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.

The oriented area $A(u,v)$ of the parallelogram spanned by vectors $u,v$ is bilinear (eg. $A(u+v,w)=A(u,w)+A(v,w)$ can be seen by adding and removing a triangle) and skew-symmetric. Hence $A(ae_1+be_2,ce_1+de_2)=(ad-bc)A(e_1,e_2)=ad-bc$. (the same works for oriented volumes in any dimension)
