In the previous section we reduced the original matrix and had an additional
0 column to address.
We multiplied the original problem times the third column and the result was zero.
This simply means the vector was in the null-space, but there's more than one point in the null-space and we're going
to figure out how to get at that here. (We're using 3 dimensions here, and so it's convenient to think of 3-space, but the method
works for n-space, but is more difficult to conceptualize.)
Our inverse (
If we think about our original equation:
A • x = b, and the usual
(A-1 • A) • x = A-1 • b, or just
x = A-1 • b, we must remember that the
(A-1 • A) was not always the identity matrix, and therefore contained more information that we'd
lose if we looked at the
x = A-1 • b only.
To get access to this additional information it's necessary to bring across the
(A-1 • A) component and
combine and reduce this to:
x = A-1 • b + (I - A-1 • A) • z
z vector is completely arbitrary.
Multiply the "inverse" times the original to get
(A-1 • A).
Subtract this from the identity matrix
I to get
(I - A-1 • A)
The next part requires an arbitrary vector
It's interesting to note that the
A matrix times this
(I - A-1 • A) matrix
And therefore the product
(I - A-1 • A) • z will be in the null-space
of the original
Let's multiply the original A matrix times this vector:
|1||2||2||-3||-4s||-8s +2s +6s||0|
The general solution (
xgs) is a combination of
the solution using
A-1 • b and the arbitrariness contained in the
(I - A-1 • A) • z.
Using this value for the
We can multiply the terms to get the full solution for
|1||2||3||a||-4s||2a +3b -4s|
|3||1||2||-2s||a +2b -2s|
If we multiply the original
A matrix times this
we get the
b vector returned.
|1||2||2||-3||2a +3b -4s||4a +6b -8s +2s -3a -6b +6s||a|
|2||-1||0||2||•||s||=||-2a -3b +4s +2a +4b -4s||=||b|
|3||a +2b -2s|
We put no restrictions on the value of the
z vector, and it could be anything.
As it turns out (in this example) only the second value (the
s) is used.
Any value for
s is legal because it drops out of the problem.
In this 3-dimensional problem, the "answer" is a line. If we had kept all 3 equations, the "answer" would've been a point. If we had removed two equations from the original set, instead of just the one, the "answer" would've been a plane.
Onward to the demonstration page.