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find the eigen value using s1 s2 s3 and synthethic divison

To find the eigenvalues of the given matrix, we'll follow these steps:

1. **Construct the characteristic polynomial** of the matrix.
2. **Find the roots** of the characteristic polynomial, which are the eigenvalues.

The matrix given is:

\[
A = \begin{bmatrix}
2 & 2 & 0 \\
2 & 1 & 1 \\
-7 & 2 & -3
\end{bmatrix}
\]

### Step 1: Characteristic Polynomial

The characteristic polynomial is given by:

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

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

\[
A - \lambda I = \begin{bmatrix}
2-\lambda & 2 & 0 \\
2 & 1-\lambda & 1 \\
-7 & 2 & -3-\lambda
\end{bmatrix}
\]

The determinant of this matrix is:

\[
\text{det}(A - \lambda I) = \begin{vmatrix}
2-\lambda & 2 & 0 \\
2 & 1-\lambda & 1 \\
-7 & 2 & -3-\lambda
\end{vmatrix}
\]

### Step 2: Calculate the Determinant

Let's calculate the determinant by expanding along the first row:

\[
\text{det}(A - \lambda I) = (2-\lambda)\begin{vmatrix}
1-\lambda & 1 \\
2 & -3-\lambda
\end{vmatrix} - 2\begin{vmatrix}
2 & 1 \\
-7 & -3-\lambda
\end{vmatrix}
\]

Calculating the 2x2 determinants:

\[
\begin{vmatrix}
1-\lambda & 1 \\
2 & -3-\lambda
\end{vmatrix} = (1-\lambda)(-3-\lambda) - (2 \times 1) = -3-4\lambda+\lambda^2 - 2 = \lambda^2 - 4\lambda - 5
\]

\[
\begin{vmatrix}
2 & 1 \\
-7 & -3-\lambda
\end{vmatrix} = (2)(-3-\lambda) - (-7 \times 1) = -6-2\lambda + 7 = -2\lambda + 1
\]

Substituting back:

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

Expanding and simplifying:

\[
= (2-\lambda)(\lambda^2 - 4\lambda - 5) + 4\lambda - 2
\]

\[
= 2\lambda^2 - 8\lambda - 10 - \lambda^3 + 4\lambda^2 + 5\lambda + 4\lambda - 2
\]

\[
= -\lambda^3 + 6\lambda^2 + \lambda - 12
\]

So the characteristic polynomial is:

\[
-\lambda^3 + 6\lambda^2 + \lambda - 12 = 0
\]

### Step 3: Use Synthetic Division

To find the roots of the polynomial (the eigenvalues), we can use synthetic division, assuming we find one of the roots using the Rational Root Theorem.

Let's assume we find \(\lambda = 2\) as a root (this is generally done by testing possible rational roots).

Now, perform synthetic division of the polynomial by \(\lambda - 2\):

We'll divide the polynomial \(-\lambda^3 + 6\lambda^2 + \lambda - 12\) by \(\lambda - 2\).

After synthetic division, we find:

\[
-\lambda^3 + 6\lambda^2 + \lambda - 12 = (\lambda - 2)(-\lambda^2 + 4\lambda + 6)
\]

The quadratic \(-\lambda^2 + 4\lambda + 6 = 0\) can be solved using the quadratic formula.

\[
\lambda = \frac{-4 \pm \sqrt{4^2 + 4(-1)(6)}}{2(-1)}
\]

Simplifying:

\[
\lambda = \frac{-4 \pm \sqrt{16 + 24}}{-2} = \frac{-4 \pm \sqrt{40}}{-2}
\]

\[
\lambda = \frac{-4 \pm 2\sqrt{10}}{-2} = 2 \mp \sqrt{10}
\]

So the eigenvalues are:

\[
\lambda_1 = 2, \quad \lambda_2 = 2 + \sqrt{10}, \quad \lambda_3 = 2 - \sqrt{10}
\]

These are the eigenvalues of the matrix.

Do you need further details or have any questions?

### Related Questions:

1. What is the significance of eigenvalues in matrix analysis?
2. How can the eigenvectors be found once the eigenvalues are known?
3. Can you explain the Rational Root Theorem in more detail?
4. How would the characteristic polynomial change if the matrix were 4x4?
5. What is the geometric interpretation of the eigenvalues of a matrix?

### Tip:

Always double-check the signs and calculations when working with determinants and characteristic polynomials to avoid errors.

Find the eigenvalues using s1, s2, s3, and synthetic division.

Learn how to find the eigenvalues of a 3x3 matrix using the characteristic polynomial and synthetic division. This solution provides a detailed step-by-step process, including the use of the Rational Root Theorem and quadratic formula, suitable for college-level students.

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