Proposition If linear system $\bf Ax = O$ ($\bf A$ is $n$ by $n$ square matrix) has a unique solution, then for arbitrary vector $\bf b$, $\bf Ax = b$ will have a unique solution.
My attempt:
Lemma 1. If $\bf Ax = O$ has a unique solution, then $\bf x = O$
Suppose $\bf Ax = O$ and $\bf x = b$ where $\bf b ≠ O$. Because $\bf AO = O$, it follows that linear system $\bf Ax = O$ has at least two possible solutions, which a contradiction.
Lemma 2. $\bf Ax = O$ has a unique solution $\bf x = O$ iff $\bf A$ is invertible
The $(\rightarrow$) side is a bit lengthy, hence for now we just assume that we've already proved lemma above.
Now to our main proposition:
Suppose $\bf Ax = O$ has a unique solution. By lemma 1 and 2, we conclude that $\bf A$ is invertible.
Consider linear system $\bf Ax = b$, where $\bf b$ is the arbitrary vector:
$$\tag 1\bf Ax = b \implies$$
$$\tag 2\bf A^{-1}x = A^{-1}b \implies$$
$$\tag 3\bf x = A^{-1}b$$
Therefore, for arbitrary vector $\bf b$, lin. system $\bf Ax = b$ will have a unique solution, namely $\bf A^{-1} b$. $\Box$
I believe that for the trivial proposition like the one above, more concise alternatives are available, but still, is my attempt correct?
Best Answer
Actually, you prove more than what is asked: you prove that if the homogeneous linear system has a unique solution, its matrix is invertible and you give the solution of the non-homogeneous equation. In my opinion, proving by contradiction lemma 1 is a bit too much.
Here is how I would prove what is asked in a short way:
Suppose we have (possibly equal) solutions: $$\mathbf{Ax}=\mathbf{b},\qquad \mathbf{Ax'}=\mathbf{b}.$$ Then, by linearity, we deduce $\;\mathbf{A}(\mathbf x-\mathbf x')=\mathbf 0$, and by hypothesis, $\mathbf{x}-\mathbf{x'}=0$, i.e. $\;\mathbf{x}=\mathbf{x'}$.