1. **Problem Statement:** Classify the matrix corresponding to the quadratic form $$3x_1^2 + 3x_2^2 - 5x_3^2 - 2x_1x_2 - 6x_3x_2 - 6x_1x_3.$$\n\n2. **Write the quadratic form in matrix notation:** The quadratic form can be expressed as $$\mathbf{x}^T A \mathbf{x},$$ where \(\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\) and \(A\) is a symmetric matrix. \n\n3. **Find matrix A:**\nThe coefficients of the quadratic terms form the diagonal of \(A\) and the coefficients of the cross-product terms are split evenly between the symmetric positions: \n$$A = \begin{bmatrix} 3 & -1 & -3 \\ -1 & 3 & -3 \\ -3 & -3 & -5 \end{bmatrix}.$$\nExplanation:\n- The coefficient of \(x_1^2\) is 3, so \(a_{11} = 3\).\n- The coefficient of \(x_2^2\) is 3, so \(a_{22} = 3\).\n- The coefficient of \(x_3^2\) is -5, so \(a_{33} = -5\).\n- The coefficient for \(x_1 x_2\) is -2, so \(a_{12} = a_{21} = -1\).\n- The coefficient for \(x_3 x_2\) is -6, so \(a_{23} = a_{32} = -3\).\n- The coefficient for \(x_1 x_3\) is -6, so \(a_{13} = a_{31} = -3\).\n\n4. **Determine the nature of matrix A:** We classify the matrix by checking the signs of its principal minors (leading principal determinants).\n\na) First principal minor \(D_1 = a_{11} = 3 > 0\)\n\nb) Second principal minor \(D_2 = \begin{vmatrix} 3 & -1 \\ -1 & 3 \end{vmatrix} = 3 \times 3 - (-1) \times (-1) = 9 -1 = 8 > 0\)\n\nc) Third principal minor \(D_3 = \det(A)\)\nCalculate \(D_3\):\n$$\det(A) = 3 \begin{vmatrix} 3 & -3 \\ -3 & -5 \end{vmatrix} - (-1) \begin{vmatrix} -1 & -3 \\ -3 & -5 \end{vmatrix} + (-3) \begin{vmatrix} -1 & 3 \\ -3 & -3 \end{vmatrix}.$$\n\nCalculate each 2x2 determinant:\n\n- $$\begin{vmatrix} 3 & -3 \\ -3 & -5 \end{vmatrix} = 3 \times (-5) - (-3) \times (-3) = -15 -9 = -24.$$\n- $$\begin{vmatrix} -1 & -3 \\ -3 & -5 \end{vmatrix} = -1 \times (-5) - (-3) \times (-3) = 5 -9 = -4.$$\n- $$\begin{vmatrix} -1 & 3 \\ -3 & -3 \end{vmatrix} = -1 \times (-3) - 3 \times (-3) = 3 +9 = 12.$$\n\nPutting these back in determinant calculation:\n$$\det(A) = 3(-24) - (-1)(-4) + (-3)(12) = -72 -4 -36 = -112.$$\n\n5. **Classification criteria:**\n- If all principal minors are positive, \(A\) is positive definite.\n- If they alternate in sign starting with negative, \(A\) is negative definite.\n- If some principal minors are positive and others negative (and determinant negative), \(A\) is indefinite.\n\nHere \(D_1 > 0, D_2 > 0, D_3 < 0\); since the last determinant is negative, matrix \(A\) is **indefinite**.\n\n**Final Answer:** The matrix corresponding to the quadratic form is indefinite.