Clasifica la siguiente forma cuadrática expresándola en formamatricial por el método de los Autovalores.Q(x1 , x2 , x3) = X1 2 + 4X3 X2 + 2X2 2 + 2X3 2

1. Break Down the Problem

To classify the quadratic form $Q(x_1, x_2, x_3) = x_1^2 + 4x_3x_2 + 2x_2^2 + 2x_3^2$, we need to express it in matrix form and analyze the eigenvalues to determine its classification.

2. Relevant Concepts

A quadratic form can be written in matrix notation as:

$Q(\mathbf{x}) = \mathbf{x}^T A \mathbf{x}$

where $\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$ and $A$ is a symmetric matrix representing the coefficients of the quadratic form.

3. Analysis and Detail

To express $Q$ in matrix form, we identify the coefficients corresponding to the variables and their interactions:

• $x_1^2$ contributes $1$ to the matrix.
• The term $4x_3x_2$ contributes $2$ to $A_{23}$ and $2$ to $A_{32}$ (since it involves both $x_2$ and $x_3$).
• The term $2x_2^2$ contributes $2$ to $A_{22}$.
• The term $2x_3^2$ contributes $2$ to $A_{33}$.

Thus, the coefficient matrix $A$ is given by:

$A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & 2 & 2 \end{pmatrix}$

4. Verify and Summarize

Next, we find the eigenvalues of matrix $A$ to classify the quadratic form. The characteristic polynomial is determined by:

$\det(A - \lambda I) = 0$

Calculating $A - \lambda I$:

$A - \lambda I = \begin{pmatrix} 1 - \lambda & 0 & 0 \\ 0 & 2 - \lambda & 2 \\ 0 & 2 & 2 - \lambda \end{pmatrix}$

Calculating the determinant:

$\det(A - \lambda I) = (1 - \lambda) \det\begin{pmatrix} 2 - \lambda & 2 \\ 2 & 2 - \lambda \end{pmatrix}$

Expanding this:

$= (1 - \lambda)((2 - \lambda)(2 - \lambda) - 4) = (1 - \lambda)((\lambda^2 - 4\lambda))$

Setting the equation to zero will yield the eigenvalues. The eigenvalues are $\lambda_1 = 1$, $\lambda_2 = 0$, and $\lambda_3 = 2$.

Final Answer

The classification of the quadratic form based on its eigenvalues is indefinite due to the presence of both positive (2) and zero (0) eigenvalues.