Question

Let $ A $ be a matrix of order $ 3 \times 3 $ and $ |A| = 5 $. If $ |2 \, \text{adj}(3A \, \text{adj}(2A))| = 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma}, \quad \alpha, \beta, \gamma \in \mathbb{N} $ then $ \alpha + \beta + \gamma $ is equal to

• A. \( 25 \)
• B. \( 26 \)
• C. \( 27 \)
• D. \( 28 \)

Hint:

To solve matrix determinant problems involving adjugates and scalar multiplication, remember the key formulas: \[ |\text{adj}(A)| = |A|^{n-1}, \quad |kA| = k^n |A| \] Also, the adjugate of a product doesn't distribute over multiplication like regular matrices, so simplify the inner terms first before applying adjugate or determinant rules.

Answer 1

The Correct Option is C

We begin with the expression: \[ |2 \, \text{adj}(3A \, \text{adj}(2A))| \] We use the following identities:

• \( |\text{adj}(M)| = |M|^{n-1} \) for an \( n \times n \) matrix.
• \( |kM| = k^n |M| \) for scalar \( k \).

Let us simplify step-by-step: \[ |2 \, \text{adj}(3A \, \text{adj}(2A))| = 2^3 \cdot |\text{adj}(3A \, \text{adj}(2A))| \] \[ = 2^3 \cdot |3A \, \text{adj}(2A)|^2 \] \[ = 2^3 \cdot \left( 3^3 \cdot |A| \cdot |\text{adj}(2A)| \right)^2 \] \[ = 2^3 \cdot \left( 3^3 \cdot |A| \cdot (2^3 \cdot |A|)^2 \right)^2 \] \[ = 2^3 \cdot \left( 2^6 \cdot 3^3 \cdot |A|^3 \right)^2 \] \[ = 2^3 \cdot 2^{12} \cdot 3^6 \cdot |A|^6 \] \[ = 2^{15} \cdot 3^6 \cdot |A|^6 \] Given \( |A| = 5 \), we substitute: \[ = 2^{15} \cdot 3^6 \cdot 5^6 \] Comparing with \( 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \), we get: \[ \alpha = 15, \quad \beta = 6, \quad \gamma = 6 \] Thus: \[ \alpha + \beta + \gamma = 15 + 6 + 6 = 27 \]