T.10. Using the Lagrange's method , reduce the quadratic form to the signed sum of squares and obtain the appropriate change of variables.Find positive and negative indices of inertia.Find out whether the form is positive definite, negative definite, or indefinite. (a) q=4x_(1)^2+3x_(2)^2-x_(3)^2-4x_(1)x_(2)+12x_(2)x_(3) (b) q=x_(1)x_(2)+x_(1)x_(3)+x_(2)x_(3)

(a) To reduce the quadratic form $q=4x_{1}^{2}+3x_{2}^{2}-x_{3}^{2}-4x_{1}x_{2}+12x_{2}x_{3}$ to the signed sum of squares using Lagrange's method, we need to diagonalize the matrix associated with the quadratic form.<br /><br />The matrix for the quadratic form is:<br />$$<br />\begin{bmatrix}<br />4 & -2 & 0 \\<br />-2 & 3 & 6 \\<br />0 & 6 & -1<br />\end{bmatrix}<br />$$<br /><br />To diagonalize this matrix, we need to find its eigenvalues and eigenvectors. The eigenvalues are $\lambda_1 = 5$, $\lambda_2 = -1$, and $\lambda_3 = -1$. The corresponding eigenvectors are $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$, $\mathbf{v}_2 = \begin{bmatrix} 1 \\ 0 \\ 3 \end{bmatrix}$, and $\mathbf{v}_3 = \begin{bmatrix} 1 \\ 0 \\ -3 \end{bmatrix}$.<br /><br />Now, we can diagonalize the matrix as:<br />$$<br />\begin{bmatrix}<br />4 & -2 & 0 \\<br />-2 & 3 & 6 \\<br />0 & 6 & -1<br />\end{bmatrix}<br />=<br />\begin{bmatrix}<br />1 & 1 & 1 \\<br />1 & 0 & 0 \\<br />0 & 3 & -3<br />\end{bmatrix}<br />\begin{bmatrix}<br />5 & 0 & 0 \\<br />0 & -1 & 0 \\<br />0 & 0 & -1<br />\end{bmatrix}<br />\begin{bmatrix}<br />1 & 1 & 1 \\<br />1 & 0 & 0 \\<br />0 & 3 & -3<br />\end{bmatrix}^{-1}<br />$$<br /><br />The positive index of inertia is 1, and the negative indices of inertia are -1 and -1. Since the matrix has both positive and negative eigenvalues, the quadratic form is indefinite.<br /><br />(b) For the quadratic form $q=x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}$, we can rewrite it in matrix form as:<br />$$<br />\begin{bmatrix}<br />0 & \frac{1}{2} & \frac{1}{2} \\<br />\frac{1}{2} & 0 & \frac{1}{2} \\<br />\frac{1}{2} & \frac{1}{2} & 0<br />\end{bmatrix}<br />\begin{bmatrix}<br />x_{1} \\<br />x_{2} \\<br />x_{3}<br />\end{bmatrix}<br />=<br />\begin{bmatrix}<br />q \\<br />0 \\<br />0<br />\end{bmatrix}<br />$$<br /><br />The matrix is symmetric and has eigenvalues $\lambda_1 = \frac{3}{2}$, $\lambda_2 = -\frac{1}{2}$, and $\lambda_3 = -\frac{1}{2}$. The corresponding eigenvectors are $\mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$, $\mathbf{v}_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$, and $\mathbf{v}_3 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$.<br /><br />We can diagonalize the matrix as:<br />$$<br />\begin{bmatrix}<br />0 & \frac{1}{2} & \frac{1}{2} \\<br />\frac{1}{2} & 0 & \frac{1}{2} \\<br />\frac{1}{2} & \frac{1}{2} & 0<br />\end{bmatrix}<br />=<br />\begin{bmatrix}<br />1 & 1 & 1 \\<br />1 & 0 & 0 \\<br />0 & 1 & 1<br />\end{bmatrix}<br />\begin{bmatrix}<br />\frac{3}{2} & 0 & 0 \\<br />0 & -\frac{1}{2} & 0 \\<br />0 & 0 & -\frac{1}{2}<br />\end{bmatrix}<br />\begin{bmatrix}<br />1 & 1 & 1 \\<br />1 & 0 & 0 \\<br />0 & 1 & 1<br />\end{bmatrix}^{-1}<br />$$<br /><br />The positive index of inertia is 1, and the negative indices of inertia are -1 and -1. Since the matrix has both positive and negative eigenvalues, the quadratic form is indefinite.