Animated Solution for Mathematics - Matrices and Determinants: 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 $M=A^{T}BA$, then the inverse of the matrix $A{M}^{2023}{A}^T$ is

Welcome, future engineer! Today, we are going to dismantle a problem that, at first glance, looks like a nightmare of matrix multiplication.

You see $M^{2023}$ and your heart might skip a beat. But in the world of JEE Advanced, whenever you see a high power of a matrix, it is never about brute force. It is always about finding a hidden structure, a pattern, or a symmetry that collapses the complexity into something beautiful.

The first step in any matrix problem is to inspect the players. We have matrix:

Those fractions might look intimidating, but they are actually a massive hint. They are normalized! Whenever you see entries like these, your internal alarm should ring: "Check for orthogonality!"

Let us calculate $A{A}^T$. When we perform the multiplication, the diagonal elements become:

The off-diagonal elements become:

We have discovered that $A{A}^T=I$. This is the "Open Sesame" of our problem. It means $A$ is an orthogonal matrix, and its inverse is simply its transpose, $A^T$.

Now, let us look at $M=A^{T}BA$. We need to find $M^{2023}$. Let us test the waters with $M^2$:

Look at the center of this expression. We have $A{A}^T$. Because $A$ is orthogonal, this middle section collapses into the identity matrix $I$.

Thus, $M^2=A^{T}BIBA=A^T{B}^{2}A$. Do you see the beauty here? The outer $A^T$ and $A$ remain untouched, while the power is transferred entirely to the inner matrix $B$.

By induction, this pattern holds for any power $k$:

This is the "telescoping" effect—the inner terms vanish, leaving us with a much simpler structure.

Our target is the inverse of $X=A{M}^{2023}{A}^T$. Let us substitute our generalized formula for $M^{2023}$ into this expression:

Now, group the terms:

Since $A{A}^T=I$, this simplifies to $X=I\cdot B^{2023}\cdot I=B^{2023}$. The entire, terrifying expression has reduced down to just $B^{2023}$.

We are left with finding $B^{2023}$ where:

Let us calculate $B^2$:

The pattern is clear! The power $k$ simply multiplies the top-right entry. Therefore:

The problem asks for the inverse of this result. For a $2\times 2$ matrix $P=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, the inverse is:

Here, the determinant is $(1)(1)-(0)(-2023i)=1$. The inverse is simply swapping the diagonal elements (which are both 1) and negating the off-diagonal elements.

Thus, the inverse of $\left[\begin{array}{cc}1& -2023i\\ 0& 1\end{array}\right]$ is:

We have arrived at the solution. It wasn't about brute force; it was about seeing the symmetry. Keep this mindset, and no matrix will ever be too complex for you!