MCQ
Let $A =\left[\begin{array}{cc}\frac{1}{\sqrt{2}} & -2 \\ 0 & 1\end{array}\right]$ and $P =\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right], \theta>0$.
If $B = PAP ^{ T }, C = P ^{\top} B ^{10} P$ and the sum of the diagonal elements of C is $\frac{ m }{ n }$, where $\operatorname{gcd}( m , n )=$ 1 , then $m + n$ is :
  • A
    65
  • B
    127
  • C
    258
  • D
    2049

Answer

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

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