Question

Let \( A \) and \( B \) be two square matrices of order 3 such that \( |A| = 3 \) and \( |B| = 2 \). Then \(|A^\top A (\text{adj}(2A))^{-1} (\text{adj}(4B)) (\text{adj}(AB))^{-1} A A^\top|\) is equal to:

• A. 64
• B. 81
• C. 32
• D. 108

Answer 1

The Correct Option is A

Given: \[ |A| = 3, \quad |B| = 2 \] \[ |A^{\top} (\text{adj}(2A))^{-1} (\text{adj}(4B)) (\text{adj}(AB))^{-1} AA^{\top}| = 3 \times 3 \times |(\text{adj}(2A))^{-1}| \times |\text{adj}(4B)| \times |(\text{adj}(AB))^{-1}| \times 3 \times 3 \]

Breaking it into steps: \[ = \frac{1}{|\text{adj}(2A)|} \times 2^{12} \times 2^2 \times \frac{1}{|\text{adj}(AB)|} \]

Now calculating the determinant of adjugates: \[ = \frac{1}{2^6 |\text{adj} A|} \times \frac{1}{2^2 \times 3^2} \quad (\text{for } |\text{adj}(2A)|) \] \[ = \frac{1}{|\text{adj} B| \times |\text{adj} A|} \quad (\text{for } |\text{adj}(AB)|) \]

Further simplification: \[ = \frac{1}{2^{12} \times 3^2} \]

Simplifying: \[ = \frac{3^4}{2^6 \times 3^2} \times \frac{2^{12} \times 2^2}{2^2 \times 3^2} \]

Combining terms: \[ = 64 \]