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If $A=\left(\begin{array}{cc}\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right), B=\left(\begin{array}{ll}1 & 0 \\ i & 1\end{array}\right), i=\sqrt{-1}$, and
$\mathrm{Q}=\mathrm{A}^{\mathrm{T}} \mathrm{BA}$, then the inverse of the matrix $\mathrm{A} \mathrm{Q}^{2021} \mathrm{~A}^{\mathrm{T}}$ is equal to :
$\left(\begin{array}{cc}\frac{1}{\sqrt{5}} & -2021 \\ 2021 & \frac{1}{\sqrt{5}}\end{array}\right)$
$\left(\begin{array}{cc}1 & 0 \\ -2021 i & 1\end{array}\right)$
$\left(\begin{array}{cc}1 & 0 \\ 2021 i & 1\end{array}\right)$
$\left(\begin{array}{cc}1 & -2021 i \\ 0 & 1\end{array}\right)$
Solution
$\mathrm{AA}^{\mathrm{T}}=\left(\begin{array}{cc}\frac{1}{5} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right)\left(\begin{array}{ll}\frac{1}{\sqrt{5}} & \frac{-2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right)$
$\mathrm{AA}^{\mathrm{T}}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)=\mathrm{I}$
$\mathrm{Q}^{2}=\mathrm{A}^{\mathrm{T}} \mathrm{B} \mathrm{A} \mathrm{A}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{BIB} \mathrm{A}$
$\Rightarrow Q^{2}=A^{T} B^{2} A$
$Q^{3}=A^{T} B^{2} A A^{T} B A \Rightarrow Q^{3}=A^{T} B^{3} A$
Similarly $: \mathrm{Q}^{2021}=\mathrm{A}^{\mathrm{T}} \mathrm{B}^{2021} \mathrm{~A} \ldots \ldots .$ (1)
Now $\mathrm{B}^{2}=\left(\begin{array}{ll}1 & 0 \\ \mathrm{i} & 1\end{array}\right)\left(\begin{array}{ll}1 & 0 \\ \mathrm{i} & 1\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 2 \mathrm{i} & 1\end{array}\right)$
$\mathrm{B}^{3}=\left(\begin{array}{ll}1 & 0 \\ 2 \mathrm{i} & 1\end{array}\right)\left(\begin{array}{ll}1 & 0 \\ \mathrm{i} & 1\end{array}\right) \Rightarrow \mathrm{B}^{3}=\left(\begin{array}{cc}1 & 0 \\ 3 \mathrm{i} & 1\end{array}\right)$
Similarly $\mathrm{B}^{2021}=\left(\begin{array}{cc}1 & 0 \\ 2021 \mathrm{i} & 1\end{array}\right)$
$\therefore \mathrm{AQ}^{2021} \mathrm{~A}^{\mathrm{T}}=\mathrm{AA}^{\mathrm{T}} \mathrm{B}^{2021} \mathrm{AA}^{\mathrm{T}}=\mathrm{IB}^{2021} \mathrm{I}$
$\Rightarrow \mathrm{AQ}^{2021} \mathrm{~A}^{\mathrm{T}}=\mathrm{B}^{2021}=\left(\begin{array}{cc}1 & 0 \\ 2021 \mathrm{i} & 1\end{array}\right)$
$\therefore\left(\mathrm{AQ}^{2021} \mathrm{~A}^{\mathrm{T}}\right)^{-1}=\left(\begin{array}{cc}1 & 0 \\ 2021 \mathrm{i} & 1\end{array}\right)^{-1}=\left(\begin{array}{cc}1 & 0 \\ -2021 \mathrm{i} & 1\end{array}\right)$
Similar Questions
Match the Statements / Expressions in Column $I$ with the Statements / Expressions in Column $II$ and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the $ORS$.
| Column $I$ | Column $II$ |
| $(A)$ The minimum value of $\frac{x^2+2 x+4}{x+2}$ is | $(p)$ $0$ |
| $(B)$ Let $A$ and $B$ be $3 \times 3$ matrices of real numbers, where $A$ is symmetric, $B$ is skewsymmetric, and $(A+B)(A-B)=(A-B)(A+B)$. If $(A B)^t=(-1)^k A B$, where $(A B)^t$ is the transpose of the matrix $A B$, then the possible values of $k$ are | $(q)$ $1$ |
| $(C)$ Let $\mathrm{a}=\log _3 \log _3 2$. An integer $\mathrm{k}$ satisfying $1<2^{\left(-k+3^{-2}\right)}<2$, must be less than | $(r)$ $2$ |
| $(D)$ If $\sin \theta=\cos \phi$, then the possible values of $\frac{1}{\pi}\left(\theta \pm \phi-\frac{\pi}{2}\right)$ are | $(s)$ $3$ |
