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:

Updated On: Nov 4, 2025
  • 243
  • 729
  • 27
  • 891
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Approach Solution - 1

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

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

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

Was this answer helpful?
0
0
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

Given matrices \( A = \begin{bmatrix} \sqrt{2} & 1 \\ -1 & \sqrt{2} \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \), \( C = ABA^\top \) and \( X = A^\top C^2 A \), find \( \det(X) \).

Concept Used:

Properties of determinants:

  • \( \det(AB) = \det(A)\det(B) \)
  • \( \det(A^\top) = \det(A) \)
  • \( \det(A^n) = (\det(A))^n \)

Step-by-Step Solution:

Step 1: Compute \( \det(A) \).

\[ \det(A) = (\sqrt{2})(\sqrt{2}) - (1)(-1) = 2 + 1 = 3 \]

Step 2: Compute \( \det(B) \).

\[ \det(B) = (1)(1) - (0)(1) = 1 \]

Step 3: Express \( \det(X) \) in terms of known determinants.

Given \( C = ABA^\top \), and \( X = A^\top C^2 A \).

\[ \det(X) = \det(A^\top C^2 A) = \det(A^\top) \cdot \det(C^2) \cdot \det(A) \] \[ = \det(A) \cdot (\det(C))^2 \cdot \det(A) = (\det(A))^2 \cdot (\det(C))^2 \]

Step 4: Compute \( \det(C) \).

\[ C = ABA^\top \Rightarrow \det(C) = \det(A) \cdot \det(B) \cdot \det(A^\top) \] \[ = \det(A) \cdot \det(B) \cdot \det(A) = (\det(A))^2 \cdot \det(B) \] \[ = (3)^2 \cdot 1 = 9 \]

Step 5: Compute \( \det(X) \).

\[ \det(X) = (\det(A))^2 \cdot (\det(C))^2 = (3)^2 \cdot (9)^2 = 9 \cdot 81 = 729 \]

Therefore, \( \det(X) = \mathbf{729} \).

Was this answer helpful?
0
0