The problem states:

"Let $A$ and $B$ be two square matrices with $\text{det}(A) = 2$ and $\text{det}(B) = 3$. Calculate $\text{det}(A^2 B^T A^{-1} B)$."

Solution:

We are given two square matrices $A$ and $B$, and we need to compute the determinant of the matrix expression $A^2 B^T A^{-1} B$.

Step-by-step approach:

1. Use the properties of determinants:

   • $\text{det}(XY) = \text{det}(X) \cdot \text{det}(Y)$ for any matrices $X$ and $Y$ of the same size.
   • $\text{det}(A^{-1}) = \frac{1}{\text{det}(A)}$.
   • $\text{det}(A^T) = \text{det}(A)$ for any square matrix $A$.

2. Break down the expression $A^2 B^T A^{-1} B$:

   • $\text{det}(A^2) = \text{det}(A)^2 = 2^2 = 4$.
   • $\text{det}(B^T) = \text{det}(B) = 3$.
   • $\text{det}(A^{-1}) = \frac{1}{\text{det}(A)} = \frac{1}{2}$.
   • $\text{det}(B) = 3$.

3. Now calculate the determinant of the full expression: $\text{det}(A^2 B^T A^{-1} B) = \text{det}(A^2) \cdot \text{det}(B^T) \cdot \text{det}(A^{-1}) \cdot \text{det}(B)$ Substituting the values we computed: $\text{det}(A^2 B^T A^{-1} B) = 4 \cdot 3 \cdot \frac{1}{2} \cdot 3 = 4 \cdot 9 = 36$

Thus, the determinant of $A^2 B^T A^{-1} B$ is $\boxed{36}$.

Let me know if you'd like more details or have any questions!

Here are five related questions:

1. What are some other properties of matrix determinants?
2. How does the determinant of a product of matrices relate to the individual determinants?
3. What is the significance of the determinant being non-zero for square matrices?
4. How does the transpose of a matrix affect its determinant?
5. How can we interpret determinants geometrically?

Tip: For inverse matrices, remember that $\text{det}(A^{-1}) = \frac{1}{\text{det}(A)}$, which is useful in complex matrix determinant calculations.