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:

Question

For the matrices \( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \) and \( B = \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix} \), if \( (A^{15} + B) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} \), then among the following which one is true?}

Answer 1

Provided matrices:

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

Determinants of given matrices:

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

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

Compute \( C = ABA^T \). The determinant of C is calculated as \( \text{det}(C) = (\text{det}(A))^2 \times \text{det}(B) \):

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

For \( X = A^T C^2 A \), the determinant is given by:

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

Using the properties \( \text{det}(A^T) = \text{det}(A) \) and \( \text{det}(C^2) = (\text{det}(C))^2 \):

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

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

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:

The Correct Option is B

Solution and Explanation

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