Write the given matrix A.

For a $3 \times 3$ matrix $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, the determinant is $a(ei - fh) - b(di - fg) + c(dh - eg)$.

$$\det(A) = 2 \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} - 1 \begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} + 2 \begin{vmatrix} 3 & 1 \\ 2 & 2 \end{vmatrix}$$ $$\det(A) = 2((1)(1) - (2)(2)) - 1((3)(1) - (2)(2)) + 2((3)(2) - (1)(2))$$ $$\det(A) = 2(1 - 4) - 1(3 - 4) + 2(6 - 2)$$ $$\det(A) = 2(-3) - 1(-1) + 2(4)$$ $$\det(A) = -6 + 1 + 8$$ $$\det(A) = 3$$

The determinant of $A$ is $\boxed{3}$.

6. (b)

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

$$C_{11} = +\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = (1)(1) - (2)(2) = 1 - 4 = -3$$ $$C_{12} = -\begin{vmatrix} 3 & 2 \\ 2 & 1 \end{vmatrix} = -((3)(1) - (2)(2)) = -(3 - 4) = -(-1) = 1$$ $$C_{13} = +\begin{vmatrix} 3 & 1 \\ 2 & 2 \end{vmatrix} = (3)(2) - (1)(2) = 6 - 2 = 4$$ $$C_{21} = -\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = -((1)(1) - (2)(2)) = -(1 - 4) = -(-3) = 3$$ $$C_{22} = +\begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix} = (2)(1) - (2)(2) = 2 - 4 = -2$$ $$C_{23} = -\begin{vmatrix} 2 & 1 \\ 2 & 2 \end{vmatrix} = -((2)(2) - (1)(2)) = -(4 - 2) = -(2) = -2$$ $$C_{31} = +\begin{vmatrix} 1 & 2 \\ 1 & 2 \end{vmatrix} = (1)(2) - (2)(1) = 2 - 2 = 0$$ $$C_{32} = -\begin{vmatrix} 2 & 2 \\ 3 & 2 \end{vmatrix} = -((2)(2) - (2)(3)) = -(4 - 6) = -(-2) = 2$$ $$C_{33} = +\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} = (2)(1) - (1)(3) = 2 - 3 = -1$$

The cofactor matrix $C$ is:

$$C = \begin{pmatrix} -3 & 1 & 4 \\ 3 & -2 & -2 \\ 0 & 2 & -1 \end{pmatrix}$$

The adjoint matrix is the transpose of the cofactor matrix.

$$\text{adj}(A) = C^T = \begin{pmatrix} -3 & 3 & 0 \\ 1 & -2 & 2 \\ 4 & -2 & -1 \end{pmatrix}$$

The formula for the inverse of a matrix is $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$.
Using $\det(A) = 3$ from part (a):

$$A^{-1} = \frac{1}{3} \begin{pmatrix} -3 & 3 & 0 \\ 1 & -2 & 2 \\ 4 & -2 & -1 \end{pmatrix}$$ $$A^{-1} = \begin{pmatrix} \frac{-3}{3} & \frac{3}{3} & \frac{0}{3} \\ \frac{1}{3} & \frac{-2}{3} & \frac{2}{3} \\ \frac{4}{3} & \frac{-2}{3} & \frac{-1}{3} \end{pmatrix}$$ $$A^{-1} = \begin{pmatrix} -1 & 1 & 0 \\ \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ \frac{4}{3} & -\frac{2}{3} & -\frac{1}{3} \end{pmatrix}$$

The inverse of $A$ is $\boxed{\begin{pmatrix} -1 & 1 & 0 \\ \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} \\ \frac{4}{3} & -\frac{2}{3} & -\frac{1}{3} \end{pmatrix}}$.