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:
Show 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.
The given information is:
\[
|A| = -2
\]
\[
\text{det}(3 \cdot \text{adj}(-6 \cdot \text{adj}(3A))) = 3^3 \cdot \text{det}(\text{adj}(-\text{adj}(3A)))
\]
This simplifies to:
\[
\text{det}(3 \cdot \text{adj}(-6 \cdot \text{adj}(3A))) = 3^3 \cdot 6^6 \cdot \text{det}(3A)^4
\]
We are given the relationship:
\[
3^{21} \cdot 2^{10} = 3^7 \cdot 2^3
\]
Solving for \( m \) and \( n \) based on this relationship, we find:
\[
m + n = 10, \quad mn = 21
\]
The values for \( m \) and \( n \) are:
\[
m = 7, \quad n = 3
\]
Consequently, \( 4m + 2n = 4 \times 7 + 2 \times 3 = 28 + 6 = 34 \).
