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:

• A. 3
• B. 2
• C. 4
• D. 6

Answer 1

Answer: (C)

4

Answer 2

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:
$m + 2n = 4$