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3 and 4 . determinant and metrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths3 and 4 . determinant and metrices1 Mark
MCQ
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 :

  • A
    $\left(\begin{array}{cc}\frac{1}{\sqrt{5}} & -2021 \\ 2021 & \frac{1}{\sqrt{5}}\end{array}\right)$
  • ✓
    $\left(\begin{array}{cc}1 & 0 \\ -2021 i & 1\end{array}\right)$
  • C
    $\left(\begin{array}{cc}1 & 0 \\ 2021 i & 1\end{array}\right)$
  • D
    $\left(\begin{array}{cc}1 & -2021 i \\ 0 & 1\end{array}\right)$
✓

Answer

Correct option: B.
$\left(\begin{array}{cc}1 & 0 \\ -2021 i & 1\end{array}\right)$
b
$\mathrm{AA}^{\mathrm{T}}=\left(\begin{array}{cc}\frac{1}{5} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right)\left(\begin{array}{ll}\frac{1}{\sqrt{5}} & \frac{-2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right)$

$\mathrm{AA}^{\mathrm{T}}=\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)=\mathrm{I}$

$\mathrm{Q}^{2}=\mathrm{A}^{\mathrm{T}} \mathrm{B} \mathrm{A} \mathrm{A}^{\mathrm{T}} \mathrm{BA}=\mathrm{A}^{\mathrm{T}} \mathrm{BIB} \mathrm{A}$

$\Rightarrow Q^{2}=A^{T} B^{2} A$

$Q^{3}=A^{T} B^{2} A A^{T} B A \Rightarrow Q^{3}=A^{T} B^{3} A$

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

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

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

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

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