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.
Take as an example finding the area of a parallelogram such as shown below.
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
Ex. 2: Find the area of the parallelogram 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 absolute. value must be taken because the determinant is negative.
Suggested Problems:
1) Find the area of a parallelogram such that 3 of its corners are given by the points:
(1,1), (2, -1) and (4, 6)
2) Consider the parallelopipped below spanned by the vectors u, v and w in 3-space.
Find the volume of this solid given that:
u = (1, 1,
3)
v = (1, 2,
-1)
w = (1, 4,
1)
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