# Solutions of Ax = b (for singular A)

1. Aug 16, 2010

### monea83

A system of linear equations, Ax = b (with A a square matrix), has a unique solution iff $$det(A) \ne 0$$. If b = 0, the system is homogeneous and can be solved using SVD (which gives the null space of A).

Now, how can the solution set be characterized for singular A and $$b \ne 0$$? If a single solution $$s$$ is known, $$s + v$$ is also a solution for all $$v$$ from the null space of A... but how is it possible to determine whether such an $$s$$ exists at all, and if so, find it?

2. Aug 16, 2010

### Eynstone

Gaussian elimination can be employed to determine whether a solution exists.

3. Aug 17, 2010

### HallsofIvy

Staff Emeritus
A singular matrix, A, will map Rn into a proper subspace of Rn. There will exist x such that Ax= b if and only if b is in that subspace. What that subspace is, and whether or not b is in it, can be determined by row reduction of A augmented by adding b as a last column. Any row that reduces to "all 0s" in the first n columns (and, since A is singuar, there will be at least one such) must also have a "0" in the last column in order that Ax= b have a solution.

Of course, if a solution exists, it is not unique. Adding any non-zero vector in the kernel of A (which is non-trivial for A singular) to a solution will give another solution.

Last edited: Aug 18, 2010
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