Question

If $A=\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}, P=\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ and $X=A P A^{T}$, then $A^{T} X^{50} A=$

• A. $ \begin{bmatrix} 0 & 1 \\ 1&0\end{bmatrix}$
• B. $ \begin{bmatrix} 2 &1 \\ 0&-1\end{bmatrix}$
• C. $ \begin{bmatrix}25 & 1\\ 1&- 25 \end{bmatrix}$
• D. $ \begin{bmatrix} 1& 50\\ 0& 1\end{bmatrix}$

Answer 1

The Correct Option is D

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