To solve the problem, let’s use properties of determinants.

Given:

• $\det(A) = 3$
• $\det(B) = -3$
• $\det(C) = 1$

We need to find $\det\left(A^4 B^T C^{-1} (B^{-1})^2\right)$.

Step-by-Step Solution

1. Determinant of Powers: For any matrix $M$, $\det(M^k) = (\det(M))^k$.

   • $\det(A^4) = (\det(A))^4 = 3^4 = 81$.
   • $\det((B^{-1})^2) = (\det(B^{-1}))^2 = \left(\frac{1}{\det(B)}\right)^2 = \left(\frac{1}{-3}\right)^2 = \frac{1}{9}$.

2. Determinant of a Transpose: For any matrix $M$, $\det(M^T) = \det(M)$.

   • Therefore, $\det(B^T) = \det(B) = -3$.

3. Determinant of an Inverse: For any invertible matrix $M$, $\det(M^{-1}) = \frac{1}{\det(M)}$.

   • So, $\det(C^{-1}) = \frac{1}{\det(C)} = \frac{1}{1} = 1$.

4. Combining Determinants: Using the property that $\det(ABC) = \det(A) \cdot \det(B) \cdot \det(C)$ for any matrices $A$, $B$, and $C$, we can now find: $\det\left(A^4 B^T C^{-1} (B^{-1})^2\right) = \det(A^4) \cdot \det(B^T) \cdot \det(C^{-1}) \cdot \det((B^{-1})^2).$

5. Substitute Values: $\det\left(A^4 B^T C^{-1} (B^{-1})^2\right) = 81 \cdot (-3) \cdot 1 \cdot \frac{1}{9}.$

6. Calculate: $= 81 \cdot (-3) \cdot \frac{1}{9} = 81 \cdot -\frac{3}{9} = 81 \cdot -\frac{1}{3} = -27.$

Final Answer

$\det\left(A^4 B^T C^{-1} (B^{-1})^2\right) = -27.$

Would you like a more detailed explanation of any step?

Here are some related questions to explore further:

1. What are the properties of determinants when involving matrix inverses?
2. How do transposes affect the determinant of a matrix?
3. How does raising a matrix to a power affect its determinant?
4. What are common mistakes to avoid when dealing with determinants of composite matrices?
5. How does the determinant property help in solving real-world matrix equations?

Tip: Always check the properties of determinants (like power, transpose, and inverse) before starting complex operations; they often simplify calculations.