Solution

1. Key Concepts and Formulas

This problem extensively uses fundamental properties of determinants for square matrices. For two $n \times n$ matrices P and Q, and a scalar $k$:

• Determinant of a Product: $|\mathrm{PQ}| = |\mathrm{P}||\mathrm{Q}|$
• Determinant of a Transpose: $|\mathrm{P}^{\mathrm{T}}| = |\mathrm{P}|$
• Determinant of an Inverse: $|\mathrm{P}^{-1}| = \frac{1}{|\mathrm{P}|}$ (provided $|\mathrm{P}| \neq 0$)
• Determinant of an Adjoint: $|\operatorname{adj}(\mathrm{P})| = |\mathrm{P}|^{\mathrm{n}-1}$
• Determinant of Scalar Multiplication: $|\mathrm{kP}| = \mathrm{k}^{\mathrm{n}}|\mathrm{P}|$

In this problem, the matrices A and B are of order 3, so $n=3$.

2. Problem Analysis and Setup

We are given two square matrices A and B of order 3, with:

• $|\mathrm{A}| = 3$
• $|\mathrm{B}| = 2$

We need to evaluate the determinant of the given matrix expression: $\left|\mathrm{A}^{\mathrm{T}} \mathrm{A}(\operatorname{adj}(2 \mathrm{~A}))^{-1}(\operatorname{adj}(4 \mathrm{~B}))(\operatorname{adj}(\mathrm{AB}))^{-1} \mathrm{AA}^{\mathrm{T}}\right|$

Let's apply the determinant properties step-by-step.

3. Step-by-Step Derivation

Step 3.1: Decompose the determinant of the product Using the property $|\mathrm{PQ}| = |\mathrm{P}||\mathrm{Q}|$ repeatedly, we can break down the determinant of the entire product into the product of individual determinants: $\left|\mathrm{A}^{\mathrm{T}}\right||\mathrm{A}|\left|(\operatorname{adj}(2 \mathrm{~A}))^{-1}\right||\operatorname{adj}(4 \mathrm{~B})|\left|(\operatorname{adj}(\mathrm{AB}))^{-1}\right||\mathrm{A}|\left|\mathrm{A}^{\mathrm{T}}\right|$

Step 3.2: Apply properties of transpose and inverse determinants

• We know $|\mathrm{P}^{\mathrm{T}}| = |\mathrm{P}|$. Thus, $|\mathrm{A}^{\mathrm{T}}| = |\mathrm{A}|$.
• We know $|\mathrm{P}^{-1}| = \frac{1}{|\mathrm{P}|}$. Thus, $\left|(\operatorname{adj}(2 \mathrm{~A}))^{-1}\right| = \frac{1}{|\operatorname{adj}(2 \mathrm{~A})|}$ and $\left|(\operatorname{adj}(\mathrm{AB}))^{-1}\right| = \frac{1}{|\operatorname{adj}(\mathrm{AB})|}$.

Substituting these into the expression: $|\mathrm{A}||\mathrm{A}| \left(\frac{1}{|\operatorname{adj}(2 \mathrm{~A})|}\right) |\operatorname{adj}(4 \mathrm{~B})| \left(\frac{1}{|\operatorname{adj}(\mathrm{AB})|}\right) |\mathrm{A}||\mathrm{A}|$ $= |\mathrm{A}|^4 \frac{|\operatorname{adj}(4 \mathrm{~B})|}{|\operatorname{adj}(2 \mathrm{~A})||\operatorname{adj}(\mathrm{AB})|}$

Step 3.3: Apply the determinant of adjoint property For a matrix P of order