DETERMINANTS AS AREAS AND VOLUMES
MATH 3144, FALL, 2015
Hx,yL
H0,yL
H0,0L
Hx,0L
Corners = {(0, 0), (x, 0), (x, y), (0, y)}; Area = det
x 0
0 y
Figure 1: Rectangle
Hax,yL
H0,0L
Hx+ax,yL
Hx,0L
Corners = {(0, 0), (x, 0), (x + ax, y), (ax, y)}; Area = det
x ax
0 y
Figure 2: Rectangle skewed in x-direction by a
a b
Let A =
be a matrix over R. We define the determinant of A by detA := ad − bc.
c d
The purpose of this discussion is to prove that the area of a parallelogram whose adjacent sides are
the vectors ~u and ~v is equal to |det[~u ~v ]|, where the vectors are written as column vectors.
It is a routine exercise to show that
Hx,y+byL
H0,yL
Hx,byL
H0,0L
Corners = {(0, 0), (x, by), (x, y + by), (0, y)}; Area = det
x 0
by y
Figure 3: Rectangle skewed in y-direction by b
1. adding a multiple of one row (column) of a matrix A to another preserves det(A);
2. switching two rows (columns) of A changes the sign of det(A); and
3. multiplying a row (column) of A by a nonzero scalar k multiplies det(A) by k
We begin with the rectangle in Figure 1 with sides of lengths x and y. It is clear that
x 0
Area(rectangle) = xy = xy − 0 · 0 = det
0 y
We use the elementary row operation of adding a times the first column to the second column, so
that
A=
x 0
0 y
1 a
0 1
=
x ax
0 y
,
det(A) = xy − 0 · ax = xy, and the area is preserved.
We use the elementary column operation of adding b times the second column to the first column,
so that
A=
x 0
0 y
1 0
b 1
=
x 0
by y
,
det(A) = xy − by · 0 = xy, and the area is preserved.
Now, let’s combine these operations and see what happens.
We first use the elementary column operation of adding a times the first column to the second
column, then use the elementary column operation of adding b times the second column to the first
column, so that
Hx+ax+abx,y+byL
Hax,yL
Hx+abx,byL
H0,0L
Corners = {(0, 0), (ax, y), (x + ax + abx, y + by), (x + abx, y + by)}; Area = det
x + abx ax
by
y
Figure 4: Rectangle skewed in x-direction by a, then in (x, by)-direction by b
A=
x 0
0 y
1 a
0 1
1 0
b y
=
x + abx ax
by
y
,
det(A) = (x + abx)y − byax = xy, and the area is preserved.
Figure 5 shows the rectangle and all three parallelograms together, so that you can see the end
result of first skewing the rectangle by a in the x direction, then skewing the resulting parallelogram
by b in the (x, by) direction by b yields the parallelogram with dashed lines. And they all have the
same area.
Finally, it is easy to see that interchanging two rows or two columns changes the sign of the
determinant, and multiplying a row or column by a scale factor k multiplies the area by k. We
conclude that our definition of the area of a parallelogram – namely, the absolute value of the
matrix whose columns are the vectors that define two adjacent sides of the figure – is correct.
Hx,y+byL
H0,yL
Hax,yL
Hx+ax+abx,y+byL
Hx,yL
Hx+ax,yL
Hx,byL
Hx+abx,byL
H0,0L
Hx,0L
Figure 5: Rectangle and three parallelograms on the same plot
Question 1. Does the figure you get depend upon the order of the skewing of the figures? That
is, does skewing by a first and then by b give the same parallelogram as skewing by b first and then
by a?
Question 2. Suppose you skew the rectangle by the same value in both directions: that is, suppose
b = a. Does the order in which you skew matter? Will you get the same parallelogram in either
case?
Question 3. How can you extend these arguments to prove that the volume of a 3-dimensional
parallelepiped generated by the vectors {~u, ~v , w,
~ } in R3 is equal to |det[~u ~v ] w|?
~ With what sort
of parallelepiped do you begin?
Question 4. Answer Question 3 for an n-dimensional parallelepiped generated by n vectors in Rn .