Can you give me characteristic equation for the matrix A=[(1,2,1),(2,2,1),(1,1,3)]

To find the characteristic equation of a matrix, we need to compute the determinant of \( A - \lambda I \), where \( A \) is the given matrix, \( \lambda \) is an eigenvalue, and \( I \) is the identity matrix of the same size as \( A \). 

Given matrix \( A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 3 \end{bmatrix} \), let's find the characteristic equation.

1. **Form \( A - \lambda I \):**

\[
A - \lambda I = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 3 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 - \lambda & 2 & 1 \\ 2 & 2 - \lambda & 1 \\ 1 & 1 & 3 - \lambda \end{bmatrix}
\]

2. **Calculate the determinant \( \det(A - \lambda I) \):**

\[
\det(A - \lambda I) = \left| \begin{matrix} 1 - \lambda & 2 & 1 \\ 2 & 2 - \lambda & 1 \\ 1 & 1 & 3 - \lambda \end{matrix} \right|
\]

We will calculate this determinant using the cofactor expansion along the first row.

\[
\det(A - \lambda I) = (1 - \lambda) \begin{vmatrix} 2 - \lambda & 1 \\ 1 & 3 - \lambda \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \\ 1 & 3 - \lambda \end{vmatrix} + 1 \begin{vmatrix} 2 & 2 - \lambda \\ 1 & 1 \end{vmatrix}
\]

Let's compute each 2x2 determinant:

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

Substitute these back into the determinant formula:

\[
\det(A - \lambda I) = (1 - \lambda)(\lambda^2 - 5\lambda + 5) - 2(5 - 2\lambda) + \lambda
\]

3. **Expand and simplify the equation:**

\[
\det(A - \lambda I) = (1 - \lambda)(\lambda^2 - 5\lambda + 5) - 10 + 4\lambda + \lambda
\]

\[
= (\lambda^2 - 5\lambda + 5) - \lambda(\lambda^2 - 5\lambda + 5) - 10 + 5\lambda
\]

Now, distribute and simplify:

\[
= \lambda^2 - 5\lambda + 5 - \lambda^3 + 5\lambda^2 - 5\lambda - 10 + 5\lambda
\]

Combine like terms:

\[
= -\lambda^3 + 6\lambda^2 - 5\lambda - 5
\]

4. **The characteristic equation:**

\[
\lambda^3 - 6\lambda^2 + 5\lambda + 5 = 0
\]

Thus, the characteristic equation for matrix \( A \) is:

\[
\boxed{\lambda^3 - 6\lambda^2 + 5\lambda + 5 = 0}
\]

Would you like to see the eigenvalues or have any questions? 

### Related Questions:
1. How do you find the eigenvalues of a matrix?
2. What are the steps to find the eigenvectors once the eigenvalues are known?
3. Can you show how to diagonalize the matrix \( A \)?
4. What are some applications of eigenvalues and eigenvectors in real-world problems?
5. How does the characteristic equation change if the matrix \( A \) is modified?

### Tip:
When calculating determinants for matrices, using cofactor expansion along a row or column with the most zeros can simplify the computation.

Learn how to find the characteristic equation of matrix A = [1, 2, 1; 2, 2, 1; 1, 1, 3] step-by-step. This problem covers matrix algebra concepts, including eigenvalues and determinants. Suitable for college-level mathematics.

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