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Here are the given vertices of a given parallelepiped... $ (-1, 0, 0), (0, 4, 0), (-3, -5, 2), (-2, 2, -1) $ I know that first, we should translate all to the origin... $ (0, 0, 0), (1, 4, 0), (-2, -5, 2), (-1, 2, -1) $ Is the matrix to evaluate the determinant by $\begin{bmatrix}1 & 4 & 0\\-2 & -5 & 2\\-1 & 2 & -1\end{bmatrix}$ or is this incorrect? |
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There are two approaches: either you transform one point to the origin, or you add a $1$ to every vector: $$\begin{vmatrix} -1 & 0 & -3 & -2\\ 0 & 4 & -5 & 2 \\ 0 & 0 & 2 & -1 \\ 1 & 1 & 1 & 1 \end{vmatrix} = -\begin{vmatrix} 1 & -2 & -1 \\ 4 & -5 & 2 \\ 0 & 2 & -1 \end{vmatrix} = 15 $$ So the two solutions differ in their sign. Which doesn't matter if you only care about absolute values, otherwise you should make sure to use one way exclusively. Of course, transforming a matrix doesn't change its determinant, so you might as well write the vectors as rows, the way you do in your question. |
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