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.

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    $\begingroup$ Note $\ $ Pick employed this and his area theorem to give a beautiful geometric proof of the Bezout linear representation of the GCD. $\endgroup$ – Bill Dubuque Mar 26 '11 at 15:45
  • $\begingroup$ Have a look at this. $\endgroup$ – Giuseppe Negro Mar 26 '11 at 16:25
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    $\begingroup$ See if your question is answered by the discussion here: math.stackexchange.com/questions/668/… . A short answer is that this should be taken (properly modified to take orientation into account) as the definition of the determinant. $\endgroup$ – Qiaochu Yuan Mar 26 '11 at 16:46
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    $\begingroup$ n.b. as pointed out by @martin-sleziak in his answer below, (3) and (4) are not (generally) parallelograms $\endgroup$ – JobJob Dec 25 '15 at 11:42
  • $\begingroup$ Going back to the determinant view of this, it might be useful for people to note that 1 and 2 can be seen to have the same area from the fact that det(A)=det(transpose(A)) $\endgroup$ – JasoonS Jun 15 '16 at 12:44

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)

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Spend a little time with this figure due to Solomon W. Golomb and enlightenment is not far off:

enter image description here

(Appeared in Mathematics Magazine, March 1985.)

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  • $\begingroup$ Can you point to the actual book this was taken from? I'd like to explore this further. = ) $\endgroup$ – Phonon Jun 6 '14 at 0:46
  • $\begingroup$ Image link is broken for me. Here's a re-paste: i.stack.imgur.com/gCaz3.png $\endgroup$ – Laser Aug 16 '15 at 5:52

enter image description here

I know I'm extremely late with my answer, but there's a pretty straightforward geometrical approach to explaining it. I'm surprised no one has mentioned it. It does have a shortcoming though - it does not explain why area flips the sign, because there's no such thing as negative area in geometry, just like you can't have a negative amount of apples(unless you are economics major).

It's basically:

  Parallelogram = Rectangle - Extra Stuff.

If you simplify $(c+a)(b+d)-2ad-cd-ab$ you will get $ad-bc$.

Also interesting to note that if you swap vectors places then you get a negative(opposite of what $ad-bc$ would produce) area, which is basically:

  -Parallelogram = Rectangle - (2*Rectangle - Extra Stuff)

Or more concretely:

$(c+a)(b+d) - [2*(c+a)(b+d) - (2ad+cd+ab)]$

Also it's $bc-ad$, when simplified.

The sad thing is that there's no good geometrical reason why the sign flips, you will have to turn to linear algebra to understand that.

Like others had noted, determinant is the scale factor of linear transformation, so a negative scale factor indicates a reflection.

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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.

Determinant 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.

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  • 1
    $\begingroup$ +1. While it's clear that $|A|\sin\alpha=|R_{90^\circ}A|\cos\beta$, it may be visibly clearER in your diagram that $|\text{altitude of parallelogram}|=|B|\sin\alpha=|B|\cos\beta$ ... which ---since $|A|=|R_{90^\circ}A|$--- works just as well in your argument. (Drawing the altitude from the tip of $B$ onto $A$ would help drive this home.) $\endgroup$ – Blue Mar 13 '12 at 3:29
  • $\begingroup$ @Don, good point. I will make the edit tonight. $\endgroup$ – Tpofofn Mar 13 '12 at 10:27
  • $\begingroup$ I posed a video on youtube explaining this answer. youtu.be/m0DrWT18gBA $\endgroup$ – Tpofofn Aug 4 '17 at 2:23

I keep forgetting the tricks to prove this so I found the only way to remember it is to stick to basic principles. We can deform the parallelogram to get a square like so. enter image description here

Now we only need to find the side lengths of this rectangle. The height is easy, it's just d. To get the base length, we find:

enter image description here

and so with a little algebra we find that the base length is $a-b\times\frac{c}{d}$.

Hence the area is just $$ d(a-\frac{bc}{d}) = ad - bc$$

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  • $\begingroup$ Very nice answer! Thank you. $\endgroup$ – Mahathi Vempati Jun 6 '17 at 4:46

Here is an animated gif I made to derive the area of a parallelogram.

wordless proof about area of parallelogram

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If you compute the cross product of (a,b,0) and (c,d,0), then you get (in the third coordinate) ad-bc. This is, up to the sign, the area of the parallelogram.

BTW I think that (3) and (4) are not parallelograms, are they?

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  • $\begingroup$ Thank you. (3) and (4) are not parallelograms. $\endgroup$ – user Mar 26 '11 at 16:56

Also, if the coordinates of any shape are transformed by a matrix, the area will be changed by a scale factor equal to the determinant.

Since the determinant is the scale factor when the unit square is transformed to a parallelogram, it will be the scale factor when any parallelogram with the origin as a vertex is transformed to any other parallelogram because the inverse matrix will transform a parallelogram back into a square and has reciprocal determinant. If there is no inverse, the determinant is 0 and the transformed shape has no area.

Any triangle with the origin as a vertex can be drawn as half of a parallelgram including the origin. Any triangle not including the origin is the area of a triangle containing the origin minus two triangles inside not containing the origin. The area of any shape can be split into triangles, although an infinite number will be required if it has curved sides.

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Wrote this for a linear algebra class of mine. The argument is predicated on using shears.

Assume you have two vectors, (a, ay) and (xd, xyd+d). Weird choice and abundance of variables to be explained in a moment. I can obviously find the determinant of this, which is ad (do it).

So if I want to prove that the determinant is an area, I need to show that these weirdo vectors share an area with (a,0) and (0,d), which also has the determinant ad. Well, it turns out what I can do is shear the matrix with (a,0) and (0,d) as columns, since a shear does not alter the area at all (show this geometrically).

Take vector (a,0) and (0,d) and apply shear matrix (1,x,0,1), followed by (1,0,y,1), which gives you the original weirdo vectors (a,ay) and (xd, xyd+d). Since the shears do not change area, and we know the area of the rectangle formed by (a,0) and (0,d), the area of two arbitrary vectors may be expressed by its determinant, which we have shown to be identical to the determinant of rectangular matrix (a,0,0,d). QED.

You can extend this argument to 3D, etc. by applying shears in any arbitrary direction. Therefore, a determinant will provide you with a volume in any R^n.

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