Given that A = [[5, 0, 8], [12, 7, 6], [6, 4, 4]] Evaluate A⁻¹

$$\det(A) = 5 \begin{vmatrix} 7 & 6 \\ 4 & 4 \end{vmatrix} - 0 \begin{vmatrix} 12 & 6 \\ 6 & 4 \end{vmatrix} + 8 \begin{vmatrix} 12 & 7 \\ 6 & 4 \end{vmatrix}$$ $$\det(A) = 5(7 \times 4 - 6 \times 4) - 0(12 \times 4 - 6 \times 6) + 8(12 \times 4 - 7 \times 6)$$ $$\det(A) = 5(28 - 24) - 0 + 8(48 - 42)$$ $$\det(A) = 5(4) + 8(6)$$ $$\det(A) = 20 + 48$$ $$\det(A) = 68$$

The cofactor $C_{ij}$ is given by $(-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$.

$$C_{11} = \begin{vmatrix} 7 & 6 \\ 4 & 4 \end{vmatrix} = 28 - 24 = 4$$ $$C_{12} = - \begin{vmatrix} 12 & 6 \\ 6 & 4 \end{vmatrix} = -(48 - 36) = -12$$ $$C_{13} = \begin{vmatrix} 12 & 7 \\ 6 & 4 \end{vmatrix} = 48 - 42 = 6$$ $$C_{21} = - \begin{vmatrix} 0 & 8 \\ 4 & 4 \end{vmatrix} = -(0 - 32) = 32$$ $$C_{22} = \begin{vmatrix} 5 & 8 \\ 6 & 4 \end{vmatrix} = 20 - 48 = -28$$ $$C_{23} = - \begin{vmatrix} 5 & 0 \\ 6 & 4 \end{vmatrix} = -(20 - 0) = -20$$ $$C_{31} = \begin{vmatrix} 0 & 8 \\ 7 & 6 \end{vmatrix} = 0 - 56 = -56$$ $$C_{32} = - \begin{vmatrix} 5 & 8 \\ 12 & 6 \end{vmatrix} = -(30 - 96) = -(-66) = 66$$ $$C_{33} = \begin{vmatrix} 5 & 0 \\ 12 & 7 \end{vmatrix} = 35 - 0 = 35$$

The cofactor matrix is:

$$C = \begin{bmatrix} 4 & -12 & 6 \\ 32 & -28 & -20 \\ -56 & 66 & 35 \end{bmatrix}$$

The adjoint of $A$ is the transpose of the cofactor matrix $C$.

$$\text{adj}(A) = C^T = \begin{bmatrix} 4 & 32 & -56 \\ -12 & -28 & 66 \\ 6 & -20 & 35 \end{bmatrix}$$

$$A^{-1} = \frac{1}{\det(A)} \text{adj}(A) = \frac{1}{68} \begin{bmatrix} 4 & 32 & -56 \\ -12 & -28 & 66 \\ 6 & -20 & 35 \end{bmatrix}$$

Distribute the scalar $\frac{1}{68}$ into the matrix and simplify the fractions:

$$A^{-1} = \begin{bmatrix} \frac{4}{68} & \frac{32}{68} & \frac{-56}{68} \\ \frac{-12}{68} & \frac{-28}{68} & \frac{66}{68} \\ \frac{6}{68} & \frac{-20}{68} & \frac{35}{68} \end{bmatrix} = \begin{bmatrix} \frac{1}{17} & \frac{8}{17} & \frac{-14}{17} \\ \frac{-3}{17} & \frac{-7}{17} & \frac{33}{34} \\ \frac{3}{34} & \frac{-5}{17} & \frac{35}{68} \end{bmatrix}$$

The inverse of matrix A is:

$$\boxed{A^{-1} = \begin{bmatrix} \frac{1}{17} & \frac{8}{17} & \frac{-14}{17} \\ \frac{-3}{17} & \frac{-7}{17} & \frac{33}{34} \\ \frac{3}{34} & \frac{-5}{17} & \frac{35}{68} \end{bmatrix}}$$