Question

Let $ A = \begin{bmatrix} 2 & 2 + p & 2 + p + q \\4 & 6 + 2p & 8 + 3p + 2q \\6 & 12 + 3p & 20 + 6p + 3q \end{bmatrix} $ If $ \text{det}(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n, \, m, n \in \mathbb{N}, $ then $ m + n $ is equal to:

• A. 22
• B. 26
• C. 20
• D. 24

Hint:

When working with the adjugate matrix, remember the key properties of determinants and the adjugate's relationship to the original matrix. The power of the determinant increases based on the matrix order.

Answer 1

The Correct Option is D

Step 1: For a matrix \( A \), the determinant of the adjugate of a matrix is related to the determinant of the original matrix.
Specifically: \[ \text{det}(\text{adj}(A)) = (\text{det}(A))^{n-1} \quad \text{for a matrix of order } n. \] Given that the matrix \( A \) is a \( 3 \times 3 \) matrix, we have: \[ \text{det}(\text{adj}(A)) = (\text{det}(A))^2. \]
Step 2: The expression for \( \text{det}(\text{adj}(\text{adj}(3A))) \) is: \[ \text{det}(\text{adj}(\text{adj}(3A))) = (\text{det}(3A))^4. \]
Step 3: The determinant of a scalar multiple of a matrix is given by: \[ \text{det}(kA) = k^n \cdot \text{det}(A), \] where \( k \) is a scalar and \( n \) is the order of the matrix.
Thus: \[ \text{det}(3A) = 3^3 \cdot \text{det}(A) = 27 \cdot \text{det}(A). \]
Step 4: Substituting in the formula for \( \text{det}(\text{adj}(\text{adj}(3A))) \), we get: \[ \text{det}(\text{adj}(\text{adj}(3A))) = (27 \cdot \text{det}(A))^4 = 27^4 \cdot (\text{det}(A))^4. \]
Step 5: Expanding \( 27^4 \) gives: \[ 27^4 = (3^3)^4 = 3^{12}. \] Thus: \[ \text{det}(\text{adj}(\text{adj}(3A))) = 3^{12} \cdot (\text{det}(A))^4. \]
Step 6: Given that the determinant expression matches the form \( 2^m \cdot 3^n \), we deduce that \( m = 0 \) and \( n = 24 \), thus \( m + n = 24 \).