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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?

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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.

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  • $\begingroup$ I honestly did not understand what you meant by: "Firstly, show that the transformation of the points of the unit square map to the parallelogram that you show." $\endgroup$ – TheNotMe Mar 21 '13 at 16:33
  • $\begingroup$ Oh. there are 4 points on the unit square (0,0), (0,1), (1,0), (1,1) Calculate what happens when the matrix multiplies each of these vectors. $\endgroup$ – muzzlator Mar 21 '13 at 16:38
  • $\begingroup$ tube.geogebra.org/m/40980 and tube.geogebra.org/m/615071 are somewhat useful illustrations. See proofwiki.org/wiki/… for the complex numbers analogy. $\endgroup$ – Nickolay Jun 12 '15 at 22:01
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Try finding the area of the parallelogram via Heron's formula and then show that it is equal to the determinant of the corresponding matrix.

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