Question

If A = $\begin{bmatrix} 0 & 0 & \sqrt{3} \\ 0 & \sqrt{3} & 0 \\ \sqrt{3} & 0 & 0 \end{bmatrix}$, then |adj A| is equal to

• A. 3
• B. 9
• C. 27
• D. 81

Hint:

For determinants of matrices with many zeros, always expand along the row or column containing the most zeros to simplify the calculation. Also, remember the key property $|\text{adj}(A)| = |A|^{n-1}$, which is frequently tested.

Answer 1

The Correct Option is C

Step 1: Concept Identification: This question assesses the relationship between the determinant of a matrix and the determinant of its adjugate. Step 2: Governing Principle: For any square matrix A of order n, the determinant of its adjugate is calculated using the formula: \[ |\text{adj}(A)| = |A|^{n-1} \] Step 3: Calculation and Derivation: Given A is a 3x3 matrix, therefore n = 3. The formula simplifies to: \[ |\text{adj}(A)| = |A|^{3-1} = |A|^2 \] First, compute the determinant of A, denoted as |A|. \[ |A| = \begin{vmatrix} 0 & 0 & \sqrt{3} \\ 0 & \sqrt{3} & 0 \\ \sqrt{3} & 0 & 0 \end{vmatrix} \] Expanding along the first row: \[ |A| = 0 \begin{vmatrix} \sqrt{3} & 0 \\ 0 & 0 \end{vmatrix} - 0 \begin{vmatrix} 0 & 0 \\ \sqrt{3} & 0 \end{vmatrix} + \sqrt{3} \begin{vmatrix} 0 & \sqrt{3} \\ \sqrt{3} & 0 \end{vmatrix} \] \[ |A| = 0 - 0 + \sqrt{3} ((0)(0) - (\sqrt{3})(\sqrt{3})) \] \[ |A| = \sqrt{3} (0 - 3) = -3\sqrt{3} \] Now, apply the formula to find \(|\text{adj}(A)|\): \[ |\text{adj}(A)| = |A|^2 = (-3\sqrt{3})^2 \] \[ |\text{adj}(A)| = (-3)^2 \cdot (\sqrt{3})^2 = 9 \cdot 3 = 27 \] Step 4: Conclusion: The determinant of the adjugate of A, \(|\text{adj}(A)|\), is 27.