If `A=[{:(1,1),(0,1):}]` and `B=[{:(sqrt(3)//2,1//2),(-1//2,sqrt(3)//2):}]`, then `("BB"^(T)A)^(5)` is equal to

Options

A. `[{:(2+sqrt(3),1),(-1,2-sqrt(3)):}]`
B. `(1)/(2)[{:(1,5),(0,1):}]`
C. `[{:(1,1),(0,1):}]`
D. `[{:(5,1),(0,1):}]`

Answer

To solve the problem, we need to compute \( (BB^T A)^5 \) where \( A \) and \( B \) are given matrices. Let's go through the steps to find the solution.

### Step 1: Define the matrices \( A \) and \( B \)

The matrices are given as:
\[ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \]
\[ B = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \]

### Step 2: Compute \( B^T \)

The transpose of matrix \( B \) is obtained by swapping rows and columns:
\[ B^T = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \]

### Step 3: Compute \( BB^T \)

Now, we multiply \( B \) by \( B^T \):
\[ BB^T = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \]

Calculating each element:
- First row, first column:
\[ \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1 \]
- First row, second column:
\[ \frac{\sqrt{3}}{2} \cdot -\frac{1}{2} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = 0 \]
- Second row, first column:
\[ -\frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = 0 \]
- Second row, second column:
\[ -\frac{1}{2} \cdot -\frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1 \]

Thus, we have:
\[ BB^T = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Step 4: Compute \( BB^T A \)

Now we multiply \( BB^T \) by \( A \):
\[ BB^T A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \]

Calculating:
\[ BB^T A = \begin{pmatrix} 1 \cdot 1 + 0 \cdot 0 & 1 \cdot 1 + 0 \cdot 1 \\ 0 \cdot 1 + 1 \cdot 0 & 0 \cdot 1 + 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = A \]

### Step 5: Compute \( (BB^T A)^5 \)

Since \( BB^T A = A \), we need to compute \( A^5 \):
\[ A^2 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \]

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

Continuing this process:
\[ A^4 = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \cdot A = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \]

\[ A^5 = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \cdot A = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} \]

### Final Answer

Thus, the final result is:
\[ (BB^T A)^5 = A^5 = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} \]