# Area of a triangle using a matrix

Find the area of a triangle whose vertices are (1,0), (2,2), and (4,3)

$$\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}1&0&1\\2&2&1\\4&3&1\end{vmatrix}$$

$$=1(-1)^2\begin{vmatrix}2&1\\3&1\end{vmatrix}+0(-1)^3\begin{vmatrix}2&1\\4&1\end{vmatrix}+1(-1)^4\begin{vmatrix}2&2\\4&3\end{vmatrix}$$ $$=1(-1)+0+1(-2)$$ $$=-3$$

Then you multiply by $-\frac{1}{2}$ to get a positive area of $\frac{3}{2}$.

My question is where does $-1$ come from in the second line of the equation and why is it being squared first, then cubed, then raised to a power of 4?

It comes from $$\det(A) = \sum_{j=1}^n (-1)^{i+j} a_{i,j} M_{i,j}$$ with $n=3$ and minor $M_{i,j}$, where $i=1$. So in this case $$\det(A) = \sum_{j=1}^3 a_{1,j}(-1)^{1+j} M_{1,j}=1(-1)^2\begin{vmatrix}2&1\\3&1\end{vmatrix}+0(-1)^3\begin{vmatrix}2&1\\4&1\end{vmatrix}+1(-1)^4\begin{vmatrix}2&2\\4&3\end{vmatrix}$$
$$A=\frac12 |\det(u,v)|$$
with, for example, $u=(2,2)-(1,0)=(1,2)$ and $v=(4,3)-(1,0)=(3,3)$ that is
$$A=\frac12\begin{vmatrix}1&2\\3&3\end{vmatrix}=\frac32$$