Question

Let \( A \) be a square matrix of order 3 such that \( \text{det}(A) = -2 \) and \( \text{det}(3 \cdot \text{adj}(-6 \cdot \text{adj}(3A))) = 2^{m+n} \cdot 3^{mn} \), where \( m > n \). Then \( 4m + 2n \) is equal to:

Hint:

For matrix determinant properties, remember that the determinant of a product is the product of determinants, and the properties of adjugates and scalar multiplication can simplify calculations.

Answer 1

Correct Answer: 34

We are given: \[ |A| = -2 \] \[ \text{det}(3 \cdot \text{adj}(-6 \cdot \text{adj}(3A))) = 3^3 \cdot \text{det}(\text{adj}(-\text{adj}(3A))) \] Simplifying this: \[ \text{det}(3 \cdot \text{adj}(-6 \cdot \text{adj}(3A))) = 3^3 \cdot 6^6 \cdot \text{det}(3A)^4 \] We know that: \[ 3^{21} \cdot 2^{10} = 3^7 \cdot 2^3 \] Thus, solving for \( m \) and \( n \), we find: \[ m + n = 10, \quad mn = 21 \] Solving for \( m \) and \( n \), we get: \[ m = 7, \quad n = 3 \] Therefore, \( 4m + 2n = 4 \times 7 + 2 \times 3 = 28 + 6 = 34 \).