1. **State the problem:** Find the determinant of matrix $$A=\begin{bmatrix} 1 & 3 & -2 \\ 3 & 0 & 1 \\ -1 & 2 & -4 \end{bmatrix}$$ using the basket method. 2. **Recall the basket method (also known as the rule of Sarrus):** For a 3x3 matrix $$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$$ the determinant is calculated as: $$\det(A) = aei + bfg + cdh - ceg - bdi - afh$$ This involves summing the products of the diagonals from left to right and subtracting the products of the diagonals from right to left. 3. **Identify the elements:** $$a=1, b=3, c=-2, d=3, e=0, f=1, g=-1, h=2, i=-4$$ 4. **Calculate the positive diagonal products:** $$aei = 1 \times 0 \times (-4) = 0$$ $$bfg = 3 \times 1 \times (-1) = -3$$ $$cdh = -2 \times 3 \times 2 = -12$$ Sum of positive diagonals: $$0 + (-3) + (-12) = -15$$ 5. **Calculate the negative diagonal products:** $$ceg = -2 \times 0 \times (-1) = 0$$ $$bdi = 3 \times 3 \times (-4) = -36$$ $$afh = 1 \times 1 \times 2 = 2$$ Sum of negative diagonals: $$0 + (-36) + 2 = -34$$ 6. **Calculate the determinant:** $$\det(A) = (-15) - (-34) = -15 + 34 = 19$$ **Final answer:** $$\boxed{19}$$