I know the geometrical meaning of the determinant of a matrix, and I know that, for example, the area of a parallelogram defined by two vectors $$ v=\begin{bmatrix}a \\b \end{bmatrix},\quad w=\begin{bmatrix}c \\d \end{bmatrix}, $$ is equal to $$ \det \begin{pmatrix} a & c \\ b & d \end{pmatrix}. $$

I know that the area of the triangle $ABC$ is equal to $$\frac{1}{2} \det\begin{pmatrix} 1 & 1 & 1\\ x_A & x_B & x_C \\ y_A & y_B & y_C \end{pmatrix}. $$

I would like to find a geometrical proof of the last formula. Why is the area of a triangle (numerically) equal to the volume of the solid generated by the three vectors $[1, x_A, y_A]$, $[1, x_B, y_B]$, $[1, x_C, y_C]$? I can verify it, but I can't see it.

  • $\begingroup$ Sorry, I forgot a fraction... thanks $\endgroup$ – zar Feb 26 at 11:00

Subtract the second and the third columns by the first (this is geometrically a shear). Now you get a block lower triangular matrix and the determinant becomes the area of a parallelogram (or strictly speaking, a slanted cylinder with a parallelogram base and unit height) with two adjacent sides $\pmatrix{x_B-x_A\\ y_B-y_A}$ and $\pmatrix{x_C-x_A\\ y_C-y_A}$. Multiply it by one half, you get the area of the triangle.

  • $\begingroup$ Thanks. I tried with a GeoGebra construction, and now I "see" it. $\endgroup$ – zar Feb 26 at 17:20

It might be easier to visualize if you move the row of $1$s to the bottom so that the triangle lies on the plane $z=1$. (This doesn’t change the value of the determinant.)

enter image description here

Together with the origin, this triangle forms a tetrahedron with altitude $1$, so its volume is numerically equal to the area of $\triangle{ABC}$. However, you can take any of the other faces as the base, say $\triangle{OBC}$. The area of this triangle is $\frac12\|B\times C\|$. (This is fundamentally the same as using a determinant in 2-D.) The altitude to $A$ from this face is the length of the projection of $A$ onto a normal $\mathbf n$ to the face, which can be computed via a dot product: $A\cdot{\mathbf n\over\|\mathbf n\|}$. For the normal, we can take $\mathbf n=B\times C$, so the volume of the tetrahedron, and hence the area of $\triangle{ABC}$, is $$\frac12\|B\times C\| \left(A\cdot{B\times C \over \|B\times C\|}\right) = \frac12A\cdot B\times C = \frac12\begin{vmatrix}x_A&x_B&x_C\\y_A&y_B&y_C\\1&1&1 \end{vmatrix}.$$


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