3 and 4 .Determinants and Matrices
hard

જો $A=\left(\begin{array}{cc}\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right), B=\left(\begin{array}{ll}1 & 0 \\ i & 1\end{array}\right), i=\sqrt{-1}$, અને 

$\mathrm{Q}=\mathrm{A}^{\mathrm{T}} \mathrm{BA}$,તો શ્રેણિક  $\mathrm{A} \mathrm{Q}^{2021} \mathrm{~A}^{\mathrm{T}}$ નો વ્યસ્ત શ્રેણિક મેળવો.

A

$\left(\begin{array}{cc}\frac{1}{\sqrt{5}} & -2021 \\ 2021 & \frac{1}{\sqrt{5}}\end{array}\right)$

B

$\left(\begin{array}{cc}1 & 0 \\ -2021 i & 1\end{array}\right)$

C

$\left(\begin{array}{cc}1 & 0 \\ 2021 i & 1\end{array}\right)$

D

$\left(\begin{array}{cc}1 & -2021 i \\ 0 & 1\end{array}\right)$

(JEE MAIN-2021)

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 $\mathrm{B}^{2021}=\left(\begin{array}{cc}1 & 0 \\ 2021 \mathrm{i} & 1\end{array}\right)$

$\therefore \mathrm{AQ}^{2021} \mathrm{~A}^{\mathrm{T}}=\mathrm{AA}^{\mathrm{T}} \mathrm{B}^{2021} \mathrm{AA}^{\mathrm{T}}=\mathrm{IB}^{2021} \mathrm{I}$

$\Rightarrow \mathrm{AQ}^{2021} \mathrm{~A}^{\mathrm{T}}=\mathrm{B}^{2021}=\left(\begin{array}{cc}1 & 0 \\ 2021 \mathrm{i} & 1\end{array}\right)$

$\therefore\left(\mathrm{AQ}^{2021} \mathrm{~A}^{\mathrm{T}}\right)^{-1}=\left(\begin{array}{cc}1 & 0 \\ 2021 \mathrm{i} & 1\end{array}\right)^{-1}=\left(\begin{array}{cc}1 & 0 \\ -2021 \mathrm{i} & 1\end{array}\right)$

Standard 12
Mathematics

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