Question

If \( A \) is a square matrix of order \( 3 \) such that \( |2(\text{adj}\,A)| = 288 \), then the possible value of the determinant \( |A| \) is:

• A. \( 144 \)
• B. \( 36 \)
• C. \( \pm 12 \)
• D. \( \pm 6 \)

Hint:

Don't forget to raise the scalar constant to the power of the matrix order (\( k^n \)) before dealing with the adjoint properties. Skipping this step is the most common cause of negative marks in matrix questions!

Answer 1

The Correct Option is D

Concept: For any square matrix \( A \) of order \( n \) and a scalar constant \( k \), the following determinant properties hold true:
• \( |k \cdot A| = k^n \cdot |A| \)
• \( |\text{adj}\,A| = |A|^{n-1} \)

Step 1: Apply the scalar multiplication property of determinants. Given that the order of the matrix \( n = 3 \), we evaluate the scalar factor \( 2 \) pulled outside the determinant brackets: \[ |2(\text{adj}\,A)| = 2^3 \cdot |\text{adj}\,A| = 8 \cdot |\text{adj}\,A| \] We are given that this expression equals \( 288 \): \[ 8 \cdot |\text{adj}\,A| = 288 \quad \Rightarrow \quad |\text{adj}\,A| = \frac{288}{8} = 36 \]

Step 2: Use the adjoint determinant property to find \( |A| \). Using the formula \( |\text{adj}\,A| = |A|^{n-1} \) with \( n = 3 \): \[ |A|^{3-1} = 36 \quad \Rightarrow \quad |A|^2 = 36 \] Taking the square root on both sides yields: \[ |A| = \pm 6 \]