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Mathematics Question on Transpose of a Matrix

If A=[10 0−1],P=[11 01]A=\begin{bmatrix}1 & 0 \\\ 0 & -1\end{bmatrix}, P=\begin{bmatrix}1 & 1 \\\ 0 & 1\end{bmatrix} and X=APATX=A P A^{T}, then ATX50A=A^{T} X^{50} A=

A

[01 10] \begin{bmatrix} 0 & 1 \\\ 1&0\end{bmatrix}

B

[21 0−1] \begin{bmatrix} 2 &1 \\\ 0&-1\end{bmatrix}

C

[251 1\-25] \begin{bmatrix}25 & 1\\\ 1&\- 25 \end{bmatrix}

D

[150 01] \begin{bmatrix} 1& 50\\\ 0& 1\end{bmatrix}

Answer

[150 01] \begin{bmatrix} 1& 50\\\ 0& 1\end{bmatrix}

Explanation

Solution

Given matrix A=[10 0−1]A=\begin{bmatrix}1 & 0 \\\ 0 & -1\end{bmatrix}
is orthogonal matrix, because AAT=IA A^{T}=I.
So, ATX50A=ATX49(APAT)AA^{T} X^{50} A=A^{T} X^{49}\left(A P A^{T}\right) A
=ATX49AP(ATA)=A^{T} X^{49} A P\left(A^{T} A\right)
=ATX49AP=A^{T} X^{49} A P
=ATX48(APAT)AP=ATX48AP2...=A^{T} X^{48}\left(A P A^{T}\right) A P=A^{T} X^{48} A P^{2}...
...=ATAP50=IP50=P50...=A^{T} A P^{50}=I P^{50}=P^{50}
∵P=[11 01]\because P=\begin{bmatrix} 1 & 1 \\\ 0 & 1 \end{bmatrix}
⇒P2=[12 01]\Rightarrow P^{2}=\begin{bmatrix} 1 & 2 \\\ 0 & 1 \end{bmatrix}
⇒P3=[13 01]...\Rightarrow P^{3}=\begin{bmatrix} 1 & 3 \\\ 0 & 1 \end{bmatrix}...
⇒P50=[150 01]\Rightarrow P^{50}=\begin{bmatrix} 1 & 50 \\\ 0 & 1 \end{bmatrix}
So, ATX50A=P50=[150 01]A^{T} X^{50} A=P^{50}=\begin{bmatrix} 1 & 50 \\\ 0 & 1 \end{bmatrix}