Solve the equations for x and y. Find the eigenvalues and eigenvectors for the given matrix A.

The real parts are $x$ and $y$.
The imaginary parts are $-2y$ (from $-i2y$) and $-3x$ (from $-i3x$).
So, the equation becomes:

$$(x+y) + i(-2y-3x) = 2 + i3$$

Equating the real parts:

$$x + y = 2 \quad \text{(Equation 1)}$$

Equating the imaginary parts:

$$-2y - 3x = 3 \quad \text{(Equation 2)}$$

From Equation 1, express $y$ in terms of $x$:

$$y = 2 - x$$

Substitute this expression for $y$ into Equation 2:

$$-2(2 - x) - 3x = 3$$ $$-4 + 2x - 3x = 3$$ $$-4 - x = 3$$ $$-x = 3 + 4$$ $$-x = 7$$ $$x = -7$$

Now substitute the value of $x$ back into the expression for $y$:

$$y = 2 - (-7)$$ $$y = 2 + 7$$ $$y = 9$$

The solutions are $\boxed{x = -7, y = 9}$.

8. b) Find the eigenvalues and eigenvectors for the given matrix A.
Given the matrix:

$$A = \begin{bmatrix} 2 & 2 & -2 \\ 1 & 3 & 2 \\ -1 & 1 & 1 \end{bmatrix}$$

$$A - \lambda I = \begin{bmatrix} 2-\lambda & 2 & -2 \\ 1 & 3-\lambda & 2 \\ -1 & 1 & 1-\lambda \end{bmatrix}$$

Calculate the determinant:

$$\det(A - \lambda I) = (2-\lambda)[(3-\lambda)(1-\lambda) - (2)(1)] - 2[(1)(1-\lambda) - (2)(-1)] + (-2)[(1)(1) - (3-\lambda)(-1)]$$ $$= (2-\lambda)[\lambda^2 - 4\lambda + 3 - 2] - 2[1-\lambda+2] - 2[1+3-\lambda]$$ $$= (2-\lambda)[\lambda^2 - 4\lambda + 1] - 2[3-\lambda] - 2[4-\lambda]$$ $$= (2\lambda^2 - 8\lambda + 2 - \lambda^3 + 4\lambda^2 - \lambda) - (6 - 2\lambda) - (8 - 2\lambda)$$ $$= -\lambda^3 + 6\lambda^2 - 5\lambda - 12$$

Set the determinant to zero:

$$-\lambda^3 + 6\lambda^2 - 5\lambda - 12 = 0$$ $$\lambda^3 - 6\lambda^2 + 5\lambda + 12 = 0$$

By testing integer factors of 12, we find that $\lambda = -1$ is a root:
$(-1)^3 - 6(-1)^2 + 5(-1) + 12 = -1 - 6 - 5 + 12 = 0$.
So, $(\lambda+1)$ is a factor. Dividing the polynomial by $(\lambda+1)$ gives $\lambda^2 - 7\lambda + 12$.
Factoring the quadratic equation $\lambda^2 - 7\lambda + 12 = 0$:

$$(\lambda-3)(\lambda-4) = 0$$

The eigenvalues are $\boxed{\lambda_1 = -1, \lambda_2 = 3, \lambda_3 = 4}$.

For $\lambda_1 = -1$:
Solve $(A - (-1)I)\mathbf{v} = \mathbf{0} \implies (A+I)\mathbf{v} = \mathbf{0}$:

$$\begin{bmatrix} 3 & 2 & -2 \\ 1 & 4 & 2 \\ -1 & 1 & 2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

From the second and third rows, we can perform row operations:
$R_2 \to 3R_2 - R_1$: $3(1)v_1 + 3(4)v_2 + 3(2)v_3 - (3v_1 + 2v_2 - 2v_3) = 0 \implies 10v_2 + 8v_3 = 0 \implies 5v_2 + 4v_3 = 0$.
$R_3 \to 3R_3 + R_1$: $3(-1)v_1 + 3(1)v_2 + 3(2)v_3 + (3v_1 + 2v_2 - 2v_3) = 0 \implies 5v_2 + 4v_3 = 0$.
Let $v_2 = 4k$. Then $5(4k) + 4v_3 = 0 \implies 20k + 4v_3 = 0 \implies v_3 = -5k$.
Substitute $v_2$ and $v_3$ into the first equation: $3v_1 + 2(4k) - 2(-5k) = 0 \implies 3v_1 + 8k + 10k = 0 \implies 3v_1 + 18k = 0 \implies v_1 = -6k$.
Choosing $k=1$, the eigenvector is $\boxed{\mathbf{v}_1 = \begin{bmatrix} -6 \\ 4 \\ -5 \end{bmatrix}}$.

For $\lambda_2 = 3$:
Solve $(A - 3I)\mathbf{v} = \mathbf{0}$:

$$\begin{bmatrix} -1 & 2 & -2 \\ 1 & 0 & 2 \\ -1 & 1 & -2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

From the second equation: $v_1 + 2v_3 = 0 \implies v_1 = -2v_3$.
Substitute $v_1$ into the first equation: $-(-2v_3) + 2v_2 - 2v_3 = 0 \implies 2v_3 + 2v_2 - 2v_3 = 0 \implies 2v_2 = 0 \implies v_2 = 0$.
Let $v_3 = k$. Then $v_1 = -2k$ and $v_2 = 0$.
Choosing $k=1$, the eigenvector is $\boxed{\mathbf{v}_2 = \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}}$.

For $\lambda_3 = 4$:
Solve $(A - 4I)\mathbf{v} = \mathbf{0}$:

$$\begin{bmatrix} -2 & 2 & -2 \\ 1 & -1 & 2 \\ -1 & 1 & -3 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

From the first equation, divide by -2: $v_1 - v_2 + v_3 = 0 \implies v_1 = v_2 - v_3$.
Substitute $v_1$ into the second equation: $(v_2 - v_3) - v_2 + 2v_3 = 0 \implies v_3 = 0$.
Substitute $v_3 = 0$ into $v_1 = v_2 - v_3$: $v_1 = v_2 - 0 \implies v_1 = v_2$.
Let $v_2 = k$. Then $v_1 = k$ and $v_3 = 0$.
Choosing $k=1$, the eigenvector is $\boxed{\mathbf{v}_3 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}}$.