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: \[ \alpha + \beta + \gamma = 15 + 6 + 6 = 27 \]

Answer 2

To solve this problem, we need to analyze the determinant expression given and calculate the required values of \( \alpha \), \( \beta \), and \( \gamma \).

We're provided with the matrix \( A \) of order \( 3 \times 3 \) and its determinant \( |A| = 5 \). We need to find the determinant of the expression \( |2 \, \text{adj}(3A \, \text{adj}(2A))| \) and write it in the form \( 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \), then find \( \alpha + \beta + \gamma \).

1. Firstly, recall the properties of the adjugate matrix. For a \( 3 \times 3 \) matrix \( B \), \(|\text{adj}(B)| = |B|^{2}\).
2. Given \( |A| = 5 \), find \(|2A|\):
3. \(|2A| = 2^3 \cdot |A| = 8 \cdot 5 = 40\)
4. Calculate \(|\text{adj}(2A)|\):
5. \(|\text{adj}(2A)| = |2A|^2 = 40^2 = 1600\)
6. Next, consider \( B = 3A \, \text{adj}(2A) \). Then, \(|B|\) is calculated as:
7. \(|B| = |3A| \cdot |\text{adj}(2A)| = (3^3 \cdot |A|) \cdot 1600 = 27 \cdot 5 \cdot 1600\)
8. Simplify:
9. \(|B| = 27 \cdot 5 \cdot 1600 = 216000\)
10. Now, calculate \(|\text{adj}(B)|\) using \( B \) of order \( 3 \times 3 \):
11. \(|\text{adj}(B)| = |B|^2 = 216000^2\)
12. Find \(|2 \, \text{adj}(B)|\):
13. \(|2 \, \text{adj}(B)| = 2^3 \cdot |\text{adj}(B)| = 8 \cdot 216000^2\)
14. Finally, express \(|2 \, \text{adj}(B)|\) as \( 2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \) and determine \(\alpha + \beta + \gamma\).
15. Simplify with the prime factorization:
16. \( 216000 = 2^5 \times 3^3 \times 5^3 \)
17. \( 216000^2 = (2^5 \times 3^3 \times 5^3)^2 = 2^{10} \times 3^6 \times 5^6 \)
18. \( 8 \times 216000^2 = 2^3 \times 2^{10} \times 3^6 \times 5^6 = 2^{13} \times 3^6 \times 5^6 \)
19. Thus, \(\alpha = 13\), \(\beta = 6\), \(\gamma = 6\), and \(\alpha + \beta + \gamma = 13 + 6 + 6 = 27\).

Therefore, the answer is 27.