Diagonalize the matrix
2 1 -1
1 1 -2
1-2 1
Find A?

To diagonalize the matrix \( A \), we need to find its eigenvalues and eigenvectors. Then, the matrix \( A \) can be diagonalized as \( A = PDP^{-1} \), where \( D \) is the diagonal matrix of eigenvalues, and \( P \) is the matrix whose columns are the corresponding eigenvectors.

Given matrix \( A \):

\[
A = \begin{pmatrix}
2 & 1 & -1 \\
1 & 1 & -2 \\
1 & -2 & 1
\end{pmatrix}
\]

### Step 1: Find the Eigenvalues
To find the eigenvalues, we solve the characteristic equation:

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

where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix.

\[
\text{det}\left(\begin{pmatrix}
2 & 1 & -1 \\
1 & 1 & -2 \\
1 & -2 & 1
\end{pmatrix} - \lambda \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}\right) = 0
\]

This simplifies to:

\[
\text{det}\begin{pmatrix}
2-\lambda & 1 & -1 \\
1 & 1-\lambda & -2 \\
1 & -2 & 1-\lambda
\end{pmatrix} = 0
\]

### Step 2: Compute the Determinant
Let's calculate the determinant of the matrix:

\[
\text{det}\begin{pmatrix}
2-\lambda & 1 & -1 \\
1 & 1-\lambda & -2 \\
1 & -2 & 1-\lambda
\end{pmatrix}
\]

I'll calculate this determinant to find the eigenvalues.The characteristic equation is:

\[
-\lambda^3 + 4\lambda^2 - \lambda - 6 = 0
\]

The eigenvalues are:

\[
\lambda_1 = -1, \quad \lambda_2 = 2, \quad \lambda_3 = 3
\]

### Step 3: Find the Eigenvectors
Now, we need to find the eigenvectors corresponding to each eigenvalue by solving \( (A - \lambda I) \mathbf{v} = 0 \) for each \( \lambda \).

I'll calculate the eigenvectors for these eigenvalues.The eigenvectors corresponding to the eigenvalues are:

1. For \( \lambda_1 = -1 \):
   \[
   \mathbf{v}_1 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}
   \]

2. For \( \lambda_2 = 2 \):
   \[
   \mathbf{v}_2 = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix}
   \]

3. For \( \lambda_3 = 3 \):
   \[
   \mathbf{v}_3 = \begin{pmatrix} -4 \\ -3 \\ 1 \end{pmatrix}
   \]

### Step 4: Form the Matrices \( P \) and \( D \)

Now, we can form the matrix \( P \) with the eigenvectors as columns and the diagonal matrix \( D \) with the eigenvalues.

\[
P = \begin{pmatrix} 
0 & 3 & -4 \\ 
1 & 1 & -3 \\ 
1 & 1 & 1 
\end{pmatrix}
\]
\[
D = \begin{pmatrix} 
-1 & 0 & 0 \\ 
0 & 2 & 0 \\ 
0 & 0 & 3 
\end{pmatrix}
\]

Thus, the matrix \( A \) can be diagonalized as \( A = PDP^{-1} \).

Do you want details on any specific part or have any questions?

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Here are some related questions you might find useful:

1. How do you compute the inverse of matrix \( P \)?
2. What does it mean for a matrix to be diagonalizable?
3. Can all matrices be diagonalized? Why or why not?
4. How do you verify that \( A = PDP^{-1} \)?
5. What are the applications of diagonalization in linear algebra?
6. How do eigenvalues and eigenvectors relate to matrix transformations?
7. What is the significance of the diagonal elements in the matrix \( D \)?
8. How does the process differ if the matrix has complex eigenvalues?

**Tip:** When finding eigenvectors, it's essential to check for linear independence to ensure that the matrix can be diagonalized.

Learn how to diagonalize the matrix with detailed steps. Find eigenvalues and eigenvectors, form the matrices P and D, and understand diagonalization theorem. Suitable for advanced undergraduate students and anyone studying linear algebra.

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