Solving Ax = b completely: one solution plus the nullspace
Slide 3 found every input that does nothing. Now the right side is a real target b. Think of walking to the bakery. You know one route that gets you there. Before you leave, add any stroll that brings you back to your own door, and you still end up at the bakery. Those strolls are the nullspace vectors. So once you have one solution, the particular solution x_p, every solution is x_p plus a nullspace vector. That is the complete solution. And if the bakery can't be reached at all, because b is not in the column space, there is no solution.
Lesson 4 · slide 4Complete solution- If Ax_p = b, then every solution is x = x_p + x_null with x_null in N(A).
- Ax = b is solvable exactly when b is in the column space C(A).
- To find x_p, eliminate on [A | b], set all free variables to 0 and read off the pivot variables.
- The four cases on this slide show how r, m and n decide what can happen: 0, 1 or infinitely many solutions.
One route plus a loop
Take A = 1224 from slide 2 and b = 48. One solution is easy to see: x_p = 40, four of column 1. Now add the nullspace vector −21:
Add it twice and you get 02, another solution: two of column 2.
Why every solution has this form
- Adding a nullspace vector keeps a solution: A(x_p + x_null) = Ax_p + Ax_null = b + 0 = b.
- Any two solutions differ by a nullspace vector: if Ax = b and Ay = b, then A(x − y) = b − b = 0. Check: 02 − 40 = −42 = 2·−21 ✓
So the solutions form a line parallel to N(A), shifted to pass through x_p. In the row picture from lesson 2 you can see that line directly:
The nullspace is the parallel line x + 2y = 0 through the origin. Slide the nullspace over by x_p and you get the solution line.
How to find x_p
Write b as an extra column after A, with a bar between them: [A | b]. This is the augmented matrix from lesson 2. Row operations act on whole rows, so they change the b column too.
For our A and b = 48: row 2 − 2·row 1 turns 24|8 into 00|0, leaving 12|4. The free variable is x₂. Set x₂ = 0, then x₁ = 4, so x_p = 40. With N(A) = all c·−21:
When there is no solution
Change b to 412. Row 2 − 2·row 1 gives 0 = 12 − 2·4 = 4. Impossible. b = 412 is not a multiple of 12, so it is off the column space line.
Keep reading with the full course
Finish lesson 4 and you will be able to: write down every solution of Ax = b, not just one.
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Examples
Example 1 · One equation, two unknowns
x + y = 2.
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