Linear Algebra Applied to Geometry

One of the more interesting applications of linear algebra is to plane and solid geometry. Most of these applications entail computing the determinant D of a matrix. To recap, given a 2 x 2 matrix say:

(a11.......a12)
(a21......a22)

The determinant D is computed:

D = (a11 x a22) - [(a12) x (a21)]

In geometric applications, one will take the absolute value [x] of the result.

Let's take as an example finding the area of a parallelogram such as shown in Graph 1. This figure is spanned by the vectors (2, 1) and (-4, 5) as shown. The area will then be the determinant D of the matrix formed. This matrix will be such that:

a11 = 2, a12 = 1, a21 = -4 and a22 = 5

Then: D = (a11 x a22) - [(a12) x (a21)] = (2 x 5) - [(1 x (-4)] = 10 + 4 = 14 sq. units

Example (2):

Find the area of the parallelogram as depicted in Graph 2.

In this case, the spanning vectors are (3,2) and (-2, -3), so we have for the elements of the matrix:

a11 = 3, a12 = 2, a21 = -2 and a22 = -3.

Then: D = (a11 x a22) - [(a12) x (a21)] = [(3 x (-3)] - [2 x (-2)] = -9 + 4 = [-5] = 5 sq. units

Note that the abs. value must be taken because the determinant is negative in Ex. (2).


Example (3):

Find the area of a parallelogram such that 3 of its corners are given by the points:

(1,1), (2, -1) and (4, 6)

We obtain the spanning vectors here by taking differences. So, let: A = (1, 1), B = (2, -1) and C = (4, 6). Then:

B - A = [(2 - 1), (-1 -1)] = (1, -2)

C - B = [(4 - 2), (6 - (-1)] = (2, 7)

Then: the matrix is: M =

(1....-2)
(2.....7)

So Det (M) = (1 x 7) - [(-2) x 2]= 7 - (-4) = 7 + 4 = 11 sq. units.

Additional Problems:

1) Find the area of a parallelogram for which three corners have the coordinates: (2, 5), (-1, 4) and (1, 2).

2) The same as (1) except the three corners have coordinates: (1,1), (1, 0) and (2, 3)

3) How would you find the determinant of a 3 x 3 matrix, say:

(u1....u2......u3)
(v1....v2.....v3)
(w1....w2....w3)

Give the value of D in terms of the given elements.