Question

Let \(A\) be a matrix of order \(3 \times 3\) and \(|A| = 5\). If \[ \left|\,2\,\text{adj}\big(3A\,\text{adj}(2A)\big)\right| = 2^{\alpha}\cdot 3^{\beta}\cdot 5^{\gamma}, \quad \alpha, \beta, \gamma \in \mathbb{N}, \] then the value of \(\alpha + \beta + \gamma\) is:

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

Hint:

For \(3 \times 3\) matrices: \[ |\text{adj}(A)| = |A|^2 \] Always apply determinant properties step by step and simplify using prime factorization.

Answer 1

The Correct Option is C

To solve the given problem, we will first analyze and use formulae related to determinants and adjugate matrices. The problem can be broken down into several key steps and concepts:

1. Understanding the Adjugate Function:
   • The adjugate of a matrix \( A \) is denoted as \( \text{adj}(A) \) and its property: \(|\text{adj}(A)| = |A|^{n-1}\) for an \( n \times n \) matrix.
2. Using the Property of Determinants:
   • The determinant of a product of matrices follows: \(|AB| = |A| \cdot |B|\).
   • For a scalar multiple of a matrix, \(|kA| = k^n \cdot |A|\) where \( k \) is a scalar and \( A \) is an \( n \times n \) matrix.

Given: \( |A| = 5 \) for a \( 3 \times 3 \) matrix, and we need to find \(|2\, \text{adj}\big(3A\, \text{adj}(2A)\big)|\).

1. Calculate Inside Terms:
   • First, compute \( |2A| \):
     • \(|2A| = 2^3 \cdot |A| = 8 \cdot 5 = 40\).
   • Next compute \(|\text{adj}(2A)| = |2A|^{2} = 40^2 = 1600\).
   • Now, compute \(|3A|\):
     • \(|3A| = 3^3 \cdot |A| = 27 \cdot 5 = 135\).
2. Find the Determinant of the Compound Matrix:
   • Calculate \(|3A \cdot \text{adj}(2A)| = |3A| \cdot |\text{adj}(2A)| = 135 \cdot 1600 = 216000\).
   • Finally, evaluate the original expression \(|2\, \text{adj}\big(3A\, \text{adj}(2A)\big)| = |2|^3 \cdot |\text{adj}(3A \cdot \text{adj}(2A))|\ = 8 \times (216000)^2.\)
3. Simplify the Expression:
   • Express \(216000\) as a product of primes:
     • \(216000 = 2^5 \times 3^3 \times 5^3\).
   • Thus, \(|(216000)^2| = (2^5 \times 3^3 \times 5^3)^2 = 2^{10} \times 3^6 \times 5^6.\)
   • The determinant becomes \(|2^3 \times 2^{10} \times 3^6 \times 5^6 = 2^{13} \times 3^6 \times 5^6.\)
4. Final Calculation and Sum of Exponents:
   • The expression simplifies to \(2^{\alpha} \times 3^{\beta} \times 5^{\gamma}\) with \(\alpha = 13\), \(\beta = 6\), \(\gamma = 6\).
   • The sum \(\alpha + \beta + \gamma = 13 + 6 + 6 = 25\).

Therefore, the value of \(\alpha + \beta + \gamma\) is 25.