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If we have a matrix $A\in M_2(\mathbb{R})$, $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then $det(A)=ad-bc$. $A$ sends the unit square to the parallelogram whose sides are the vectors $(a, c)$ and $(c, d)$. Since the area of a parallelogram is the product of its two side lengths, the area of this parallelogram is $\|(a, c)\|\|(b,d)\|=\sqrt{a^2+c^2}\sqrt{b^2+d^2}=\sqrt{(ab)^2+(ad)^2+(cb)^2+(cd)^2}$. If this is the absolute value of $det(A)$, then it should equal $\mid ad-bc \mid$. But I just don't see how it does . . .

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    $\begingroup$ The area of a parallelogram is the length times the height - not the product of the two sides - that only holds for the square. See en.wikipedia.org/wiki/Parallelogram#Area_formula $\endgroup$ – Winther Jan 2 '17 at 21:07
  • $\begingroup$ Well, also for rectangles which are not squares, but I take your point, that is indeed what was wrong with the proof. Thanks! $\endgroup$ – Rasputin Jan 2 '17 at 21:22
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The area of a parallelogram isn't the product of the two side lengths, it's the product of its base and height. If $\theta$ is the angle between adjacient sides, then the area becomes $||(a,c)||\, ||(b,d)||\sin(\theta)$.

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  • $\begingroup$ Ah yeah that makes sense. And if $\theta=90$ degrees, then WLOG we can say $b, c=0$, in which case the expression I wrote above is equal to $\sqrt{(ad)^2}=ad=ad-bc$. $\endgroup$ – Rasputin Jan 2 '17 at 21:20

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