determinantgeometry
Given a Parallelogram with the co-ordinates: $(a+c, b+d), (c,d), (a, b)$ and $(0, 0)$
I have to prove that the area of the Parallelogram is:
$|ad-bc|$ as in the determinant of:
$$\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}$$
How do I even begin using the concept of determinants for this geometrical question?
Spend a little time with this figure due to Solomon W. Golomb and enlightenment is not far off:
(Appeared in Mathematics Magazine, March 1985.)
The determinant $$ \det(A) = \det(a_1, a_2, a_3) $$ is an alternating multi linear form, this means $$ \det(a_1+ \alpha a_2, a_2, a_3) = \det(a_1, a_2, a_3) + \alpha \det(a_2, a_2, a_3) $$ because $\det$ is linear in the first argument as multi linear form.
Because it is an alternating form, if we exchange the first two arguments, the sign changes and we have $$ \det(a_2, a_2, a_3) = - \det(a_2, a_2, a_3) $$ which means $\det(a_2,a_2,a_3)$ vanishes. This leads to $$ \det(a_1+ \alpha a_2, a_2, a_3) = \det(a_1, a_2, a_3) $$ and similar for the other arguments.
We can use this property to simplify your determinant: \begin{align} d &= \begin{vmatrix} y + z & x & x \\ y & z + x & y \\ z & z & x + y \end{vmatrix} = \begin{vmatrix} 0 & -2z & -2y \\ y & z + x & y \\ z & z & x + y \end{vmatrix} = 2 \begin{vmatrix} 0 & -z & -y \\ y & z + x & y \\ z & z & x + y \end{vmatrix} \\ &= 2 \begin{vmatrix} 0 & -z & -y \\ y & x & 0 \\ z & 0 & x \end{vmatrix} = -2 \begin{vmatrix} 0 & z & y \\ y & x & 0 \\ z & 0 & x \end{vmatrix} = 2 \begin{vmatrix} 0 & y & z \\ y & 0 & x \\ z & x & 0 \end{vmatrix} \end{align} If $x = y = z = 0$ then $d = 0$ and the proposed formula holds. If two of the variables $x,y,z$ are zero, then one of the columns is zero and we have e.g. $$ \det(0, a_2, a_3) = \det(0 + a_2, a_2, a_3) = \det(a_2, a_2, a_3) = 0 $$ If e.g. $x = 0$ we have $$ d = 2 \begin{vmatrix} 0 & y & z \\ y & 0 & 0 \\ z & 0 & 0 \end{vmatrix} = 2 y z \begin{vmatrix} 0 & 1 & 1 \\ y & 0 & 0 \\ z & 0 & 0 \end{vmatrix} = 2 y z \begin{vmatrix} 0 & 0 & 1 \\ y & 0 & 0 \\ z & 0 & 0 \end{vmatrix} = -2 y z \begin{vmatrix} 0 & 0 & 1 \\ 0 & y & 0 \\ 0 & z & 0 \end{vmatrix} = 0 $$ and similar if just $y$ or $z$ vanishes.
If $x,y,z$ do not vanish each, we have \begin{align} d &= 2 \begin{vmatrix} 0 & y & z \\ y & 0 & x \\ z & x & 0 \end{vmatrix} = \frac{2}{xyz} \begin{vmatrix} 0 & xy & xz \\ yz & 0 & xz \\ yz & xy & 0 \end{vmatrix} = \frac{2}{xyz} \begin{vmatrix} 0 & xy & xz \\ yz & 0 & xz \\ yz & 0 & -xz \end{vmatrix} \\ &= \frac{2}{xyz} \begin{vmatrix} 0 & xy & xz \\ 0 & 0 & 2xz \\ yz & 0 & -xz \end{vmatrix} = \frac{4}{y} \begin{vmatrix} 0 & xy & xz \\ 0 & 0 & 1 \\ yz & 0 & -xz \end{vmatrix} = \frac{4}{y} \begin{vmatrix} 0 & xy & 0 \\ 0 & 0 & 1 \\ yz & 0 & 0 \end{vmatrix} \\ &= 4xyz \begin{vmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{vmatrix} = -4xyz \begin{vmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 4xyz \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ &= 4xyz \end{align}
Best Answer
Firstly, show that the transformation of the points of the unit square map to the parallelogram that you show. Secondly, calculate the area of a parallelogram using some basic symmetries of the shape and show it is $|a d - b c|$. This is in fact the basic principle behind determinants, they were invented to see how the area of shapes change under a matrix transformation.
How you show the area depends on what you already know. If you know about complex numbers, think about what the imaginary part of $\bar{z_1} z_2$ represents.