Let $\mathrm{A}=\left[\begin{array}{cc}\frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{cc}1 & -\mathrm{i} \\ 0 & 1\end{array}\right]$, where $\mathrm{i}=\sqrt{-1}$. If $\mathrm{M}=\mathrm{A}^{\mathrm{T}} \mathrm{BA}$, then the inverse of the matrix $\mathrm{AM}^{2023} \mathrm{~A}^{\mathrm{T}}$ is

(D) $\left[\begin{array}{cc}1 & 2023 i \\ 0 & 1\end{array}\right]$

(A) $\left[\begin{array}{cc}1 & -2023 i \\ 0 & 1\end{array}\right]$

(B) $\left[\begin{array}{ll}1 & 0 \\ -2023 i & 1\end{array}\right]$

(C) $\left[\begin{array}{ll}1 & 0 \\ 2023 i & 1\end{array}\right]$

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Matrices

Let

A = [\frac{1}{10} \frac{- 3}{10} \frac{3}{10} \frac{1}{10}]

and

B = [10 - i 1]

, where

i = - 1

. If

M = A^{T} BA

, then the inverse of the matrix

AM^{2023} A^{T}

Answer: (D)

The rows of

A

are orthonormal, so

A A^{T} = I

. With

M = A^{T} B A

, repeated cancellation gives

M^{2023} = A^{T} B^{2023} A

, and therefore

A M^{2023} A^{T} = (A A^{T}) B^{2023} (A A^{T}) = B^{2023}

. For the unitriangular matrix

B = [10 - i 1]

one has

B^{n} = [10 - ni 1]

, so

B^{2023} = [10 - 2023 i 1]

. Its inverse reverses the sign of the off-diagonal entry, giving

[10 2023 i 1]

JEE Main · 2023 · 25 Jan 2023 · Shift 2 · Single correct option