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

Options

A. ` [{:( (sqrt(3))/(2) , ( 1)/(2) ),(1,0):}]`
B. ` [{:( 1,0),( (sqrt(3))/(2) ,1) :}]`
C. ` [{:( 1,(sqrt3)/(2) ),(0,3):}]`
D. ` [{:(1,3),( 0,1):}]`

Answer

To solve the problem step by step, we need to compute \( A^T C^3 A \) where \( C = ABA^T \). Let's break it down:

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

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

### Step 3: Compute \( C = ABA^T \)
Now we need to compute \( C \):
1. First, calculate \( AB \):
\[ AB = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} + \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} + \frac{\sqrt{3}}{2} \end{pmatrix} \]
Calculating the entries:
- First row: \( \left(\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2} + \frac{1}{2}\right) \)
- Second row: \( \left(-\frac{1}{2}, -\frac{1}{2} + \frac{\sqrt{3}}{2}\right) \)

Thus,
\[ AB = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} + \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} + \frac{\sqrt{3}}{2} \end{pmatrix} \]

2. Now compute \( C = AB A^T \):
\[ C = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2} + \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} + \frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \]
Calculating the entries of \( C \):
- First row, first column: \( \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} + \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right) \cdot \frac{1}{2} \)
- First row, second column: \( \frac{\sqrt{3}}{2} \cdot -\frac{1}{2} + \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right) \cdot \frac{\sqrt{3}}{2} \)
- Second row, first column: \( -\frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}\right) \cdot \frac{1}{2} \)
- Second row, second column: \( -\frac{1}{2} \cdot -\frac{1}{2} + \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}\right) \cdot \frac{\sqrt{3}}{2} \)

### Step 4: Compute \( C^3 \)
To compute \( C^3 \), we first need \( C^2 \) and then multiply by \( C \) again.

### Step 5: Compute \( A^T C^3 A \)
Finally, we compute \( A^T C^3 A \).

### Final Result
After performing all calculations, we find that:
\[ A^T C^3 A = B^3 \]
Where \( B^3 \) is computed by multiplying \( B \) with itself three times.