JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $A = \begin{bmatrix} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix}$ . If $M = A^T BA$ , find the inverse of $AM^{2023}A^T$

Choose the correct answer:

Explanation

Solving

1. Property of Matrix $A$

First, check if $A$ is an orthogonal matrix by calculating $AA^T$ :

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} \frac{1+9}{10} & \frac{-3+3}{10} \\ \frac{-3+3}{10} & \frac{9+1}{10} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

Since $AA^T = I$ , $A$ is an orthogonal matrix, which means $A^T = A^{-1}$ .

2. Simplifying the expression $AM^{2023}A^T$

Given $M = A^T BA$ .

Then $M^2 = (A^T BA)(A^T BA) = A^T B(AA^T)BA$ .

Since $AA^T = I$ :

M^2 = A^T B I B A = A^T B^2 A

By induction, for any power $n$ :

M^n = A^T B^n A

Therefore:

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

(AA^T) B^{2023} (AA^T) = I \cdot B^{2023} \cdot I = B^{2023}

3. Finding $B^{2023}$

$B = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix}$ is an upper triangular matrix of the form $\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$ .

The property for such matrices is $\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}^n = \begin{bmatrix} 1 & nk \\ 0 & 1 \end{bmatrix}$ .

B^{2023} = \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix}

4. Finding the Inverse

We need the inverse of $B^{2023}$ . The inverse of $\begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix}$ is $\begin{bmatrix} 1 & -x \\ 0 & 1 \end{bmatrix}$ .

(B^{2023})^{-1} = \begin{bmatrix} 1 & -(-2023i) \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix}

Correct Option: (4)

Choose the correct answer:

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