reduce the quadratic form to cannonical form by orthogonal reduction for 3x1square+2x2square+3x2square-2x1x2-2x2x3 and find its rank,index,signature and nature

### Problem:
We are given a quadratic form:

\[ Q(x_1, x_2, x_3) = 3x_1^2 + 2x_2^2 + 3x_2^2 - 2x_1x_2 - 2x_2x_3 \]

We need to:
1. **Reduce the quadratic form to canonical form** using orthogonal reduction.
2. Find its **rank**, **index**, **signature**, and **nature**.

---

### Step 1: Write the quadratic form in matrix notation

We can express the quadratic form as:

\[
Q(x_1, x_2, x_3) = \mathbf{x}^T \mathbf{A} \mathbf{x}
\]

where \( \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \), and \( \mathbf{A} \) is the symmetric matrix representing the quadratic form.

The given quadratic form can be rearranged as:

\[
Q(x_1, x_2, x_3) = 3x_1^2 + (2+3)x_2^2 - 2x_1x_2 - 2x_2x_3
\]

This simplifies to:

\[
Q(x_1, x_2, x_3) = 3x_1^2 + 5x_2^2 - 2x_1x_2 - 2x_2x_3
\]

So, the matrix \( \mathbf{A} \) is:

\[
\mathbf{A} = \begin{pmatrix} 
3 & -1 & 0 \\
-1 & 5 & -1 \\
0 & -1 & 0 
\end{pmatrix}
\]

---

### Step 2: Diagonalization of the symmetric matrix \( \mathbf{A} \)

We aim to diagonalize \( \mathbf{A} \) using orthogonal transformation, which involves finding the eigenvalues and eigenvectors of \( \mathbf{A} \).

#### Eigenvalues:
The eigenvalues of \( \mathbf{A} \) can be found by solving the characteristic equation:

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

Let's compute the eigenvalues.

The eigenvalues of matrix \( \mathbf{A} \) are:

\[
\lambda_1 = 5.57, \quad \lambda_2 = 2.64, \quad \lambda_3 = -0.20
\]

These eigenvalues help in transforming the quadratic form into its **canonical form**, which is:

\[
Q'(x_1, x_2, x_3) = 5.57x_1'^2 + 2.64x_2'^2 - 0.20x_3'^2
\]

This is the canonical form, where the quadratic form has been diagonalized.

---

### Step 3: Rank, Index, Signature, and Nature

- **Rank**: The rank is the number of non-zero eigenvalues. Since all the eigenvalues are non-zero, the rank is \( 3 \).
  
- **Index**: The index is the number of negative eigenvalues. Since there is one negative eigenvalue (\( \lambda_3 = -0.20 \)), the index is \( 1 \).

- **Signature**: The signature is a triplet \( (p, q, r) \), where:
  - \( p \) is the number of positive eigenvalues.
  - \( q \) is the number of negative eigenvalues.
  - \( r \) is the number of zero eigenvalues.

  In this case, \( p = 2 \) (for \( \lambda_1 \) and \( \lambda_2 \)), \( q = 1 \) (for \( \lambda_3 \)), and \( r = 0 \) (no zero eigenvalues). Thus, the signature is \( (2, 1, 0) \).

- **Nature**: Since the quadratic form has both positive and negative eigenvalues, the quadratic form is **indefinite**.

---

### Summary:

- **Canonical form**: \( 5.57x_1'^2 + 2.64x_2'^2 - 0.20x_3'^2 \)
- **Rank**: \( 3 \)
- **Index**: \( 1 \)
- **Signature**: \( (2, 1, 0) \)
- **Nature**: Indefinite

Would you like more details or have any questions? 

Here are 5 relative questions you might find helpful:
1. How do eigenvectors transform the original variables in the quadratic form?
2. What are the steps involved in finding eigenvectors for matrix diagonalization?
3. How does orthogonal reduction preserve the structure of the quadratic form?
4. Can we determine the type of conic sections using this method for 2D quadratic forms?
5. What role do eigenvalues play in determining the nature of quadratic surfaces?

**Tip:** In quadratic forms, the sign of the eigenvalues directly influences the shape of the surfaces defined by the form.

Learn how to reduce a quadratic form 3x1^2 + 2x2^2 + 3x2^2 - 2x1x2 - 2x2x3 to its canonical form using orthogonal reduction. This problem covers matrix diagonalization, eigenvalues, and key concepts such as rank, index, and signature of the quadratic form. Suitable for undergraduate linear algebra students.

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