1. ## Invertible matrix proof

Hello,

Problem statement: show that if Ax = b has a unique solution, then A is invertible.

I have a hint given to me also: suppose Ax = b has a solution and consider that Ay = 0. Deduce y = 0.

I'm not sure how to incorporate this hint into the proof. If someone can help me out, that would be really great. I haven't done mathematical proofs in years while I was working, and now I'm back at school. Any general words of advice on how to start/set up a proof will be greatly appreciated, too! Thanks!

2. If A is invertible, then the system, Ax=b has only one solution.

Namely, that solution is $\displaystyle x=A^{-1}b$.

Since $\displaystyle A(A^{-1}b)=b$, then it follows that

$\displaystyle x=A^{-1}b$ is a solution to Ax=b.

Assume that $\displaystyle x_{0}$ is an arbitrary solution and then show that

$\displaystyle x_{0}$ must be a solution to $\displaystyle A^{-1}b$.

If $\displaystyle x_{0}$ is any solution, then $\displaystyle Ax_{0}=b$.

Multiply both sides by $\displaystyle A^{-1}$, and we get:

$\displaystyle x_{0}=A^{-1}b$

3. Hey Galactus, I need to assume that there is a unique solution first and prove that A is invertible. I think in your proof you assume A is invertible and showed that there is one solution. Could you give it another try and solve the other way using the hint, "suppose Ax = b has a solution and consider that Ay = 0. Deduce y = 0?"

Or.. I might not be seeing something here. Thanks regardless.

4. While it's not the only way to prove it, I like to prove by contradiction. So assume A is not invertible. Then that must mean that A is not 1:1, which means that its kernel contains a non-zero element.

This is where your hint comes in. If Ay=0, what is A(x+y)?