If \( A = \begin{bmatrix} \frac{1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ -\frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 0 \\ i & 1 \end{bmatrix} \), \( i = \sqrt{-1} \), and \( Q = A^T B A \), then the inverse of the matrix \( A Q^{2021} A^T \) is equal to :
Show Hint
\lambda & 1 \end{bmatrix}\), the powers follow the pattern \(B^n = \begin{bmatrix} 1 & 0
n\lambda & 1 \end{bmatrix}\) and the inverse is \(B^{-1} = \begin{bmatrix} 1 & 0
-\lambda & 1 \end{bmatrix}\).
- \(\begin{bmatrix} 1 & 0 \\ -2021i & 1 \end{bmatrix}\)
- \(\begin{bmatrix} \frac{1}{\sqrt{5}} & -2021 \\ 2021 & \frac{1}{\sqrt{5}} \end{bmatrix}\)
- \(\begin{bmatrix} 1 & -2021i \\ 0 & 1 \end{bmatrix}\)
- \(\begin{bmatrix} 1 & 0 \\ 2021i & 1 \end{bmatrix}\)
The Correct Option is A
Solution and Explanation
Notice that matrix \( A \) is an orthogonal matrix, as \( A A^T = I \). For any matrix \( M = A^T B A \), we have \( M^n = A^T B^n A \).
Step 2: Key Formula or Approach:
1. Identify \( A^T A = I \).
2. \( Q^n = (A^T B A)^n = A^T B^n A \).
3. Calculate \( B^n \) for a given triangular matrix.
Step 3: Detailed Explanation:
First, calculate \( A A^T \):
\[ A A^T = \begin{bmatrix} 1/\sqrt{5} & 2/\sqrt{5} \\ -2/ \sqrt{5} & 1/\sqrt{5} \end{bmatrix} \begin{bmatrix} 1/\sqrt{5} & -2/\sqrt{5} \\ 2/\sqrt{5} & 1/\sqrt{5} \end{bmatrix} = \begin{bmatrix} 1/5 + 4/5 & -2/5 + 2/5 \\ -2/5 + 2/5 & 4/5 + 1/5 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \]
Now, evaluate the matrix expression:\[ A Q^{2021} A^T = A (A^T B^{2021} A) A^T = (A A^T) B^{2021} (A A^T) = I \cdot B^{2021} \cdot I = B^{2021} \]
Next, find \( B^{2021} \):\[ B^2 = \begin{bmatrix} 1 & 0 \\ i & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ i & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 2i & 1 \end{bmatrix}, \quad B^3 = \begin{bmatrix} 1 & 0 \\ 3i & 1 \end{bmatrix}, \dots, B^n = \begin{bmatrix} 1 & 0 \\ ni & 1 \end{bmatrix} \]
Thus, \( A Q^{2021} A^T = \begin{bmatrix} 1 & 0 \\ 2021i & 1 \end{bmatrix} \).The inverse of \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is \( \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \). For \( B^{2021} \), the determinant is \( 1 \cdot 1 - 0 = 1 \).
Inverse = \( \begin{bmatrix} 1 & 0 \\ -2021i & 1 \end{bmatrix} \).
Step 4: Final Answer:
The inverse is \( \begin{bmatrix} 1 & 0 \\ -2021i & 1 \end{bmatrix} \).
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