Let A = [ cases 1 10 & 3 10 - 3 10 & 1 10 cases ] and B = [ cases 1 & - i 0 & 1 cases ] , where i = - 1 . If M = A ^ T BA , then the inverse of the matrix AM ^ 2023 A ^ T is

Correct answer: D. [ cases 1 & 2023i 0 & 1 cases ] Given, A = [ cases 1 10 & 3 10 - 3 10 & 1 10 cases ], B = [ cases 1 & - i 0 & 1 cases ] AA ^ T = [ cases 1 10 & 3 10 - 3 10 & 1 10 cases ][ cases 1 10 & - 3 10 3 10 & 1 10 cases ] = [ cases 1 & 0 0 & 1 cases ] B ^ 2 = [ cases 1 & - i 0 & 1 cases ][ cases 1 & - i 0 & 1 cases ] = [ cases 1 & - 2i 0 & 1 cases ] B ^ 3 = [ cases 1 & - 2i 0 & 1 cases ][ cases 1 & - i 0 & 1 cases ] = [ cases 1 & - 3i 0 & 1 cases ] So, we can write B ^ 2023 = [ cases 1 & - 2023i 0 & 1 cases ] M = A ^ T BA (given) M = AA ^ T B = [ cases 1 & 0 0 & 1 cases ][ cases 1 & - i 0 & 1 cases ] = [ cases 1 & - i 0 & 1 cases ] M ^ 2 = M · M = A ^ T BAA ^ T BA = A A ^ T . AA ^ T . B ^ 2 = AA ^ T B ^ 2 M ^ 3 = AA ^ T B ^ 3 M ^ 2023 = AA ^ T B ^ 2023 = B ^ 2023 ( ∵ A A ^ T is an identity matrix) ∴ M ^ 2023 = B ^ 2023 = [ cases 1 & - 2023i 0 & 1 cases ] Inverse of ( AM ^ 2023 A ^ T ) is [ cases 1 & 2023i 0 & 1 cases ]

JEE Main 202325 JanuaryEvening Shift

Exam: JEE Main
Year: 2023
Date: 25 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2023 · 25 January · Evening Shift · Mathematics · Matrices

Let A = [ cases 1 10 & 3 10 - 3 10 & 1 10 cases ] and B = [ cases 1...

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

Options

A $\left[$ \begin{cases}1 & - 2023i \\ 0 & 1\end{cases} $\right]$
B $\left[$ \begin{cases}1 & 0 \\ - 2023i & 1\end{cases} $\right]$
C $\left[$ \begin{cases}1 & 0 \\ 2023i & 1\end{cases} $\right]$
D $\left[$ \begin{cases}1 & 2023i \\ 0 & 1\end{cases} $\right]$

Correct answer

D. $\left[$ \begin{cases}1 & 2023i \\ 0 & 1\end{cases} $\right]$

Step-by-step solution

Given,

$A = \left[$ \begin{cases}\dfrac{1}{\sqrt{10}} & \dfrac{3}{\sqrt{10}} \\ \dfrac{- 3}{\sqrt{10}} & \dfrac{1}{\sqrt{10}}\end{cases} $\right], B = \left[$ \begin{cases}1 & - i \\ 0 & 1\end{cases} $\right]$

${AA}^{T} = \left[$ \begin{cases}\dfrac{1}{\sqrt{10}} & \dfrac{3}{\sqrt{10}} \\ \dfrac{- 3}{\sqrt{10}} & \dfrac{1}{\sqrt{10}}\end{cases} $\right]\left[$ \begin{cases}\dfrac{1}{\sqrt{10}} & \dfrac{- 3}{\sqrt{10}} \\ \dfrac{3}{\sqrt{10}} & \dfrac{1}{\sqrt{10}}\end{cases} $\right] = \left[$ \begin{cases}1 & 0 \\ 0 & 1\end{cases} $\right]$

${B}^{2} = \left[$ \begin{cases}1 & - i \\ 0 & 1\end{cases} $\right]\left[$ \begin{cases}1 & - i \\ 0 & 1\end{cases} $\right] = \left[$ \begin{cases}1 & - 2i \\ 0 & 1\end{cases} $\right]$

${B}^{3} = \left[$ \begin{cases}1 & - 2i \\ 0 & 1\end{cases} $\right]\left[$ \begin{cases}1 & - i \\ 0 & 1\end{cases} $\right] = \left[$ \begin{cases}1 & - 3i \\ 0 & 1\end{cases} $\right]$

So, we can write

${B}^{2023} = \left[$ \begin{cases}1 & - 2023i \\ 0 & 1\end{cases} $\right]$

$M = {A}^{T}BA$ (given)

$M = {AA}^{T}B = \left[$ \begin{cases}1 & 0 \\ 0 & 1\end{cases} $\right]\left[$ \begin{cases}1 & - i \\ 0 & 1\end{cases} $\right] = \left[$ \begin{cases}1 & - i \\ 0 & 1\end{cases} $\right]$

${M}^{2} = M \cdot M = {A}^{T}{BAA}^{T}BA = A{A}^{T} . {AA}^{T} . {B}^{2}$

$= {AA}^{T}{B}^{2}$

${M}^{3} = {AA}^{T}{B}^{3}$

${M}^{2023} = {AA}^{T}{B}^{2023} = {B}^{2023}$ ( $∵ A{A}^{T}$ is an identity matrix)

$∴ {M}^{2023} = {B}^{2023} = \left[$ \begin{cases}1 & - 2023i \\ 0 & 1\end{cases} $\right]$

Inverse of \left({AM}^{2023}{A}^{T}\right) $is \left[$ \begin{cases}1 & 2023i \\ 0 & 1\end{cases} $\right]$

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