& 1\end{array}\right]\text{ (Given) }\\ & \Rightarrow A^2=\left[\begin{array}{cc}\frac{1}{2}& -\sqrt{2}-2\\ 0& 1\end{array}\right]\end{array}$ Similarly check $A^3$ and so on since $C=A^{10}$ $\Rightarrow$ Sum of diagonal elements of C is ${\left(\frac{1}{\sqrt{2}}\right)}^{10}+1$ $\begin{array}{rl}& =\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}}\\ & g\; cd(\mathrm{m},\mathrm{n})=1(\text{ Given })\\ & \Rightarrow \mathrm{m}+\mathrm{n}=65\end{array}$