Why determinant of a 2 by 2 matrix is the area of a parallelogram? Let $A=\begin{bmatrix}a & b\\ c & d\end{bmatrix}$.
How could we show that $ad-bc$ is the area of a parallelogram with vertices $(0, 0),\ (a, b),\ (c, d),\ (a+b, c+d)$?
Are the areas of the following parallelograms the same? 
$(1)$ parallelogram with vertices $(0, 0),\ (a, b),\ (c, d),\ (a+c, b+d)$.
$(2)$ parallelogram with vertices $(0, 0),\ (a, c),\ (b, d),\ (a+b, c+d)$.
$(3)$ parallelogram with vertices $(0, 0),\ (a, b),\ (c, d),\ (a+d, b+c)$.
$(4)$ parallelogram with vertices $(0, 0),\ (a, c),\ (b, d),\ (a+d, b+c)$.
Thank you very much.

Also, if the coordinates of any shape are transformed by a matrix, the area will be changed by a scale factor equal to the determinant.
Since the determinant is the scale factor when the unit square is transformed to a parallelogram, it will be the scale factor when any parallelogram with the origin as a vertex  is transformed to any other parallelogram because the inverse matrix will transform a parallelogram back into a square and has reciprocal determinant. If there is no inverse, the determinant is 0 and the transformed shape has no area. 
Any triangle with the origin as a vertex can be drawn as half of a parallelgram including the origin. Any triangle not including the origin is the area of a triangle containing the origin minus two triangles inside not containing the origin. The area of any shape can be split into triangles, although an infinite number will be required if it has curved sides.
