# Given a parallelepiped, how do I find the determinant given vertices?

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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It looks right. –  Aleks Vlasev Oct 17 '12 at 5:18
Yes it is. The volume is 15. –  copper.hat Oct 17 '12 at 5:25
Great! Thank you –  Grace C Oct 17 '12 at 5:57

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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