Question

If \( A = \begin{bmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \), \( C = ABA^\top \) and \( X = A^\top C^2 A \), then \( \det (X) \) is equal to:

• A. 243
• B. 729
• C. 27
• D. 891

Answer 1

The Correct Option is B

Given:

\( A = \begin{bmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)

First, find:

\( \text{det}(A) = (\sqrt{2}) \times (\sqrt{2}) - (1) \times (-1) = 3 \)

\( \text{det}(B) = 1 \)

Now, compute \( C = ABA^T \). Since \( \text{det}(C) = (\text{det}(A))^2 \times \text{det}(B) \):

\( \text{det}(C) = 3^2 \times 1 = 9 \)

For \( X = A^T C^2 A \), we use:

\( \text{det}(X) = [\text{det}(A^T)] \times [\text{det}(C^2)] \times [\text{det}(A)] \)

Since \( \text{det}(A^T) = \text{det}(A) \) and \( \text{det}(C^2) = (\text{det}(C))^2 \):

\( \text{det}(X) = (\text{det}(A)) \times (\text{det}(C))^2 \times (\text{det}(A)) \)

\( \text{det}(X) = 3 \times 9^2 \times 3 = 729 \)