The problem asks to solve the expression:

$\left( \left( \frac{1}{2} A \right)^2 C (2B)^T \right)^{-1}$

Given the matrices $A^{-1}, B^{-1}, C^{-1}$, we first need to derive $A, B, C$ by inverting these matrices, then compute the expression step by step.

Matrices given:

1. $A^{-1} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ -1 & -1 & 1 \end{bmatrix}$
2. $B^{-1} = \begin{bmatrix} 1 & 0 & -1 \\ -2 & -1 & 0 \\ -2 & -1 & -1 \end{bmatrix}$
3. $C^{-1} = \begin{bmatrix} 1 & 1 & 0 \\ -2 & 2 & -2 \\ 0 & 1 & -2 \end{bmatrix}$

Steps:

1. Find $A$, $B$, and $C$:

   • The inverse of $A^{-1}$, $B^{-1}$, and $C^{-1}$ will give us $A$, $B$, and $C$, respectively.

2. Compute $\frac{1}{2} A$:

   • After obtaining $A$, multiply it by $\frac{1}{2}$.

3. Square $\left( \frac{1}{2} A \right)^2$:

   • Matrix multiplication of $\frac{1}{2} A$ with itself.

4. Find the transpose of $2B$:

   • After obtaining $B$, multiply by 2 and find the transpose.

5. Multiply $\left( \frac{1}{2} A \right)^2$, $C$, and $(2B)^T$:

   • Multiply these matrices in the order provided.

6. Take the inverse of the final result.

Let me perform these steps now.The solution to the matrix expression $\left( \left( \frac{1}{2} A \right)^2 C (2B)^T \right)^{-1}$ is: