We start with the given determinant expression:
\[ |3 \, \text{adj}(2 \, \text{adj}(|A|A))| \]
Now simplifying step-by-step:
\[ |3 \, \text{adj}(2|A|^2 \, \text{adj}A)| \]
\[ |3 \cdot 2^2 \cdot |A|^4 \, \text{adj}(\text{adj}A)| \]
We know that \( |\text{adj}A| = |A|^{n-1} \). Hence, substituting this property, we get:
\[ 3^3 \cdot 2^6 \cdot |A|^{12} \cdot |A|^2 = 3^{-13} \cdot 2^{-10} \]
From this,
\[ |A|^{16} = 3^{-16} \cdot 2^{-16} \]
\[ |A| = \frac{1}{6} \]
Now consider:
\[ |3 \, \text{adj}(2A)| = |3 \cdot 2^2 \, \text{adj}A| = 3^3 \cdot 2^6 |A|^2 \]
\[ = 3 \cdot 2^4 \]
Therefore,
\[ M = 4, \quad n = 1 \]
Hence,
\[ |3m + 2n| = 14 \]
Step 1: Simplify \( |3\operatorname{adj}(2\operatorname{adj}(|A|A))| \)
Using determinant properties:
\[ |3\operatorname{adj}(2\operatorname{adj}(|A|A))| = |3| \times |\operatorname{adj}(2)| \times |\operatorname{adj}(|A|A)|. \]
1. Simplify \( |\operatorname{adj}(|A|A)| \):
Using the property \( |\operatorname{adj}(A)| = |A|^{n-1} \), where \( n = 4 \):
\[ |\operatorname{adj}(|A|A)| = |A|^4. \]
2. Simplify \( |\operatorname{adj}(2)| \):
Using \( \operatorname{adj}(kA) = k^{n-1}\operatorname{adj}(A) \):
\[ |\operatorname{adj}(2)| = 2^{n-1} = 2^3. \]
Thus:
\[ |3\operatorname{adj}(2\operatorname{adj}(|A|A))| = 3^3 \times 2^3 \times |A|^4 \times |A|^4 = 3^3 \times 2^3 \times |A|^8. \]
Given:
\[ |3\operatorname{adj}(2\operatorname{adj}(|A|A))| = 2^{-10} \times 3^{-13}. \]
Equating powers of \( 2 \) and \( 3 \):
\[ 2^3 \times |A|^8 = 2^{-10}, \quad 3^3 \times |A|^8 = 3^{-13}. \]
Solve for \( |A| \):
\[ |A| = 2^{-1} \times 3^{-1}. \]
Step 2: Simplify \( |3\operatorname{adj}(2A)| \)
\[ |3\operatorname{adj}(2A)| = |3| \times |\operatorname{adj}(2A)| = 3^2 \times |\operatorname{adj}(2)|^2 \times |A|^2. \] \[ |3\operatorname{adj}(2A)| = 3^2 \times 2^2 \times (2^{-1} \times 3^{-1})^2 = 2^4 \times 3^1. \]
Step 3: Compute \( |3m + 2n| \)
Given \( |A| = 2^m \cdot 3^n \), substitute \( m = -4 \), \( n = -1 \):
\[ 3m + 2n = 3(-4) + 2(-1) = -12 - 2 = -14. \] \[ |3m + 2n| = 14. \]
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