1. Tardigrade
  2. Question
  3. Mathematics
  4. Let A=[ beginarraycc(1/√10) (3/√10) (-3/√10) (1/√10) endarray] and B=[ beginarraycc1 -i 0 1 endarray], where i=√-1. If M = A T B A, then the inverse of the matrix AM 2023 A T is

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

JEE MainJEE Main 2023Determinants

Solution:

$ AA ^{ T }=\begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}}\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} $
$ B ^2=\begin{bmatrix}1 & - i \\ 0 & 1\end{bmatrix} \begin{bmatrix}1 & - i \\ 0 & 1\end{bmatrix}=\begin{bmatrix} 1 & -2 i \\ 0 & 1\end{bmatrix} $
$ B ^3= \begin{bmatrix}1 & -3 i \\ 0 & 1\end{bmatrix}$
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:
$ B ^{2023}=\begin{bmatrix}1 & -2023 i \\ 0 & 1\end{bmatrix} $
$ M = A ^{ T } B A $
$ M ^2= M \cdot M = A ^{ T } BA A ^{ T } BA = A ^{ T } B ^2 A$
$ M ^3= M ^2 \cdot M = A ^{ T } B ^2 AA ^{ T } BA = A ^{ T } B ^3 A $
:
:
$M ^{2023}=\ldots \ldots \ldots \ldots . . A ^{ T } B ^{2023} A$
$AM ^{2023} A ^{ T }= AA ^{ T } B ^{2023} AA ^{ T }= B ^{2023}$
$=\begin{bmatrix} 1 & -2023 i \\0 & 1\end{bmatrix}$
Inverse of $\left( AM ^{2023} A ^{ T }\right)$ is $\begin{bmatrix}1 & 2023 i \\ 0 & 1\end{bmatrix}$