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

Concept:
For a matrix \(A\) of order \(n\):
\(|kA| = k^{n}|A|\)
\(|\text{adj}(A)| = |A|^{\,n-1}\)
\(|AB| = |A|\cdot|B|\) Here, \(n = 3\).
Step 1: Evaluate \(|\text{adj}(2A)|\). \[ |2A| = 2^3|A| = 8 \times 5 = 40 \] \[ |\text{adj}(2A)| = |2A|^{2} = 40^2 = 1600 \]
Step 2: Evaluate \(|3A\,\text{adj}(2A)|\). \[ |3A| = 3^3|A| = 27 \times 5 = 135 \] \[ |3A\,\text{adj}(2A)| = |3A|\cdot|\text{adj}(2A)| = 135 \times 1600 = 216000 \]
Step 3: Evaluate \(\left|\text{adj}\big(3A\,\text{adj}(2A)\big)\right|\). \[ \left|\text{adj}(M)\right| = |M|^{2} \Rightarrow \left|\text{adj}\big(3A\,\text{adj}(2A)\big)\right| = (216000)^2 \]
Step 4: Multiply by the scalar \(2\). \[ \left|2\,\text{adj}\big(3A\,\text{adj}(2A)\big)\right| = 2^3 \times (216000)^2 \] Now, \[ 216000 = 2^6 \cdot 3^3 \cdot 5^3 \] \[ (216000)^2 = 2^{12}\cdot 3^6 \cdot 5^6 \] Thus, \[ 2^3 \times (216000)^2 = 2^{15}\cdot 3^6 \cdot 5^6 \]
Step 5: Identify powers. \[ \alpha = 15,\quad \beta = 6,\quad \gamma = 6 \] \[ \alpha + \beta + \gamma = 15 + 6 + 6 = \boxed{27} \]