If `A=[(sqrt2,1),(-1,sqrt2)],B=[(1,0),(0,1)], C=ABA^T,X=A^TC^2A` then det (x) is equal to

Options

A. 729
B. 726
C. 728
D. 723

Answer

To solve the problem, we need to find the determinant of the matrix \( X = A^T C^2 A \), where \( C = A B A^T \) and \( A \) and \( B \) are given matrices.

### Step-by-Step Solution:

1. **Define the Matrices**:
Given:
\[ A = \begin{pmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \quad (B \text{ is the identity matrix}) \]

2. **Calculate \( C \)**:
\[ C = A B A^T = A A^T \]
First, we need to find \( A^T \):
\[ A^T = \begin{pmatrix} \sqrt{2} & -1 \\ 1 & \sqrt{2} \end{pmatrix} \]

Now, calculate \( A A^T \):
\[ A A^T = \begin{pmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{pmatrix} \begin{pmatrix} \sqrt{2} & -1 \\ 1 & \sqrt{2} \end{pmatrix} \]
Performing the multiplication:
- First row, first column: \( \sqrt{2} \cdot \sqrt{2} + 1 \cdot 1 = 2 + 1 = 3 \)
- First row, second column: \( \sqrt{2} \cdot -1 + 1 \cdot \sqrt{2} = -\sqrt{2} + \sqrt{2} = 0 \)
- Second row, first column: \( -1 \cdot \sqrt{2} + \sqrt{2} \cdot 1 = -\sqrt{2} + \sqrt{2} = 0 \)
- Second row, second column: \( -1 \cdot -1 + \sqrt{2} \cdot \sqrt{2} = 1 + 2 = 3 \)

Thus,
\[ C = A A^T = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = 3I \]
where \( I \) is the identity matrix.

3. **Calculate \( C^2 \)**:
\[ C^2 = (3I)^2 = 9I \]

4. **Calculate \( X \)**:
\[ X = A^T C^2 A = A^T (9I) A = 9 A^T A \]

5. **Calculate \( A^T A \)**:
We already have \( A^T \):
\[ A^T A = \begin{pmatrix} \sqrt{2} & -1 \\ 1 & \sqrt{2} \end{pmatrix} \begin{pmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{pmatrix} \]
Performing the multiplication:
- First row, first column: \( \sqrt{2} \cdot \sqrt{2} + (-1) \cdot (-1) = 2 + 1 = 3 \)
- First row, second column: \( \sqrt{2} \cdot 1 + (-1) \cdot \sqrt{2} = \sqrt{2} - \sqrt{2} = 0 \)
- Second row, first column: \( 1 \cdot \sqrt{2} + \sqrt{2} \cdot (-1) = \sqrt{2} - \sqrt{2} = 0 \)
- Second row, second column: \( 1 \cdot 1 + \sqrt{2} \cdot \sqrt{2} = 1 + 2 = 3 \)

Thus,
\[ A^T A = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} = 3I \]

6. **Final Calculation of \( X \)**:
\[ X = 9 A^T A = 9 (3I) = 27I \]

7. **Calculate the Determinant of \( X \)**:
The determinant of a scalar multiple of the identity matrix is given by:
\[ \det(X) = \det(27I) = 27^2 \cdot \det(I) = 27^2 \cdot 1 = 729 \]

### Final Answer:
\[ \det(X) = 729 \]