Solving Simple Problems In Linear Algebra (3)

 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 Problem:

Find the area of the parallelogram in Graph 2:

Solution:

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 absolute. value must be taken because the determinant is negative.

Suggested Problem:

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

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