# Elemetary matrices proof problem PLS help

1. Feb 13, 2008

### jmcasall

If A is an n x n matrix and Ax = 0 has many solutions, then

which ones are true??

A. Ax = b has either no solutions or many solutions
B. Ax = b has a unique solution for some b
C. Ax = b has many solutions for any b
D. Ax = b has no solutions for some b

or are all of the above false??

thanks

2. Feb 13, 2008

### Vid

Think about what it means for Ax = 0 to have many solutions. If the 0's in the augmented matrix of linear equations were changed to some constants, would that change how A reduces?

3. Feb 14, 2008

### HallsofIvy

Staff Emeritus
Unfortunately, I don't know what you have to work with. If Ax= 0 has more than one solution, then the kernel of A is non-trivial and has non-zero dimension. That, in turn, means that the image of A is not all of Rn.

Note also that is Ax= b and v is in the kernel of A, A(x+ v)= Ax+ Av= b+ 0= b.

4. Feb 15, 2008

### jmcasall

well this is what i kinda know

Ax=0
if you solve for x then x=0,
which will make A invertible
So, what does that say about Ax=B

5. Feb 15, 2008

### HallsofIvy

Staff Emeritus
If x= 0 is the only solution to Ax= 0, then A is invertible. If A has an inverse, how would you solve Ax= B?

However, that is completely irrelevant to your question since the question specifically says that A has MORE than one solution.

I can only repeat: if v is any vector such that Av= 0, and x is a solution to Ax= b, then A(x+ v)= Ax+ Av= Ax+ 0= Ax= b. If there are many solutions to Ax= 0, what does that tell you about the number of solutions to Ax=b?