TITLE: Show that the area of a triangle is given by this determinant
QUESTION [28 upvotes]: I'm not sure how to solve this problem. Can you guys provide some input/hints? 

Let $A=(x_1,y_1)$, $B=(x_2,y_2)$ and $C=(x_3,y_3)$ be three points in $\mathbb{R}^{2}$.
Show that the area of $\Delta ABC$ is given by
  $\frac{1}{2}\left| \det(M) \right|$, where
  $$M = \begin{pmatrix} 1 & 1
 & 1\\  x_1 & x_2 & x_3\\  y_1 & y_2 & y_3 \end{pmatrix}$$

REPLY [2 votes]: I saw you guys' proofs. 
To be honest, it is a very simple linear algebra problem, and you only need to know very shallow knowledge of the determinant properties to solve it, which determinants work as area or volume in many calculations. And some theorem of the determinants.
First, you need to know $\frac{1}{2}\det[\mathbf{v_1},\mathbf{v_2}]$ is the area formula of the triangle whose vertices are $\mathbf{v_1},\mathbf{v_2},\mathbf{0}$ (in $R^2$), which is very easy to proof. If you still do not know it yet, remember it.
Next, translating the vertices points by the same vector at the same time does not change the area of the triangle (if the world is sane), which means if you have a triangle whose vertices are $\mathbf{a_1}, \mathbf{a_2},\mathbf{a_3}$, you can translate $\mathbf{a_1}, \mathbf{a_2}, \mathbf{a_3}$ by $\mathbf{-a_1}$ and you will get new vertice vectors, a zero vector and $\mathbf{a_2-a_1}, \mathbf{a_3-a_1}$. $\frac{1}{2}\det[\mathbf{a_2-a_1}, \mathbf{a_3-a_1}]$ is the area of the triangle.
Finally, By transpose theorem, if A is an $n\times n$ matrix, $\det A=\det A^T$, and row operation theorem, If a multiple of one row of A is added to another row to produce a matrix B,
then $\det B = \det A$.
Transpose the given matrix, and then substract the first row from the rest, expand the cofactors at column 1. Then transpose back, you will get it. (I do not really want to write the details down because it is so simple)