Sunday, January 8, 2012

Linear Algebra Solutions

We had the problems from the previous blog, given again with their respective solutions:

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

Solution:

Let A = (2, 5), B = (-1, 4) and C = (1, 2)

Then:

B - A = [(-1 -2), (4 - 5)] = (-3, -1)

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

Yielding the matrix A =

(-3......-1)
(2.......-2)

And Det M = (-3 x -2) - [(-1 ) x 2] = 6 - (-2) = 6 + 2 = 8

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

Solution:

Let: A = (1, ), B = (1, 0) and C = (2, 3)

Then:

B - A = [(1 - 1), (0 - 1)] = (0, -1)

C - B = [(2 - 1), (3 - 0)] = (1, 3)

Then the matrix is M =

(0.....-1)
(1......3)

So: Det M = (0 x 3) - [(-1) x 1] = 0 - (-1) = 1

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.

Solution:

This is easily done if one bears in mind the 3 x 3 form will be resolved into separate 2 x 2 matrices each with their own value and with alternting +/- signs.

Thus:

Det M = u1[v2 w3 - v3 w2] - u2 [v1 w3 - v3 w1] + u3 [v1 w2 - v2 w1]

This form will be found useful when we come to solid geometry applications!