Tuesday, May 30, 2023

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)

No comments: