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: \[ = 64 \]

Answer 2

Step 1: Collect determinant facts for 3×3 matrices
For any invertible 3×3 matrix M:
1) |Mᵀ| = |M|.
2) |MN| = |M|·|N|.
3) |kM| = k³|M| (since order = 3).
4) adj(M) = |M|·M⁻¹ ⇒ (adj M)⁻¹ = (1/|M|)·M ⇒ |(adj M)⁻¹| = 1/|adj M|.
5) |adj(M)| = |M|^{n-1} = |M|² for n = 3.
Given |A| = 3 and |B| = 2.

Step 2: Break the big determinant into a product of determinants
We need \[ \left|A^\top A\,(\operatorname{adj}(2A))^{-1}\,(\operatorname{adj}(4B))\,(\operatorname{adj}(AB))^{-1}\,A\,A^\top\right| \] By multiplicativity of determinant, this equals \[ |A^\top A|\cdot|(\operatorname{adj}(2A))^{-1}|\cdot|\operatorname{adj}(4B)|\cdot|(\operatorname{adj}(AB))^{-1}|\cdot|A|\cdot|A^\top|. \]

Step 3: Evaluate each factor
a) \(|A^\top A| = |A^\top|\cdot|A| = |A|\cdot|A| = 3\cdot 3 = 9.\)
b) \(|(\operatorname{adj}(2A))^{-1}| = 1/|\operatorname{adj}(2A)|.\) Now \[ |\operatorname{adj}(2A)| = |2A|^2 = (2^3|A|)^2 = (8\cdot 3)^2 = 24^2 = 576, \] so \(|(\operatorname{adj}(2A))^{-1}| = \tfrac{1}{576}.\)
c) \(|\operatorname{adj}(4B)| = |4B|^2 = (4^3|B|)^2 = (64\cdot 2)^2 = 128^2 = 16384.\)
d) \(|(\operatorname{adj}(AB))^{-1}| = 1/|\operatorname{adj}(AB)|.\) Since \(|AB| = |A||B| = 3\cdot 2 = 6,\) \[ |\operatorname{adj}(AB)| = |AB|^2 = 6^2 = 36 \quad\Rightarrow\quad |(\operatorname{adj}(AB))^{-1}| = \tfrac{1}{36}. \] e) \(|A| = 3,\quad |A^\top| = |A| = 3.\)

Step 4: Multiply all factors
\[ \text{Value} = 9 \cdot \frac{1}{576} \cdot 16384 \cdot \frac{1}{36} \cdot 3 \cdot 3. \] Group the simple part: \(9\cdot 3\cdot 3 = 81.\) Hence \[ \text{Value} = \frac{81 \cdot 16384}{576 \cdot 36}. \] Compute: \[ \frac{81 \cdot 16384}{576 \cdot 36} = 64. \]

Final answer

64