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: Understanding the Concept:
This question tests the property relating the determinant of the adjugate of a matrix to the determinant of the matrix itself.
Step 2: Key Formula or Approach:
For any square matrix A of order n, the determinant of its adjugate is given by the formula:
\[ |\text{adj}(A)| = |A|^{n-1} \] Step 3: Detailed Explanation:
The given matrix A is a 3x3 matrix, so n = 3. The formula becomes:
\[ |\text{adj}(A)| = |A|^{3-1} = |A|^2 \] First, we need to calculate the determinant of A, |A|.
\[ |A| = \begin{vmatrix} 0 & 0 & \sqrt{3}
0 & \sqrt{3} & 0
\sqrt{3} & 0 & 0 \end{vmatrix} \] We can expand the determinant 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, we can find \(|\text{adj}(A)|\) using the formula:
\[ |\text{adj}(A)| = |A|^2 = (-3\sqrt{3})^2 \] \[ |\text{adj}(A)| = (-3)^2 \cdot (\sqrt{3})^2 = 9 \cdot 3 = 27 \] Step 4: Final Answer:
The value of \(|\text{adj}(A)|\) is 27.