Let A=[array|c|c| 1√2 & -2 0 & 1 array] and P=[array|c|c| & - & array], 0.

Question

Let $A=[\begin{array}{|c|c|} \frac{1}{\sqrt{2}} & -2 \ 0 & 1 \end{array}]$ and $P=[\begin{array}{|c|c|} \cos 	heta  & -\sin 	heta \ \sin 	heta  & \cos 	heta \end{array}],	heta >0$. If $B=PAP{}^{},C={P}^{}{B}^{10}P$ and the sum of the diagonal elements of $C$ is $\frac{m}{n}$, where $gcd(m,n)=1$, then $m+n$ is :

Accepted Answer

$$\begin{aligned}
& \mathrm{P}=\left[\[\begin{array}{cc} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{array}\end{aligned}$$\right] \
\& \because \mathrm{P}\^{\mathrm{T}} \mathrm{P}=\mathrm{I} \
\& \mathrm{~B}=\mathrm{PAPT}
\]

Pre multiply by $\mathrm{P}^{\mathrm{T}}$ ( Given)

$$\mathrm{P}^{\mathrm{T}} \mathrm{~B}=\mathrm{P}^{\mathrm{T}} \mathrm{PA} \mathrm{P}^{\mathrm{T}}=\mathrm{AP}^{\mathrm{T}}$$

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{aligned}
& A=\left[\[\begin{array}{cc} \frac{1}{\sqrt{2}} & -2 \ 0 & 1 \end{array}\end{aligned}$$\right] \text { (Given) } \
\& \Rightarrow A\^2=\left[$$\begin{array}{cc} \frac{1}{2} & -\sqrt{2}-2 \ 0 & 1 \end{array}$$\right]
\]

Similarly check $\mathrm{A}^3$ and so on since $\mathrm{C}=\mathrm{A}^{10}$

$\Rightarrow$ Sum of diagonal elements of C is $\left(\frac{1}{\sqrt{2}}\right)^{10}+1$

$$\begin{aligned}
& =\frac{1}{32}+1=\frac{33}{32}=\frac{\mathrm{m}}{\mathrm{n}} \
& \mathrm{~g} \mathrm{~cd}(\mathrm{~m}, \mathrm{n})=1 \text { (Given) } \
& \Rightarrow \mathrm{m}+\mathrm{n}=65
\end{aligned}$$

Solutions · 1