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A basic fact about $3$ dimensional vectors is that the quantity

$\pm\det\left( \begin{array}{ccc} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \\ \end{array} \right)$

is equal to the volume of the parallelepiped determined by the vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ where $\vec{a} = \langle a_1, a_2, a_3 \rangle$, $\vec{b} = \langle b_1, b_2, b_3\rangle$, and $\vec{c} = \langle c_1, c_2, c_3\rangle$. From e.g. the explicit formula for the determinant of a matrix in terms of its entries it is evident that this is the same as

$\pm\det\left( \begin{array}{ccc} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \\ \end{array} \right)$

so that the volume of of the parallelepiped determined by $\vec{a}$, $\vec{b}$ and $\vec{c}$ is the same as the the volume of parallelepiped determined by $\langle a_1, b_1, c_1\rangle$, $\langle a_2, b_2, c_2\rangle$, and $\langle a_3, b_3, c_3\rangle$.

My question is:

Is there a geometric proof that the volumes of the two parallelepipeds are the same?

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    I'm confused. Didn't you just give a geometric proof? You can't be too rigorous with such proofs?2011-11-04
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    @Alex: Suppose you let $A$'s columns represent vectors in space defining a parallelepiped. Then $A$ and its transpose $A^T$ represent two different parallelepipeds; you can show these two figures have the same volume using algebra but OP wants to know if there's a geometric proof.2011-11-04
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    I would start with $2 \times 2$ matrices. If you can understand that situation, you can probably work up to the $3 \times 3$.2011-11-04

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