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

Answer: (A)

64

Answer 2

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: $= 64$