If \(A=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right], P=\left[\begin{array}{ll}1 & 1 \\ 0 &…
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 \(A^T X^{50} A=\)
- \(\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\)
- \(\left[\begin{array}{cc}2 & 1 \\ 0 & -1\end{array}\right]\)
- \(\left[\begin{array}{cc}25 & 1 \\ 1 & -25\end{array}\right]\)
- \(\left[\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}\right]\)
Solution
Given matrix \(A=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]\) is orthogonal matrix, because \(A A^T=I\).
\(\begin{aligned}
& \text { So, } A^T X^{50} A=A^T X^{49}\left(A P A^T\right) A=A^T X^{49} A P\left(A^T A\right) \\
& =A^T X^{49} A P \\
& =A^T X^{48}\left(A P A^T\right) A P=A^T X^{48} A P^2 \ldots \ldots \\
& \ldots \ldots =A^T A P^{50}=I P^{50}=P^{50} \\
& \because \quad P=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \Rightarrow P^2=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right] \\
& \Rightarrow \quad P^3=\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right] \ldots \ldots \\
& \Rightarrow \quad P^{50}=\left[\begin{array}{cc}
1 & 50 \\
0 & 1
\end{array}\right] \\
\end{aligned}\)
So, \(\quad A^T X^{50} A=P^{50}=\left[\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}\right]\)
Hence, option (4) is correct.
Asked in: AP EAMCET 2019 (20 Apr Shift 1)
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