Area of a Triangle (and the beginning of barycentric coordinates) Let the dimension be $n=2$ and $T$ a Triangle with positive area $|T|$ and the corners $P_1, P_2, P_3 \in \mathbb{R}^2$. How do I show:
$2|T|=\det
    \begin{pmatrix}
        1 & 1  & 1 \\
        P_1(1) & P_2(1) & P_3(1) \\
         P_1(2) & P_2(2) & P_3(2)
    \end{pmatrix}$
With $P_k(j)$ being the $j$th component of the vectors $P_k\in \mathbb{R}^2$
 A: There are quite a few ways to derive this formula, but thinking of the determinant as a volume has a straightforward geometric appeal.  
Place the triangle $\triangle{ABC}$ on the plane $x=1$ in $\mathbb R^3$ and consider the pyramid with this triangle as its base and the origin for its apex: 

This pyramid’s altitude is $1$, so its volume is equal to the area of its base, $\triangle{ABC}$. If we glue a reflected copy of this pyramid to its base, we get a parallelepiped whose volume is given by the triple product $$\overrightarrow{OA}\times\overrightarrow{OB}\cdot\overrightarrow{OC} = \det\begin{bmatrix}\overrightarrow{OA} & \overrightarrow{OB} & \overrightarrow{OC}\end{bmatrix} = \det\begin{bmatrix}1&1&1 \\ A(1)&B(1)&C(1) \\ A(2)&B(2)&C(2) \end{bmatrix}.$$ The volume of the pyramid, and hence also the area of the triangle, is half of this value.  
We can also construct this geometrically in $\mathbb R^2$, but it requires some extra work to pass to a $3\times3$ determinant. Similarly to the above construction, we adjoin a reflected version of the triangle to form a parallelogram:  

The area of this paralellogram is equal to the value of the determinant $$\det\begin{bmatrix}\overrightarrow{CA}&\overrightarrow{CB}\end{bmatrix} = \det\begin{bmatrix}A(1)-C(1) & B(1)-C(1) \\ A(2)-C(2) & B(2)-C(2)\end{bmatrix}.$$ Embedding this into a $3\times3$ matrix and using some properties of determinants, $$\begin{align} \begin{vmatrix}A(1)-C(1) & B(1)-C(1) \\ A(2)-C(2) & B(2)-C(2)\end{vmatrix}
 &= \begin{vmatrix}0&0&1 \\ A(1)-C(1) & B(1)-C(1) & 0 \\ A(2)-C(2) & B(2)-C(2) & 0\end{vmatrix} \\
 &= \begin{vmatrix}0&0&1 \\ A(1)-C(1) & B(1)-C(1) & C(1) \\ A(2)-C(2) & B(2)-C(2) & C(2)\end{vmatrix} \\
 &= \begin{vmatrix}1&1&1 \\ A(1) & B(1) & C(1) \\ A(2) & B(2) & C(2)\end{vmatrix} \end{align}.$$ The second step is valid because if we expand the determinant along the top row, only the lower-left $2\times2$ submatrix contributes to its value, so the entries below the $1$ in the last column can be anything. I prefer the derivation as a volume, though, because it doesn’t require any algebraic trickery.
