If $\mathrm{A}=\left(\begin{array}{cc}\frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}}\end{array}\right), \mathrm{B}=\left(\begin{array}{cc}1 & 0 \\ \mathrm{i} & 1\end{array}\right), \mathrm{i}=\sqrt{-1}$, and $\mathrm{Q}=\mathrm{A}^{\top} \mathrm{BA}$, then the inverse of the matrix $\mathrm{AQ}^{2021} \mathrm{~A}^{\top}$ is equal to

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

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

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

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

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Matrices

A = (\frac{1}{5} \frac{- 2}{5} \frac{2}{5} \frac{1}{5}), B = (1 i 01), i = - 1

, and

Q = A^{⊤} BA

, then the inverse of the matrix

AQ^{2021} A^{⊤}

is equal to

Answer: (C)

The columns of

A

are orthonormal, so

A^{⊤} A = A A^{⊤} = I

.
From

Q = A^{⊤} B A

, each adjacent pair

A A^{⊤}

in the product collapses to the identity, so

Q^{n} = A^{⊤} B^{n} A

and therefore

A Q^{2021} A^{⊤} = (A A^{⊤}) B^{2021} (A A^{⊤}) = B^{2021} .

Writing

B = I + N

with

N = (0 i 00)

and

N^{2} = O

, the binomial expansion terminates after two terms:

B^{n} = I + n N = (1 ni 01), B^{2021} = (1 2021 i 01) .

A lower triangular matrix of this form is inverted by reversing the sign of the off-diagonal entry, so the required inverse is

(1 - 2021 i 01)

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option