Find the eigenvalues and corresponding vectors of the following matrix, and diagonalise if possible. Find the matrix which diagonolise it.  A= 1 -1 2 2 -2 4 3 -3 6 

We are given a matrix \( A \) and we need to find its eigenvalues, eigenvectors, and diagonalize it if possible.

The matrix \( A \) is:

\[
A = \begin{pmatrix}
1 & -1 & 2 \\
2 & -2 & 4 \\
3 & -3 & 6
\end{pmatrix}
\]

### Step 1: Find the eigenvalues
To find the eigenvalues of matrix \( A \), we solve the characteristic equation:

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

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

The characteristic matrix is:

\[
A - \lambda I = \begin{pmatrix}
1 - \lambda & -1 & 2 \\
2 & -2 - \lambda & 4 \\
3 & -3 & 6 - \lambda
\end{pmatrix}
\]

We now compute the determinant of \( A - \lambda I \).

### Step 2: Solve for eigenvectors
Once we have the eigenvalues, we solve for the eigenvectors by solving:

\[
(A - \lambda I) v = 0
\]

for each eigenvalue \( \lambda \), where \( v \) is the corresponding eigenvector.

### Step 3: Diagonalization
If the matrix has enough independent eigenvectors, we can diagonalize it. The matrix that diagonalizes \( A \) is:

\[
P = \begin{pmatrix} v_1 & v_2 & v_3 \end{pmatrix}
\]

where \( v_1, v_2, v_3 \) are the eigenvectors corresponding to the eigenvalues, and the diagonal matrix \( D \) is:

\[
D = \begin{pmatrix}
\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3
\end{pmatrix}
\]

Let me calculate the eigenvalues and eigenvectors for you.### Eigenvalues:
The eigenvalues of matrix \( A \) are approximately:

\[
\lambda_1 = 5, \quad \lambda_2 \approx 0, \quad \lambda_3 \approx 0
\]

### Eigenvectors:
The corresponding eigenvectors are:

\[
v_1 = \begin{pmatrix} 0.2673 \\ 0.5345 \\ 0.8018 \end{pmatrix}, \quad
v_2 = \begin{pmatrix} 0.3537 \\ -0.7538 \\ -0.5538 \end{pmatrix}, \quad
v_3 = \begin{pmatrix} -0.4987 \\ -0.8489 \\ -0.1751 \end{pmatrix}
\]

### Diagonalization:
Since we have enough independent eigenvectors, the matrix \( A \) can be diagonalized.

The matrix \( P \), which diagonalizes \( A \), is formed by the eigenvectors:

\[
P = \begin{pmatrix}
0.2673 & 0.3537 & -0.4987 \\
0.5345 & -0.7538 & -0.8489 \\
0.8018 & -0.5538 & -0.1751
\end{pmatrix}
\]

The diagonal matrix \( D \) is:

\[
D = \begin{pmatrix}
5 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\]

### Final Result:
The matrix \( A \) can be diagonalized as:

\[
A = P D P^{-1}
\]

Let me know if you'd like further clarification!

### Follow-up Questions:
1. What does it mean for a matrix to be diagonalizable?
2. How do we interpret eigenvalues and eigenvectors geometrically?
3. Can all matrices be diagonalized?
4. How do we compute the inverse of a matrix in this context?
5. What happens when eigenvalues are repeated?

### Tip:
Diagonalization is particularly useful for simplifying matrix powers, making complex calculations much easier.

Find the eigenvalues and corresponding vectors of the following matrix, and diagonalise if possible. Find the matrix which diagonalises it. A = 1 -1 2 2 -2 4 3 -3 6

This problem involves finding the eigenvalues and corresponding eigenvectors of matrix A and diagonalizing it. Explore detailed steps including solving the characteristic equation det(A - λI) = 0 and using the eigenvectors to form the matrix P that diagonalizes A. Suitable for undergraduate students.

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