We are given the matrix $A = \begin{pmatrix} 0 & 0 & -2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{pmatrix}$.

Step 1: Find the Eigenvalues

To find the eigenvalues of the matrix $A$, we need to solve the characteristic equation:

$\text{det}(A - \lambda I) = 0$

where $\lambda$ represents the eigenvalues and $I$ is the identity matrix. The matrix $A - \lambda I$ is:

$A - \lambda I = \begin{pmatrix} 0 & 0 & -2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0-\lambda & 0 & -2 \\ 1 & 2-\lambda & 1 \\ 1 & 0 & 3-\lambda \end{pmatrix}$

The determinant of $A - \lambda I$ is:

$\text{det}(A - \lambda I) = \begin{vmatrix} -\lambda & 0 & -2 \\ 1 & 2-\lambda & 1 \\ 1 & 0 & 3-\lambda \end{vmatrix}$

Now, expand this determinant:

$= -\lambda \begin{vmatrix} 2-\lambda & 1 \\ 0 & 3-\lambda \end{vmatrix} - 0 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 3-\lambda \end{vmatrix} + (-2) \begin{vmatrix} 1 & 2-\lambda \\ 1 & 0 \end{vmatrix}$

Let's simplify each part:

1. $\begin{vmatrix} 2-\lambda & 1 \\ 0 & 3-\lambda \end{vmatrix} = (2-\lambda)(3-\lambda) = 6 - 5\lambda + \lambda^2$
2. $\begin{vmatrix} 1 & 2-\lambda \\ 1 & 0 \end{vmatrix} = - (2-\lambda)$

Thus, the determinant becomes:

$-\lambda \left( 6 - 5\lambda + \lambda^2 \right) + (-2)(-(2-\lambda)) = -\lambda(\lambda^2 - 5\lambda + 6) + 2(2-\lambda)$

Simplifying:

$-\lambda^3 + 5\lambda^2 - 6\lambda + 4 - 2\lambda = -\lambda^3 + 5\lambda^2 - 8\lambda + 4$

We set this equal to 0 to get the characteristic equation:

$-\lambda^3 + 5\lambda^2 - 8\lambda + 4 = 0$

Step 2: Solve the Characteristic Equation

We solve this cubic equation for $\lambda$. Using trial and error or synthetic division, we find that $\lambda = 1$ is a root. Dividing $-\lambda^3 + 5\lambda^2 - 8\lambda + 4$ by $\lambda - 1$, we get:

$-\lambda^3 + 5\lambda^2 - 8\lambda + 4 = (\lambda - 1)(-\lambda^2 + 6\lambda - 4)$

Now, solve $-\lambda^2 + 6\lambda - 4 = 0$ using the quadratic formula:

$\lambda = \frac{-6 \pm \sqrt{6^2 - 4(-1)(-4)}}{2(-1)} = \frac{-6 \pm \sqrt{36 - 16}}{-2} = \frac{-6 \pm \sqrt{20}}{-2} = \frac{-6 \pm 2\sqrt{5}}{-2}$

Thus, the eigenvalues are:

$\lambda_1 = 1, \quad \lambda_2 = 3 - \sqrt{5}, \quad \lambda_3 = 3 + \sqrt{5}$

Step 3: Find the Eigenvectors

For each eigenvalue $\lambda$, solve $(A - \lambda I)v = 0$ to find the corresponding eigenvector $v$.

Eigenvalue $\lambda_1 = 1$:

Solve $(A - I)v = 0$:

$\begin{pmatrix} -1 & 0 & -2 \\ 1 & 1 & 1 \\ 1 & 0 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$

This gives a system of equations. Solving this system yields the eigenvector corresponding to $\lambda_1 = 1$.

Similarly, you can solve for the eigenvectors corresponding to $\lambda_2 = 3 - \sqrt{5}$ and $\lambda_3 = 3 + \sqrt{5}$.

Would you like further steps on calculating the eigenvectors? Let me know if you want more details!

Related Questions:

1. How do we determine the nature of eigenvalues (real or complex)?
2. What are the properties of matrices with real eigenvalues?
3. How do we use eigenvalues to diagonalize a matrix?
4. What is the geometric interpretation of eigenvectors and eigenvalues?
5. How can we solve the characteristic equation for more complex matrices?

Tip:

For 3x3 matrices, trial and error or using known factoring techniques can help solve cubic characteristic equations more efficiently.