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.

For the matrix $\left[\begin{array}{cc}
a & c \\
b & d \\
\end{array}\right]$ let 
$$A = \left[\begin{array}{c}
a \\
b \\
\end{array}\right]
\;\text{and}\; B = \left[\begin{array}{c}
c \\
d \\
\end{array}\right]$$
as shown in the following figure.  

Then the height of the parallelogram is 
$$\text{height} = |B|\sin\alpha = |B|\cos\beta.$$ 
If we rotate $A$ by 90 degrees in the CCW direction as follows:
$$R_{90º}A = 
\left[\begin{array}{cc}
0 &-1 \\
1 &0 \\
\end{array}\right]
\left[\begin{array}{c}
a \\
b \\
\end{array}\right] =
\left[\begin{array}{c}
-b \\
a \\
\end{array}\right],$$
maintaining the magnitude of the base as 
$$\text{base} = |A| = |R_{90º}A|,$$
then it is clear that the area of the parallelogram is therefore 
$$
\text{base}\times\text{height}=(|A|)(|B|\sin\alpha) = |R_{90º}A|\;|B|\cos\beta = (R_{90º}A)\cdot B = 
\left[\begin{array}{c}
-b \\
a \\
\end{array}\right]
\cdot
\left[\begin{array}{c}
c \\
d \\
\end{array}\right]
=
ad-bc.
$$
Q.E.D.
