Question

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

• A. $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$
• B. $\left[\begin{array}{cc}2& 1\\ 0& -1\end{array}\right]$
• C. $\left[\begin{array}{cc}25& 1\\ 1& -25\end{array}\right]$
• D. $\left[\begin{array}{cc}1& 50\\ 0& 1\end{array}\right]$

Answer 1

Answer: (D)

Given matrix $A=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$
is orthogonal matrix, because $A{A}^T=I$.
So, $A^T{X}^{50}A=A^T{X}^{49}\left(AP{A}^T\right)A$
$=A^T{X}^{49}AP\left(A^{T}A\right)$
$=A^T{X}^{49}AP$
$=A^T{X}^{48}\left(AP{A}^T\right)AP=A^T{X}^{48}A{P}^2...$
$...=A^{T}A{P}^{50}=I{P}^{50}=P^{50}$
$\because P=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$
$\Rightarrow P^2=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$
$\Rightarrow P^3=\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]...$
$\Rightarrow P^{50}=\left[\begin{array}{cc}1& 50\\ 0& 1\end{array}\right]$
So, $A^T{X}^{50}A=P^{50}=\left[\begin{array}{cc}1& 50\\ 0& 1\end{array}\right]$