Q48 / 100 · BITSAT 2024 (Memory Based Paper 1) · Mathematics
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 $A^T X^{50} A=$
- A$\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$
- B$\left[\begin{array}{cc}2 & 1 \\ 0 & -1\end{array}\right]$
- C$\left[\begin{array}{cc}25 & 1 \\ 1 & -25\end{array}\right]$
- Correct answerD$\left[\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}\right]$
Solution
Since $A A^T=I$ therefore matrix $A$ is orthogonal matrix.
Now, $A^T X^{50} A=A^T X^{49}\left(A P A^T\right) A$
$=A^T X^{49} A P\left(A^T A\right)=A^T X^{49} A P$
$=A^T X^{48}\left(A P A^T\right) A P=A^T X^{48} A P^2 \ldots \ldots$.
$=A^T A P^{50}=I P^{50}=P^{50}$
$\because \quad P=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right] \Rightarrow P^2=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$
$\Rightarrow P^3=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right] \ldots .$.
$\Rightarrow \quad P^{50}=\left[\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}\right]$
So, $A^T X^{50} A=P^{50}=\left[\begin{array}{cc}1 & 50 \\ 0 & 1\end{array}\right]$
Hence, the correct answer is option: 'D'.