If $A=\begin{bmatrix}0&0&\sqrt{7}\\0&\sqrt{7}&0\\\sqrt{7}&0&0\end{bmatrix}$, then |adj A| is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 343
Given matrix:
$A = \begin{bmatrix} 0 & 0 & \sqrt{7} \\ 0 & \sqrt{7} & 0 \\ \sqrt{7} & 0 & 0 \end{bmatrix}$
Step 1: Compute $\det(A)$
$\det(A) = 0 \cdot (\sqrt{7} \cdot 0 - 0 \cdot 0) - 0 \cdot (0 \cdot 0 - \sqrt{7} \cdot \sqrt{7}) + \sqrt{7} \cdot (0 \cdot 0 - \sqrt{7} \cdot \sqrt{7})$
$= 0 - 0 + \sqrt{7} \cdot (-7) = -7\sqrt{7}$
Step 2: Use the identity $\det(\text{adj } A) = (\det A)^{n - 1}$ for a matrix of order $n = 3$
$\det(\text{adj } A) = (-7\sqrt{7})^2 = 49 \cdot 7 = 343$