## xkehaulanix 4 years ago True or False (and explain)? If the equation Ax = b (assuming A is a square matrix of size nxn) has at least one solution for each b in Rn, then the solution is unique for each b.

1. xkehaulanix

Ah, I think I figured it out. If Ax = b has one solution for every b in Rn, then the matrix A must be invertible by the Invertible Matrix Theorem. Then since A is invertible, the equation Ax = b has the unique solution $x = A ^{-1}b$

2. Zarkon

it says' at least one solution' not one solution suppose that Ax=b has more than one solution then $$Ax_1=b$$ and $$Ax_2=b$$ with $$x_1\ne x_2$$ then $A(x_1-x_2)=0$ let $$x_3=x_1-x_2$$ so $Ax_3=0$ thus the nullspace in nontrivial...thus the matrix dones not have full rank...hence the columns can't span all of $$\mathbb{R}^n$$ thus for some $$c\in \mathbb{R}^n$$ $Ax=c$ will not have a solution ... a contradiction...thus the solutions are unique.

3. xkehaulanix

Ah, that was a typo on my part. The theorem applies when Ax = b has at least one solution. But thank you for the proof!