Step 1: Understanding the Concept:
This problem requires the combination of two different determinant properties:
1. Scalar Multiplication: For any square matrix M of order \(n\), the determinant of a scalar multiple of the matrix is given by \(|kM| = k^n |M|\).
2. Adjoint Property: For any square matrix A of order \(n\), \(|\text{adj } A| = |A|^{n-1}\).
Since we are given a \(3 \times 3\) matrix (\(n=3\)), we must apply these properties carefully in sequence.
Step 2: Key Formula or Approach:
We start with the given equation: \(|2(\text{adj } A)| = 288\).
1. Use the scalar property: pull the '2' out of the determinant.
2. Use the adjoint property: replace \(|\text{adj } A|\) with \(|A|^2\).
3. Solve the resulting equation for \(|A|\).
Step 3: Detailed Explanation:
Let's break down the calculation:
Step 1: Apply the scalar property.
The matrix inside the determinant is \((2 \times \text{adj } A)\). Since \(\text{adj } A\) is a matrix of order 3:
\[ |2(\text{adj } A)| = 2^3 \times |\text{adj } A| \]
\[ 8 \times |\text{adj } A| = 288 \]
Step 2: Simplify for \(|\text{adj } A|\).
\[ |\text{adj } A| = \frac{288}{8} \]
\[ |\text{adj } A| = 36 \]
Step 3: Apply the adjoint property for order \(n = 3\).
\[ |\text{adj } A| = |A|^{3-1} = |A|^2 \]
Substituting our calculated value:
\[ |A|^2 = 36 \]
Step 4: Solve for \(|A|\).
\[ |A| = \pm \sqrt{36} \]
\[ |A| = \pm 6 \]
Step 4: Final Answer:
The determinant of matrix A can be either 6 or -6.
This corresponds to Option (D).