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(D) Given $A=\left[\begin{array}{cc}1 & 0 \ 0 & -1\end{array}ight]$. Note that $A^T = A$ and $A^2 = I$,so $A^T A = A A = I$.We are given $X = A P A

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$\left[\begin{array}{cc}1 & 50 \ 0 & 1\end{array}ight]$

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(D) Given $A=\left[\begin{array}{cc}1 & 0 \ 0 & -1\end{array}ight]$. Note that $A^T = A$ and $A^2 = I$,so $A^T A = A A = I$.We are given $X = A P A^T$.Then $X^2 = (A P A^T)(A P A^T) = A P (A^T A) P A^T = A P I P A^T = A P^2 A^T$.By induction,$X^n = A P^n A^T$.Therefore,$X^{50} = A P^{50} A^T$.Now,$A^T X^{50} A = A^T (A P^{50} A^T) A = (A^T A) P^{50} (A^T A) = I P^{50} I = P^{50}$.Given $P = \left[\begin{array}{ll}1 & 1 \ 0 & 1\end{array}ight]$,we observe $P^2 = \left[\begin{array}{ll}1 & 2 \ 0 & 1\end{array}ight]$,$P^3 = \left[\begin{array}{ll}1 & 3 \ 0 & 1\end{array}ight]$,and in general $P^n = \left[\begin{array}{ll}1 & n \ 0 & 1\end{array}ight]$.Thus,$P^{50} = \left[\begin{array}{cc}1 & 50 \ 0 & 1\end{array}ight]$.Hence,$A^T X^{50} A = \left[\begin{array}{cc}1 & 50 \ 0 & 1\end{array}ight]$.

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 $A^T X^{50} A=$

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