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. ` [{:( 1,5),( 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.

### Step 1: Define the matrices
Given:
\[ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \]
\[ B = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \]

### Step 2: Calculate \(B^T\)
The transpose of \(B\) is:
\[ B^T = \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \]

### Step 3: Calculate \(BB^T\)
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 the elements:
- 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:
\[ BB^T = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \]

### Step 4: Calculate \(BB^T A\)
Now we multiply \(BB^T\) by \(A\):
\[ BB^T A = I A = A \]

### Step 5: Calculate \((BB^T A)^5\)
Since \(BB^T A = A\), we have:
\[ (BB^T A)^5 = A^5 \]

### Step 6: Calculate \(A^5\)
To find \(A^5\), we first need to find a pattern in the powers of \(A\).

Calculating \(A^2\):
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \]

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

Calculating \(A^4\):
\[ A^4 = A^3 \cdot A = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix} \]

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

### Final Result
Thus, we find:
\[ (BB^T A)^5 = A^5 = \begin{pmatrix} 1 & 5 \\ 0 & 1 \end{pmatrix} \]