To determine if the matrix $A$ is diagonalizable, we'll need to check if the algebraic multiplicity of each eigenvalue matches its geometric multiplicity.

The matrix given is: $A = \begin{bmatrix} -1 & 2 & -2 \\ 1 & 2 & 1 \\ -1 & -1 & 0 \end{bmatrix}$

Let's start by finding the eigenvalues of $A$. The eigenvalues are the solutions to the characteristic polynomial equation $\det(A - \lambda I) = 0$, where $I$ is the identity matrix and $\lambda$ are the eigenvalues.

We'll calculate the characteristic polynomial next.

1. First, construct the matrix $A - \lambda I$: $A - \lambda I = \begin{bmatrix} -1-\lambda & 2 & -2 \\ 1 & 2-\lambda & 1 \\ -1 & -1 & -\lambda \end{bmatrix}$

2. Compute the determinant of $A - \lambda I$: $\text{det}(A - \lambda I) = (-1-\lambda)((2-\lambda)(-\lambda) + 1) - 2(1 \cdot (-\lambda) + 1) + (-2)(-1 + (2-\lambda))$

Let's perform these computations and solve for $\lambda$.The eigenvalues of the matrix $A$ are $\lambda = 1$, $\lambda = -\sqrt{5}$, and $\lambda = \sqrt{5}$.

Next, we'll find the eigenvectors for each eigenvalue to determine the geometric multiplicity. If the geometric multiplicity of each eigenvalue matches its algebraic multiplicity (which is 1 for each eigenvalue since they are all distinct), then the matrix is diagonalizable.

Let's start by finding the eigenvectors for each eigenvalue:

1. For $\lambda = 1$
2. For $\lambda = -\sqrt{5}$
3. For $\lambda = \sqrt{5}$

I'll compute the eigenvectors for $\lambda = 1$ first.For $\lambda = 1$, the eigenvector is: