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.)
| 1 | 2 | 3 | 1 | 1 | |||||||||
| 1 | 2 | 2 | -3 | -4 | 0 | ||||||||
| 2 | -1 | 0 | 2 | • | 1 | = | 0 | ||||||
| 3 | -2 |
Our inverse (
A-1
) was:
| 1 | 2 | |||
| 1 | 2 | 3 | ||
| 2 | 0 | 0 | ||
| 3 | 1 | 2 |
If we think about our original equation:
A • x = b
, and the usual solution:
(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
Here the
z
vector is completely arbitrary.
Multiply the "inverse" times the original to get
(A-1 • A)
.
| 1 | 2 | 1 | 2 | 3 | 1 | 2 | 3 | |||||||||
| 1 | 2 | 3 | 2 | 2 | -3 | 1 | 4 | 0 | ||||||||
| 2 | 0 | 0 | • | -1 | 0 | 2 | = | 0 | 0 | 0 | ||||||
| 3 | 1 | 2 | 0 | 2 | 1 |
Subtract this from the identity matrix
I
to get
(I - A-1 • A)
| 1 | 2 | 3 | |||
| 1 | 0 | -4 | 0 | ||
| 2 | 0 | 0 | 0 | ||
| 3 | 0 | -2 | 0 |
The next part requires an arbitrary vector
z
.
| 1 | |||
| 1 | r | ||
| 2 | s | ||
| 3 | t |
It's interesting to note that the
A
matrix times this
(I - A-1 • A)
matrix is zero.
| 1 | 2 | 3 | 1 | 2 | 3 | 1 | 2 | 3 | |||||||||
| 1 | 2 | 2 | -3 | 0 | -4 | 0 | 0 | 0 | 0 | ||||||||
| 2 | -1 | 0 | 2 | • | 0 | 1 | 0 | = | 0 | 0 | 0 | ||||||
| 3 | 0 | -2 | 0 |
And therefore the product
(I - A-1 • A) • z
will be in the null-space of the original
A
matrix.
| 1 | 2 | 3 | 1 | 1 | |||||||||
| 1 | 0 | -4 | 0 | r | -4s | ||||||||
| 2 | 0 | 1 | 0 | • | s | = | s | ||||||
| 3 | 0 | -2 | 0 | t | -2s |
Let's multiply the original A matrix times this vector:
| 1 | 2 | 3 | 1 | 1 | 1 | ||||||||||||
| 1 | 2 | 2 | -3 | -4s | -8s +2s +6s | 0 | |||||||||||
| 2 | -1 | 0 | 2 | • | s | = | 4s -4s | = | 0 | ||||||||
| 3 | -2s |
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
b
vector:
| 1 | |||
| 1 | a | ||
| 2 | b |
We can multiply the terms to get the full solution for
xgs
:
| 1 | 2 | 1 | 1 | 1 | |||||||||||||
| 1 | 2 | 3 | a | -4s | 2a +3b -4s | ||||||||||||
| 2 | 0 | 0 | • | b | + | s | = | s | |||||||||
| 3 | 1 | 2 | -2s | a +2b -2s |
If we multiply the original
A
matrix times this
xgs
, we get the
b
vector returned.
| 1 | 2 | 3 | 1 | 1 | 1 | ||||||||||||
| 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.