Solution: Unraveling Matrix Powers and Inverses

This problem tests our understanding of matrix properties, specifically orthogonal matrices, matrix exponentiation, and the inverse of a 2x2 matrix. The key to solving it efficiently lies in recognizing patterns and leveraging matrix identities to simplify complex expressions.

---

1. Key Concepts and Formulas

Before diving into the solution, let's recall some essential concepts:

• Orthogonal Matrix: A square matrix $A$ is called orthogonal if its transpose is equal to its inverse, i.e., $A^T = A^{-1}$. This implies $A A^T = I$ and $A^T A = I$, where $I$ is the identity matrix.
• Inverse of a 2x2 Matrix: For a matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, its inverse $M^{-1}$ is given by: $M^{-1} = \frac{1}{\det(M)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ where $\det(M) = ad - bc$.
• Matrix Exponentiation: For a matrix $M$, $M^n$ means multiplying $M$ by itself $n$ times. Often, a pattern emerges for $M^n$ after calculating the first few powers ($M^2, M^3, \dots$).

---

2. Step-by-Step Solution

Our goal is to find the inverse of the matrix $X = A Q^{2021} A^T$. We will achieve this by simplifying $X$ first, then finding its inverse.

Step 1: Analyze Matrix A and its Properties

First, let's examine matrix $A$ and calculate the product $A A^T$. Given: A = \left( {\matrix{ {{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \cr {{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr } } \right) The transpose of $A$ is: A^T = \left( {\matrix{ {{1 \over {\sqrt 5 }}} & {{{ - 2} \over {\sqrt 5 }}} \cr {{2 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr } } \right) Now, let's calculate $A A^T$: A A^T = \left( {\matrix{ {{1 \over {\sqrt 5 }}} & {{2 \over {\sqrt 5 }}} \cr {{{ - 2} \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr } } \right) \left( {\matrix{ {{1 \over {\sqrt 5 }}} & {{{ - 2} \over {\sqrt 5 }}} \cr {{2 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} \cr } } \right) A A^T = \left( {\matrix{ {({1 \over {\sqrt 5 }})({1 \over {\sqrt 5 }}) + ({2 \over {\sqrt 5 }})({2 \over {\sqrt 5 }})} & {({1 \over {\sqrt 5 }})({{-2} \over {\sqrt 5 }}) + ({2 \over {\sqrt 5 }})({1 \over {\sqrt 5 }})} \cr {({{-2} \over {\sqrt 5 }})({1 \over {\sqrt 5 }}) + ({1 \over {\sqrt 5 }})({2 \over {\sqrt 5 }})} & {({{-2} \over {\sqrt 5 }})({{-2} \over {\sqrt 5 }}) + ({1 \over {\sqrt 5 }})({1 \over {\sqrt 5 }})} \cr } } \right) A A^T = \left( {\matrix{ {{1 \over 5} + {4 \over 5}} & {{{ - 2} \over 5} + {2 \over 5}} \cr {{{ - 2} \over 5} + {2 \over 5}} & {{4 \over 5} + {1 \over 5}} \cr } } \right) = \left( {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right) = I Since $A A^T = I$, matrix $A$ is an orthogonal matrix. This implies $A^{-1} = A^T$. This property is extremely useful for simplifying matrix expressions, as we will see in the next steps.

Step 2: Find a Pattern for Powers of Q

We are given $Q = A^T B A$. We need to find $Q^{2021}$. Let's calculate the first few powers of $Q$: $Q^1 = A^T B A$ $Q^2 = Q \cdot Q = (A^T B A) (A^T B A)$ Since $A A^T = I$ (from Step 1), we can substitute $I$ in the middle: $Q^2 = A^T B (A A^T) B A = A^T B I B A = A^T B^2 A$ Now let's calculate $Q^3$: $Q^3 = Q^2 \cdot Q = (A^T B^2 A) (A^T B A)$ Again, using $A A^T = I$: $Q^3 = A^T B^2 (A A^T) B A = A^T B^2 I B A = A^T B^3 A$ We observe a clear pattern: $Q^n = A^T B^n A$. Applying this pattern for $n=2021$: ${Q^{2021}} = A^T B^{2021} A$ This simplification is crucial as it transforms the problem of finding powers of $Q$ into finding powers of $B$, which is a simpler matrix.

Step 3: Find a Pattern for Powers of B

Now we need to find $B^{2021}$. Let's calculate the first few powers of $B$: Given: B = \left( {\matrix{ 1 & 0 \cr i & 1 \cr } } \right) B^2 = B \cdot B = \left( {\matrix{ 1 & 0 \cr i & 1 \cr } } \right) \left( {\matrix{ 1 & 0 \cr i & 1 \cr } } \right) = \left( {\matrix{ {1 \cdot 1 + 0 \cdot i} & {1 \cdot 0 + 0 \cdot 1} \cr {i \cdot 1 + 1 \cdot i} & {i \cdot 0 + 1 \cdot 1} \cr } } \right) = \left( {\matrix{ 1 & 0 \cr {2i} & 1 \cr } } \right) B^3 = B^2 \cdot B = \left( {\matrix{ 1 & 0 \cr {2i} & 1 \cr } } \right) \left( {\matrix{ 1 & 0 \cr i & 1 \cr } } \right) = \left( {\matrix{ {1 \cdot 1 + 0 \cdot i} & {1 \cdot 0 + 0 \cdot 1} \cr {2i \cdot 1 + 1 \cdot i} & {2i \cdot 0 + 1 \cdot 1} \cr } } \right) = \left( {\matrix{ 1 & 0 \cr {3i} & 1 \cr } } \right) A clear pattern emerges for $B^n$: {B^n} = \left( {\matrix{ 1 & 0 \cr {ni} & 1 \cr } } \right) This type of matrix, called a unipotent matrix (a triangular matrix with 1s on the diagonal), often exhibits such a simple pattern for its powers. Using this pattern for $n=2021$: {B^{2021}} = \left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right)

Step 4: Simplify the Expression $A Q^{2021} A^T$

Now we substitute the expression for $Q^{2021}$ (from Step 2) into the target expression: $A Q^{2021} A^T = A (A^T B^{2021} A) A^T$ We can rearrange the terms and use the property $A A^T = I$: $A Q^{2021} A^T = (A A^T) B^{2021} (A A^T)$ $A Q^{2021} A^T = I B^{2021} I$ $A Q^{2021} A^T = B^{2021}$ Now, substitute the value of $B^{2021}$ from Step 3: A Q^{2021} A^T = \left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right)

Step 5: Calculate the Inverse

Finally, we need to find the inverse of the simplified expression $A Q^{2021} A^T$, which is $B^{2021}$. Let M = B^{2021} = \left( {\matrix{ 1 & 0 \cr {2021i} & 1 \cr } } \right). Using the formula for the inverse of a 2x2 matrix: $\det(M) = (1)(1) - (0)(2021i) = 1 - 0 = 1$. The inverse is: {(A Q^{2021} A^T)^{ - 1}} = M^{-1} = \frac{1}{\det(M)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \frac{1}{1} \left( {\matrix{ 1 & 0 \cr { - 2021i} & 1 \cr } } \right) {(A Q^{2021} A^T)^{ - 1}} = \left( {\matrix{ 1 & 0 \cr { - 2021i} & 1 \cr } } \right)

Comparing this result with the given options, we find it matches option (B).

---

3. Tips and Common Mistakes

• Recognize Orthogonal Matrices: Always check if $A A^T = I$ or $A^T A = I$. This property simplifies many matrix problems involving powers or inverses. It's a common trick in JEE problems.
• Don't Rush to Multiply: Before performing lengthy matrix multiplications, look for patterns or identities that can simplify the expressions, especially when high powers are involved. Direct calculation of $Q^{2021}$ would be extremely tedious and error-prone.
• Matrix Power Patterns: For matrices with simple structures (like diagonal, triangular, or those with only a few non-zero entries), calculating $M^2, M^3, \dots$ often reveals a clear pattern for $M^n$. Be on the lookout for such patterns.
• Inverse of a Triangular Matrix: For a lower triangular matrix like $B^{2021} = \begin{pmatrix} a & 0 \\ c & d \end{pmatrix}$, its inverse is $\begin{pmatrix} 1/a & 0 \\ -c/(ad) & 1/d \end{pmatrix}$. Notice how the sign of the off-diagonal element changes. This can be a quick check or direct calculation.

---

4. Conclusion and Key Takeaway

This problem demonstrates how a seemingly complex matrix expression can be simplified significantly by:

1. Identifying special matrix types: Recognizing $A$ as an orthogonal matrix ($A A^T = I$).
2. Leveraging matrix identities: Using $A A^T = I$ to simplify powers of $Q$ ($Q^n = A^T B^n A$) and the final expression $A Q^{2021} A^T$.
3. Pattern recognition for matrix powers: Deriving a general form for $B^n$.

The ability to strategically apply these concepts is vital for solving matrix problems efficiently in competitive exams.