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

Question

Let the foot of the perpendicular from the point $ (1, 2, 4) $ on the line $ \frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3} $ be $ P $. Then the distance of $ P $ from the plane $ 3x + 4y + 12z + 23 = 0 $ is:

Answer 1

To find the inverse of the matrix $ AM^{2023} A^T $, where $ M = A^T B A $, we need to analyze the given matrices and determine the properties of $ M $.

First, we need to calculate $ M = A^T B A $.

A = \left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}\right] B = \left[\begin{array}{cc} 1 & -i \\ 0 & 1 \end{array}\right]

When calculating $ A^T $, we get: A^T = \left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}\right]
Calculate $ M = A^T B A $:
Multiply $ A^T $ with $ B $:
A^T B = \left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}\right] \cdot \left[\begin{array}{cc} 1 & -i \\ 0 & 1 \end{array}\right] = \left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{-i}{\sqrt{10}} - \frac{3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{3i}{\sqrt{10}} + \frac{1}{\sqrt{10}} \end{array}\right]
Then, multiply the result with $ A $:
M = \left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{-i}{\sqrt{10}} - \frac{3}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} & \frac{3i}{\sqrt{10}} + \frac{1}{\sqrt{10}} \end{array}\right] \cdot \left[\begin{array}{cc} \frac{1}{\sqrt{10}} & \frac{3}{\sqrt{10}} \\ \frac{-3}{\sqrt{10}} & \frac{1}{\sqrt{10}} \end{array}\right] \end{align*}
After solving the matrix multiplication, you will find that: M = B

Now, $ M = B $, and since the exponential form of matrix $ M $ is $ M^{2023} = B^{2023} $.

Because $ B $ is an upper triangular matrix of the form:

M = \begin{bmatrix} 1 & -i \\ 0 & 1 \end{bmatrix}

Hence:

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

Thus, the expression $ AM^{2023} A^T $ simplifies to:

A \begin{bmatrix} 1 & -2023i \\ 0 & 1 \end{bmatrix} A^T \end{align*}

It's clear that the inverse of this matrix is itself, except that the sign of the complex number changes in the upper diagonal.

So, the inverse of $ AM^{2023} A^T $ is:

\begin{bmatrix} 1 & 2023i \\ 0 & 1 \end{bmatrix}

The correct option is:

$\begin{bmatrix}1 & 2023 i \\ 0 & 1\end{bmatrix}$

Answer 2