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

Answer: (B)

729

Answer 2

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 $\text{det}(C^2) = (\text{det}(C))^2$:

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

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