JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
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 \(\left(\begin{array}{cc}\frac{1}{\sqrt{5}} & -2021 \\ 2021 & \frac{1}{\sqrt{5}}\end{array}\right)\)
- 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 \(\left(\begin{array}{cc}1 & -2021 i \\ 0 & 1\end{array}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\begin{array}{cc}1 & 0 \\ -2021 i & 1\end{array}\right)\)
Step-by-step Solution
Detailed explanation
\(\mathrm{AA}^{\mathrm{T}}=\left(\begin{array}{cc}\frac{1}{5} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right)\left(\begin{array}{ll}\frac{1}{\sqrt{5}} & \frac{-2}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right)\)…
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