Let for A = \(\begin{bmatrix} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}\), |A| = 2. If |2adj (2adj (2A))| = 32n , then 3n + α is equal to
- 9
- 10
- 11
- 12
The Correct Option is C
Solution and Explanation
Step 1: Using the given condition \( |A| = 2 \). We know that: \[ |kA| = k^m \cdot |A| \quad \text{where } m = \text{order of the matrix}. \] From the given, \( |2A| = 2^2 |A| = 4 |A| \), so \[ |2 \, adj(2A)| = |2^7 \, adj(A)| = 2^7 \cdot |adj(A)|. \]
Step 2: Simplifying the expression. Since we know that \( |adj(A)| = |A|^{m-1} \), we can substitute \( |A| = 2 \): \[ |adj(A)| = |A|^2 = 2^2 = 4. \] Thus, \[ |2 \, adj(2 \, adj(2A))| = 2^7 \cdot 4 = 2^7 \cdot 2^2 = 2^{9} = 32^n. \] Equating powers of 2, we get \( n = 5 \).
Step 3: Solving for \( \alpha \). We now use the equation for \( |A| = 2 \): \[ (6-1) - 2(2\alpha - 1) + 3(\alpha - 3) = 2. \] Simplifying: \[ 5 - 4\alpha + 2 + 3\alpha - 9 = 2 \quad \Rightarrow \quad 5 - 4\alpha + 2 + 3\alpha - 9 = 2 \quad \Rightarrow \quad \alpha = -4. \]
Step 4: Final Calculation. Finally, we calculate \( 3n + \alpha \): \[ 3(5) + (-4) = 15 - 4 = 11. \]
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