If A and B are two square matrices of order `2xx2` such that `A=[(-1,0),(0,1)] , B=[(1,0),(i,1)]` (where `i=sqrt(-1)`) and `A^TB^2021A=Q` then value of `AQA^T` is

Options

A. `[(1,0),(i,1)]`
B. `[(1,0),(2021i,1)]`
C. `[(1,0),(1,1)]`
D. `[(1,0),(2020i,1)]`

Answer

To solve the problem, we need to find the value of \( AQA^T \) given that \( A^T B^{2021} A = Q \), where \( A \) and \( B \) are defined as follows:

\[ A = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\ i & 1 \end{pmatrix} \]

### Step 1: Calculate \( A^T \)

The transpose of matrix \( A \) is:

\[ A^T = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 2: Find \( B^{2021} \)

To find \( B^{2021} \), we first observe the pattern in the powers of \( B \):

1. **Calculate \( B^2 \)**:

\[ B^2 = B \cdot B = \begin{pmatrix} 1 & 0 \\ i & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ i & 1 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 0 \cdot i & 1 \cdot 0 + 0 \cdot 1 \\ i \cdot 1 + 1 \cdot i & i \cdot 0 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2i & 1 \end{pmatrix} \]

2. **Calculate \( B^3 \)**:

\[ B^3 = B^2 \cdot B = \begin{pmatrix} 1 & 0 \\ 2i & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ i & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 3i & 1 \end{pmatrix} \]

From this, we can see a pattern. It appears that:

\[ B^n = \begin{pmatrix} 1 & 0 \\ ni & 1 \end{pmatrix} \]

Thus, for \( n = 2021 \):

\[ B^{2021} = \begin{pmatrix} 1 & 0 \\ 2021i & 1 \end{pmatrix} \]

### Step 3: Substitute \( B^{2021} \) into \( A^T B^{2021} A \)

Now we substitute \( B^{2021} \) into the equation \( A^T B^{2021} A \):

\[ A^T B^{2021} A = A^T \begin{pmatrix} 1 & 0 \\ 2021i & 1 \end{pmatrix} A \]

Calculating \( A^T B^{2021} \):

\[ A^T B^{2021} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 2021i & 1 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 2021i \end{pmatrix} \]

Now, we multiply this result by \( A \):

\[ A^T B^{2021} A = \begin{pmatrix} -1 & 0 \\ 0 & 2021i \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 2021i \end{pmatrix} \]

This gives us \( Q \):

\[ Q = \begin{pmatrix} 1 & 0 \\ 0 & 2021i \end{pmatrix} \]

### Step 4: Calculate \( AQA^T \)

Now we need to find \( AQA^T \):

\[ AQA^T = A \begin{pmatrix} 1 & 0 \\ 0 & 2021i \end{pmatrix} A^T \]

Calculating \( A \begin{pmatrix} 1 & 0 \\ 0 & 2021i \end{pmatrix} \):

\[ A \begin{pmatrix} 1 & 0 \\ 0 & 2021i \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 2021i \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 2021i \end{pmatrix} \]

Now, multiplying by \( A^T \):

\[ AQA^T = \begin{pmatrix} -1 & 0 \\ 0 & 2021i \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 2021i \end{pmatrix} \]

### Final Result

Thus, the value of \( AQA^T \) is:

\[ AQA^T = \begin{pmatrix} 1 & 0 \\ 0 & 2021i \end{pmatrix} \]