If $A=\left[\begin{array}{cc}1 & 0 \ 0 & -1\end{array}\right], P=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}\right]$ and $X=A P A^T$, then $A^T X^{50} A=$

Question

If $A=\left[\begin{array}{cc}1 & 0 \ 0 & -1\end{array}\right], P=\left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}\right]$ and $X=A P A^T$, then $A^T X^{50} A=$, $\left[\begin{array}{ll}0 & 1 \ 1 & 0\end{array}\right]$, $\left[\begin{array}{cc}2 & 1 \ 0 & -1\end{array}\right]$, $\left[\begin{array}{cc}25 & 1 \ 1 & -25\end{array}\right]$, $\left[\begin{array}{cc}1 & 50 \ 0 & 1\end{array}\right]$

MARKS App · Accepted Answer

Given matrix $A=\left[\begin{array}{cc}1 & 0 \ 0 & -1\end{array}\right]$ is orthogonal matrix, because $A A^T=I$. $\begin{aligned} & \text { So, } A^T X^{50} A=A^T X^{49}\left(A P A^T\right) A=A^T X^{49} A P\left(A^T A\right) \ & =A^T X^{49} A P \ & =A^T X^{48}\left(A P A^T\right) A P=A^T X^{48} A P^2 \ldots \ldots \ & \ldots \ldots =A^T A P^{50}=I P^{50}=P^{50} \ & \because \quad P=\left[\begin{array}{ll} 1 & 1 \ 0 & 1 \end{array}\right] \Rightarrow P^2=\left[\begin{array}{ll} 1 & 2 \ 0 & 1 \end{array}\right] \ & \Rightarrow \quad P^3=\left[\begin{array}{ll} 1 & 3 \ 0 & 1 \end{array}\right] \ldots \ldots \ & \Rightarrow \quad P^{50}=\left[\begin{array}{cc} 1 & 50 \ 0 & 1 \end{array}\right] \ \end{aligned}$ So, $\quad A^T X^{50} A=P^{50}=\left[\begin{array}{cc}1 & 50 \ 0 & 1\end{array}\right]$ Hence, option (4) is correct.

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