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 break this down step by step.

### Step 1: Define the Matrices
The matrices are defined as follows:
\[ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \]
\[ B = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \]

### 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 will 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 the elements:
1. First row, first column:
\[ \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1 \]
2. First row, second column:
\[ \frac{\sqrt{3}}{2} \cdot -\frac{1}{2} + \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = 0 \]
3. Second row, first column:
\[ -\frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = 0 \]
4. 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:
\[ BB^T = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \]

### Step 4: Compute \(BB^T A\)
Now we multiply \(BB^T\) by \(A\):
\[ BB^T A = I A = A \]
So,
\[ BB^T A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \]

### Step 5: Compute \((BB^T A)^5\)
Finally, we need to compute \((A)^5\):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \]
\[ A^3 = A^2 \cdot A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \]
\[ A^4 = A^3 \cdot A = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \]
\[ A^5 = A^4 \cdot A = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} \]

Thus, \((BB^T A)^5 = A^5 = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}\).

### Final Answer
\[ \boxed{\begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix}} \]