Find area of pentagon using determinants 
Find the area of the pentagon of the five vertices $(1,2), (4,1), (5,3), (3,7),  (2,6)$ . Please, use the way of using determinant.

My idea is to cut the pentagon into some triangles, then calculate each triangle, then sum them. I wonder if there is some other way to directly calculate it using a bigger matrix calculating its determinant?

If you label your points such that they form a clockwise rotation about a central point you can use the following shoelace formula:
$$A =\frac12\bigg| \sum_{i=1}^{n} x_iy_{i+1} - x_{i+1}y_i \bigg| = \frac12\bigg|\sum_{i=1}^n\det\bigg(\begin{matrix} x_i & x_{i+1} \\ y_i & y_{i+1} \end{matrix}\bigg)\bigg|$$
Where $x_0 = x_n,\ x_{n+1} = x_1$ and similarly, $y_0 = y_n,\ y_{n+1} = y_1$
