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

Question

cdquestions Admin · Accepted Answer

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

Answer

$ \begin{bmatrix} 0 & 1 \ 1&0\end{bmatrix}$

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$ \begin{bmatrix} 2 &1 \ 0&-1\end{bmatrix}$

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$ \begin{bmatrix}25 & 1\ 1&- 25 \end{bmatrix}$

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$ \begin{bmatrix} 1& 50\ 0& 1\end{bmatrix}$

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

The Correct Option is D

Solution and Explanation

Top Questions on Transpose of a Matrix

Questions Asked in AP EAPCET exam

If A=[100−1],P=[1101]A=\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}, P=\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}A=[10​0−1​],P=[10​11​] and X=APATX=A P A^{T}X=APAT, then ATX50A=A^{T} X^{50} A=ATX50A=

The Correct Option is D

Solution and Explanation

Top Questions on Transpose of a Matrix

Questions Asked in AP EAPCET exam

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