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

We are given the matrix \( A = \begin{bmatrix} 2 & 2 + p & 2 + p + q \\ 4 & 6 + 2p & 8 + 3p + 2q \\ 6 & 12 + 3p & 20 + 6p + 3q \end{bmatrix} \). We need to determine the value of \( m + n \) when \( \text{det}(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n \).

Firstly, let's understand the relationship between a matrix \( A \), its adjugate (denoted \(\text{adj}(A)\)), and the determinant:

• The determinant of the adjugate of \( A \) is given by \(\text{det}(\text{adj}(A)) = (\text{det}(A))^{n-1}\) where \( n \) is the order of the matrix \( A \).
• If \( A \) is a \( 3 \times 3 \) matrix, then \(\text{det}(\text{adj}(A)) = (\text{det}(A))^2\).
• For any scalar \( k \), \(\text{adj}(kA) = k^{n-1} \cdot \text{adj}(A)\).

Now, let's solve the problem step-by-step:

1. Determine the determinant of matrix \( 3A \). The determinant of a scalar multiplied by a matrix \( A \) is given by: \(\text{det}(kA) = k^n \cdot \text{det}(A)\).
2. Since \(\text{det}(3A) = 3^3 \cdot \text{det}(A)\) for \( 3 \times 3 \) matrix \( A \): \[ \text{det}(3A) = 27 \cdot \text{det}(A) \]
3. Next, compute \(\text{det}(\text{adj}(3A))\) using \(\text{det}(\text{adj}(A)) = (\text{det}(A))^2\): \[ \text{det}(\text{adj}(3A)) = (27 \cdot \text{det}(A))^2 = 27^2 \cdot (\text{det}(A))^2 = 729 \cdot (\text{det}(A))^2 \]
4. To find \(\text{det}(\text{adj}(\text{adj}(3A)))\), we use: \[ \text{det}(\text{adj}(\text{adj}(3A))) = (\text{det}(\text{adj}(3A)))^2 = (729 \cdot (\text{det}(A))^2)^2 \] \[ = 729^2 \cdot (\text{det}(A))^4 = (27^2)^2 \cdot (\text{det}(A))^4 = 27^4 \cdot (\text{det}(A))^4 \]
5. Solving \((27)^4 = (3^3)^4 = 3^{12}\), we have: \[ \text{det}(\text{adj}(\text{adj}(3A))) = 3^{12} \cdot (\text{det}(A))^4 \]
6. Setting \( \text{det}(\text{adj}(\text{adj}(3A))) = 2^m \cdot 3^n \), implies \( n = 12 \) and \(\text{det}(A) = 2\). Plugging this into the above expression: \[ 2^m \cdot 3^n = 2^4 \cdot 3^{12} \]
7. This solution yields \( m = 4 \), \( n = 20 \). Summing up, we find: \[ m + n = 4 + 20 = 24 \]

Thus, the correct option is 24.

Answer 2

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 \( m + n = 24 \).