If A = [[1/√5, 2/√5], [-2/√5, 1/√5]], B = [[1, 0], [i, 1]], i = √-1, and Q = Aᵀ B A, then the invers

Q: Wait, why can't we just calculate $Q$ directly and then raise it to the power of 2021?

That is a classic trap! Calculating $Q$ directly would be a nightmare, and raising a matrix to the power of 2021 is computationally impossible by hand. In JEE Advanced, whenever you see a high power like 2021, it is a massive hint that there is a pattern or a structural simplification waiting to be discovered. Always look for the 'hidden' structure first!

Q: How do I know if a matrix is orthogonal without being told?

Great question! A matrix $A$ is orthogonal if $A A^T = I$. Whenever you see entries involving square roots and fractions like $\frac{1}{\sqrt{5}}$ and $\frac{2}{\sqrt{5}}$, it is almost always a sign that the matrix is related to rotation or orthogonality. Always perform the quick check $A A^T$ to see if it simplifies to the identity matrix.

Q: Why does the expression $A Q^{2021} A^T$ simplify so easily?

It simplifies because of the associative property of matrix multiplication and the fact that $A A^T = I$. When you substitute $Q^{2021} = A^T B^{2021} A$ into the expression, you get $A (A^T B^{2021} A) A^T$. By regrouping, you get $(A A^T) B^{2021} (A A^T)$, which is just $I B^{2021} I = B^{2021}$. It is a beautiful example of how structure can make complex problems vanish!

Question

If $A = \begin{pmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \ i & 1 \end{pmatrix}$, $i = \sqrt{-1}$, and $Q = A^T B A$, then the inverse of the matrix $A Q^{2021} A^T$ is equal to :

Options:

(A) $\begin{pmatrix} \frac{1}{\sqrt{5}} & -2021 \ 2021 & \frac{1}{\sqrt{5}} \end{pmatrix}$

(B) $\begin{pmatrix} 1 & 0 \ -2021i & 1 \end{pmatrix}$

(C) $\begin{pmatrix} 1 & 0 \ 2021i & 1 \end{pmatrix}$

(D) $\begin{pmatrix} 1 & -2021i \ 0 & 1 \end{pmatrix}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the base of a mountain, looking up at a problem that seems insurmountable. You are given:

$$A = \begin{pmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \ \frac{-2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \ i & 1 \end{pmatrix}$$

You are asked to find the inverse of $A Q^{2021} A^T$, where $Q = A^T B A$. In the world of JEE Advanced, complexity is often just a mask for elegance.

Let us start by examining matrix $A$. Whenever you see entries like $\frac{1}{\sqrt{5}}$ and $\frac{2}{\sqrt{5}}$, your mathematical intuition should scream **orthogonality**.

Let us test this hypothesis by calculating $A A^T$. When we multiply $A$ by its transpose, we find that the result is the identity matrix:

$$I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}$$

Because $A$ is orthogonal, we know that $A^T = A^{-1}$. This property is the key that will unlock the entire problem.

### The Power of Similarity: Why $Q^n$ is Beautiful

Now, let us look at $Q = A^T B A$. We need to find $Q^{2021}$. Let us test the pattern for $Q^2$:

$$Q^2 = (A^T B A)(A^T B A)$$

Because matrix multiplication is associative, we can regroup the middle terms:

$$Q^2 = A^T B (A A^T) B A$$

Since $A A^T = I$, this simplifies to $A^T B I B A$, which is $A^T B^2 A$. If we continue this logic for $Q^3$, we get $Q^3 = A^T B^3 A$.

By induction, we have discovered a powerful general rule:

$$Q^n = A^T B^n A$$

This is not just a calculation; it is a fundamental property of similar matrices. We have successfully reduced the power of $Q$ to the power of $B$.

### The Great Collapse: Simplifying the Expression

Now, let us return to our target expression: $X = A Q^{2021} A^T$. Substituting our new rule for $Q^{2021}$, we get:

$$X = A (A^T B^{2021} A) A^T$$

Again, using the associative property, we group the terms:

$$X = (A A^T) B^{2021} (A A^T)$$

Since $A A^T = I$, the expression collapses into $X = I B^{2021} I = B^{2021}$. Just like that, the entire complex expression involving $A$ and $Q$ has vanished, leaving us with the simple task of finding the inverse of $B^{2021}$.

### The Pattern Hunt: Mastering Matrix $B$

Finally, we need to find $B^{2021}$. Let us look at $B = \begin{pmatrix} 1 & 0 \ i & 1 \end{pmatrix}$. Calculating $B^2$:

$$B^2 = \begin{pmatrix} 1 & 0 \ i & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \ i & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 2i & 1 \end{pmatrix}$$

Calculating $B^3$:

$$B^3 = B^2 B = \begin{pmatrix} 1 & 0 \ 2i & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \ i & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 3i & 1 \end{pmatrix}$$

The pattern is undeniable:

$$B^n = \begin{pmatrix} 1 & 0 \ ni & 1 \end{pmatrix} \implies B^{2021} = \begin{pmatrix} 1 & 0 \ 2021i & 1 \end{pmatrix}$$

To find the inverse, we use the formula $M^{-1} = \frac{1}{|M|} 	ext{adj}(M)$. The determinant $|B^{2021}| = (1)(1) - (0)(2021i) = 1$.

The inverse is simply:

$$\mathbf{\begin{pmatrix} 1 & 0 \ -2021i & 1 \end{pmatrix}}$$

We have reached the summit! The problem that seemed impossible is now solved, revealing the elegance hidden beneath the surface.

Animated Solution for Mathematics - Matrices and Determinants: If $A = (\frac{1}{5} \frac{- 2}{5} \frac{2}{5} \frac{1}{5})$ , $B = (1 i 01)$ , $i = - 1$ , and $Q = A^{T} B A$ , then the inverse of the matrix $A Q^{2021} A^{T}$ is equal to :

Select Answer:

Visualized Solution

Matrices $A$ and $B$

Checking Orthogonality of $A$

Calculating $A A^{T}$

Simplifying $Q^{2}$

Generalizing $Q^{n}$

Expanding $A Q^{2021} A^{T}$

Reduction to $B^{2021}$

Finding Powers of $B$

Generalizing $B^{n}$

Inverse of a $2 \times 2$ Matrix

Final Inverse Calculation

Key Takeaways

The Sigma Insight: Algebraic Operations on Matrices

Analyzing the Setup

The Power of Similarity

The Great Collapse

The Pattern Hunt

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Visualized Solution

Matrices .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Checking Orthogonality of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AAT

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q2

Generalizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Qn

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AQ2021AT

Reduction to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B2021

Finding Powers of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Generalizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bn

Inverse of a .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2×2 Matrix