Positive Definite D60Dd3

1. **State the problem:** We are given a quadratic form with matrix $$A=\begin{bmatrix} 1 & -2 & 2 \\ -2 & 2 & 6 \\ 0 & -4 & 3 \end{bmatrix}$$ and asked to determine if the quadratic function is positive definite by using the discriminants (leading principal minors). 2. **Recall the rule:** A symmetric matrix is positive definite if all its leading principal minors (determinants of top-left submatrices) are positive. 3. **Calculate the first leading principal minor:** $$D_1 = |1| = 1 > 0$$ 4. **Calculate the second leading principal minor:** $$D_2 = \begin{vmatrix} 1 & -1 \\ -1 & 6 \end{vmatrix} = (1)(6) - (-1)(-1) = 6 - 1 = 5 > 0$$ 5. **Calculate the third leading principal minor:** $$D_3 = \begin{vmatrix} 1 & 6 & -2 \\ -2 & 3 & -1 \\ 0 & 3 & 0 \end{vmatrix}$$ Calculate this determinant step-by-step: $$D_3 = 1 \times \begin{vmatrix} 3 & -1 \\ 3 & 0 \end{vmatrix} - 6 \times \begin{vmatrix} -2 & -1 \\ 0 & 0 \end{vmatrix} + (-2) \times \begin{vmatrix} -2 & 3 \\ 0 & 3 \end{vmatrix}$$ Calculate each minor: $$\begin{vmatrix} 3 & -1 \\ 3 & 0 \end{vmatrix} = (3)(0) - (3)(-1) = 0 + 3 = 3$$ $$\begin{vmatrix} -2 & -1 \\ 0 & 0 \end{vmatrix} = (-2)(0) - (0)(-1) = 0$$ $$\begin{vmatrix} -2 & 3 \\ 0 & 3 \end{vmatrix} = (-2)(3) - (0)(3) = -6$$ Substitute back: $$D_3 = 1 \times 3 - 6 \times 0 + (-2) \times (-6) = 3 + 0 + 12 = 15 > 0$$ 6. **Conclusion:** Since all leading principal minors $$D_1=1$$, $$D_2=5$$, and $$D_3=15$$ are positive, the matrix $$A$$ is positive definite. Therefore, the quadratic function is positive definite.