# Is there a geometric proof that the determinant of a 3x3 matrix is invariant under switching rows and columns?

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?

• I'm confused. Didn't you just give a geometric proof? You can't be too rigorous with such proofs? – Alex Youcis Nov 4 '11 at 5:07
• @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. – anon Nov 4 '11 at 5:19
• I would start with $2 \times 2$ matrices. If you can understand that situation, you can probably work up to the $3 \times 3$. – Dan Drake Nov 4 '11 at 6:56

If you're willing to posit that the determinant is linear, here's something for the $2 \times 2$ case: consider $\left(\begin{smallmatrix} a & b \\ 1 & 0 \end{smallmatrix}\right)$. Rowwise, that gives us a parallelogram with vertices $(a,b)$, $(1,0)$, and $(a+1,b)$. (And the origin, of course, but I won't bother writing that.) That's pretty clearly a parallelogram with base 1 and height $b$.
After transposing, $\left(\begin{smallmatrix} a & 1 \\ b & 0 \end{smallmatrix}\right)$ gives a parallelogram with vertices $(a,1)$, $(b,0)$, and $(a+b,1)$; now the height is 1 and the base is $b$.
Geometrically, you can slide the top of one of those until you get a rectangle, reflect about $y=x$, and slide the top again (all area-preserving transformations) to turn one into the other.
Scaling the $(1,0)$ row/column obviously just scales the resulting parallelograms. The $\left(\begin{smallmatrix} a & b \\ 0 & 1 \end{smallmatrix}\right)$ case works out the same. So if the linearity of the determinant is okay, you've got the $2 \times 2$ case.
For $3 \times 3$, my guess is that it suffices to do $\left(\begin{smallmatrix} a & b & c \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{smallmatrix}\right)$.