Question:

If $A$ is a square matrix of order 3 such that \[ \det(A) = 3 \] and \[ \det(\text{adj}(-4 \, \text{adj}(-3 \, \text{adj}(3 \, \text{adj}((2A)^{-1}))))) = 2^{m^3 n}, \] then $m + 2n$ is equal to:

Updated On: Nov 26, 2024
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The Correct Option is C

Solution and Explanation

Given \(|A| = 3\), we start with:
\[\left| \text{adj} \left( -4 \, \text{adj} - 3 \, \text{adj} \left( 3 \, \text{adj} \left( 2A^{-1} \right) \right) \right) \right|\]
Step 1: Simplify the innermost expression:
\[= \left| -4 \, \text{adj} \left( -3 \, \text{adj} \left( 3 \, \text{adj} \left( 2A^{-1} \right) \right) \right) \right|^2\]
Step 2: Expand the outer term:
\[= 4^5 \, \left| \text{adj} \left( -3 \, \text{adj} \left( 3 \, \text{adj} \left( 2A^{-1} \right) \right) \right) \right|^2\]
Step 3: Replace the outermost adj with its expression:
\[= 2^{12} \cdot 3^{12} \cdot \left| 3 \, \text{adj} \left( 2A^{-1} \right) \right|^8\]
Step 4: Simplify the term inside the absolute value:
\[= 2^{12} \cdot 3^{12} \cdot 3^8 \cdot \left| \text{adj} \left( 2A^{-1} \right) \right|^8\]
Step 5: Use the property of adjugates:
\[= 2^{12} \cdot 3^{20} \cdot \left| 2A^{-1} \right|^{16}\]
Step 6: Replace \(\left| 2A^{-1} \right|^{16}\) with its determinant form:
\[= 2^{12} \cdot 3^{20} \cdot \frac{1}{|2A|^{16}}\]
Step 7: Substitute \(|2A|^{16} = 2^{16} \cdot |A|^{16}\):
\[= 2^{12} \cdot 3^{20} \cdot \frac{1}{2^{48} \cdot |A|^{16}}\]
Step 8: Replace \(|A| = 3\):
\[= 2^{12} \cdot 3^{20} \cdot \frac{1}{2^{48} \cdot 3^{16}}\]
Step 9: Simplify powers of 2 and 3:
\[= \frac{2^{12}}{2^{48}} \cdot \frac{3^{20}}{3^{16}} = \frac{1}{2^{36}} \cdot 3^4\]
Step 10: Further simplify:
\[= 2^{-36} \cdot 3^4\]
Step 11: Combine the terms:
\[m = -36, \quad n = 20\]
Step 12: Final result:
\[m + 2n = 4\]

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