Show that if A has a right-inverse, then $Ax = b$ has at least one solution for every choice of b in $R^n$ Show that if A has a right-inverse, then $Ax = b$ has at least one solution for every choice of b in $R^n$.
 A: Suppose $A$ has a right inverse, so there is a $B$ such that $AB=I$. Then $ABb=Ib=b$ for any $b\in{\mathbb R}^n$. Letting $x=Bb$ gives us the result.
I guess it is worth mentioning that we really didn't use any linear algebra. A similar argument gives that if $A$ has a left inverse, then the null space of $A$ is trivial (${0}$). In fact, these two facts hold for arbitrary functions, not just linear maps (in the second case, the general statement is that $A$ is injective). 
A: HINT $\ $ Right-invertible maps $\rm\:f\:$ are surjective since $\rm\ f\: (f^{-1}(x)) = x\:.\ $ The converse holds true too - it is equivalent to the axiom of choice.
A: Let $A \in \mathbb{R}^{m \times n}$. $A$ has a right-inverse if $A$ is full rank and is a square or fat matrix ($m \leq n$). Let $A^{\dagger} \in \mathbb{R}^{n \times m}$ be the right inverse of $A$ i.e. $A \times A^{\dagger} = I_{m \times m}$.
Take any vector $b \in \mathbb{R}^{m \times 1}$. The claim is $x = A^{\dagger} b$ is always a solution to the system $Ax = b$. 
This can be easily seen since $$Ax = A \times (A^{\dagger} b) = (A \times A^{\dagger})b = Ib =b$$ where the intermediate step follows from associativity of matrix multiplication.
Note that if $A$ is a strictly fat full rank matrix, this is not the unique solution since $x = A^{\dagger}b + z$ where $z \in \text{Nullspace}(A)$ is also solution. And since $A$ is a strictly fat matrix, we are assured to have a non-trivial Nullspace.
Also, for a strictly fat full rank matrix, there will be more than one right inverse. So in fact $A^{\dagger}$ itself is not unique.
