If A is singular; solution space of Ax=b

[Bipolarity]

Suppose it is known that A is singular. Then the system Ax=0 has infinitely many solutions by the Invertible Matrix theorem.

I am curious about the system Ax=b, for any column vector b. In general, i.e. for all vectors b, will this system be inconsistent, or will it have infinitely many solutions?

Surely there exists a vector b for which this system is inconsistent. For otherwise, if it were consistent for every vector b, it would necessarily be invertible (again by the IM theorem), but by assumption it is not.

So here are a few questions I have begun to think about, but not fully able to explain:

Given that A is a singular square matrix:

1) Does there necessarily exist a vector b for which Ax=b has infinitely many solutions?
2) Does there necessarily exist a vector b for which Ax=b has a unique solution?
3) If b is a certain column vector, can one determine simply by inspection whether Ax=b is inconsistent, has a unique solution, or has infinitely many solutions?

I would appreciation an answer to these questions. I prefer to do the "proofs" myself, so don't give anything away. Thanks much!

BiP

 

[Michael Redei]

1) Yes.
2) No.
3) Yes.

 

[AlephZero]

3):
"simply" - yes
"by inspection" - try it on a 100 x 100 matrix and see if you still think the answer is still "yes".

 

[Erland]

Think of what happends when Gauss-Jordan elimination is applied to the augmented matrix of the system, and compare with the coefficient matrix. What happends with pivot positions, free variables, etc.?

 

[Bipolarity]

Hey all, I have been able to figure out why the answer to the second question is "No".
Also, I have been able to figure out how you can determine whether the solution set is null or whether it is infinite once the matrix A has been reduced to row-echelon form.

But is it possible to determine the solution space of Ax=b without applying Gaussian elimination?

BiP