To diagonalize a matrix $A$, we need to:

1. Find the eigenvalues of $A$ by solving the characteristic equation $\det(A - \lambda I) = 0$, where $\lambda$ is the eigenvalue and $I$ is the identity matrix.
2. Find the corresponding eigenvectors for each eigenvalue.

The matrix $A$ is given as:

$A = \begin{pmatrix} 2 & 2 & 0 \\ 2 & 1 & 1 \\ -7 & 2 & -3 \end{pmatrix}$

Step 1: Find Eigenvalues

The characteristic equation is $\det(A - \lambda I) = 0$, where $I$ is the identity matrix.

Let's calculate this determinant first:

$\det \left( \begin{pmatrix} 2 & 2 & 0 \\ 2 & 1 & 1 \\ -7 & 2 & -3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right) = 0$

This results in:

$\det \begin{pmatrix} 2 - \lambda & 2 & 0 \\ 2 & 1 - \lambda & 1 \\ -7 & 2 & -3 - \lambda \end{pmatrix} = 0$

Now, let me calculate the determinant and find the eigenvalues.The eigenvalues of the matrix $A$ are:

$\lambda_1 = -4, \quad \lambda_2 = 3, \quad \lambda_3 = 1$

The corresponding eigenvectors are:

v_2 = \begin{pmatrix} -0.6667 \\ -0.3333 \\ 0.6667 \end{pmatrix}, \quad v_3 = \begin{pmatrix} -0.4364 \\ 0.2182 \\ 0.8729 \end{pmatrix}$$ ### Conclusion: The matrix $$A$$ can be diagonalized as $$A = PDP^{-1}$$, where $$D$$ is the diagonal matrix with eigenvalues $$\lambda_1, \lambda_2, \lambda_3$$, and the columns of $$P$$ are the corresponding eigenvectors. Would you like more details on the diagonalization process or on any other step? ### Follow-up questions: 1. How do you calculate the determinant of a 3x3 matrix? 2. How can we verify if a given vector is an eigenvector of a matrix? 3. What are the steps for finding $$P^{-1}$$ once you have $$P$$? 4. Can the matrix $$A$$ have repeated eigenvalues, and what does that imply? 5. How does the matrix's diagonal form help in solving differential equations? **Tip:** Eigenvectors corresponding to distinct eigenvalues are linearly independent, which ensures the diagonalization of the matrix is possible.