# Why determinant of a 2 by 2 matrix is the area of a parallelogram?

Let $A=\begin{bmatrix}a & b\\ c & d\end{bmatrix}$ be a two by two matrix where the first row of $A$ is $a, b$ and the second row of $A$ is $c, d$. How could we show that $ad-bc$ is the area of a parallelogram with vertex $(0, 0),\ (a, b),\ (c, d),\ (a+b, c+d)$? Are the areas of the following parallelograms the same?

$(1)$ parallelogram with vertex $(0, 0),\ (a, b),\ (c, d),\ (a+c, b+d)$

$(2)$ parallelogram with vertex $(0, 0),\ (a, c),\ (b, d),\ (a+b, c+d)$

$(3)$ parallelogram with vertex $(0, 0),\ (a, b),\ (c, d),\ (a+d, b+c)$

$(4)$ parallelogram with vertex $(0, 0),\ (a, c),\ (b, d),\ (a+d, b+c)$

Thank you very much.

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Note $\$ Pick employed this and his area theorem to give a beautiful geometric proof of the Bezout linear representation of the GCD. –  Bill Dubuque Mar 26 '11 at 15:45
Have a look at this. –  Giuseppe Negro Mar 26 '11 at 16:25
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. –  Qiaochu Yuan Mar 26 '11 at 16:46

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:

(Appeared in Mathematics Magazine, March 1985.)

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Can you point to the actual book this was taken from? I'd like to explore this further. = ) –  Phonon Jun 6 at 0:46

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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Thanks for the edits xD –  sarella Dec 26 '13 at 23:54

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.

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+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.) –  Blue Mar 13 '12 at 3:29
@Don, good point. I will make the edit tonight. –  Tpofofn Mar 13 '12 at 10:27

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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Since the area changes by the determinant of a linear map between two regions.

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This is more of a restatement of the question than an explanation. –  Qiaochu Yuan Mar 26 '11 at 16:44
@Qiaochu: in defense, what is an 'explanation'? It could be a formal proof, a picture, a restatement in other terms, an informal proof, anything that psychologically gives us trust. Granted if you understood the above restatement, you'd presumably already understand how the determinant and area correspond. –  Mitch Mar 26 '11 at 16:57

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