Relation between Ax=0 and Ax=b 
I'm not sure how to solve this. My teacher told me that if $A\vec{x}=0$ has one unique solution (the trivial) then $A\vec{x}=\vec{b}$ has only one unique solution. But I don't know how to prove this.
 A: I will use boldface for vectors.
We have the following:

Theorem. Let $A$ be an $m\times n$ matrix, and $\mathbf{b}$ an $m\times 1$ vector. If $\mathbf{y}_0$ is a solution to $A\mathbf{x}=\mathbf{b}$, then every solution of $A\mathbf{x}=\mathbf{b}$ is of the form $\mathbf{y}_0+\mathbf{s}$, where $\mathbf{s}$ is a solution to $A\mathbf{x}=\mathbf{0}$, and every such vector is a solution to $A\mathbf{x}=\mathbf{b}$. In other words, the complete list of solutions to $A\mathbf{x}=\mathbf{b}$ is given by finding a particular solution $\mathbf{y}_0$ to $A\mathbf{x}=\mathbf{b}$, and adding the solutions to $A\mathbf{x}=\mathbf{0}$.

(This is one reason homogeneous systems play such an important role in the theory of systems of linear equations).
Proof. First, if $\mathbf{y}_0$ is a solution to $A\mathbf{x}=\mathbf{b}$, and $\mathbf{s}$ is a solution to $A\mathbf{x}=\mathbf{0}$, then
$$A(\mathbf{y}_0+\mathbf{s}) = A\mathbf{y}_0 + A\mathbf{s} = \mathbf{b}+\mathbf{0}=\mathbf{b}.$$
So all such vectors are also solutions. Conversely, if $\mathbf{y}_1$ is also a solution to $A\mathbf{x}=\mathbf{b}$, then $\mathbf{s}=\mathbf{y}_1-\mathbf{y}_0$ is a solution to $A\mathbf{x}=\mathbf{0}$, since
$$A\mathbf{s} = A(\mathbf{y}_1-\mathbf{y}_0) = A\mathbf{y}_1 - A\mathbf{y}_0 = \mathbf{b}-\mathbf{b}=\mathbf{0}.$$
And $\mathbf{y}_1 = \mathbf{y}_0+\mathbf{s}$, so $\mathbf{y}_1$ is of the desired form. $\Box$
This tells you that if there is at least one solution to $A\mathbf{x}=\mathbf{b}$, then there are exactly as many solutions to $A\mathbf{x}=\mathbf{b}$ as there are to $A\mathbf{x}=\mathbf{0}$. It does not tell you that there definitely is a solution to $A\mathbf{x}=\mathbf{b}$, though. It could be that $A\mathbf{x}=\mathbf{b}$ is inconsistent, so that it has no solutions.
When you know that $A\mathbf{x}=\mathbf{0}$ has exactly one solution, you actually can say more, though you may not have the tools to justify it right now. It turns out that in this situation, the matrix $A$ must have $m\geq n$ (if $m\leq n$, more columns than rows, then $A\mathbf{x}=\mathbf{0}$ will always have more than one solution). And if $m=n$, then one can prove that the system $A\mathbf{x}=\mathbf{b}$ always has at least (and hence exactly) one solution.
And in fact you would know that the solution to $A\mathbf{x}=\mathbf{0}$ has to be $\mathbf{s}=\mathbf{0}$. If you did not realize this, note that if $\mathbf{s}$ is a solution to $A\mathbf{x}=\mathbf{0}$, then using the theorem with $\mathbf{b}=\mathbf{0}$ tells you that $\mathbf{s}+\mathbf{s}=2\mathbf{s}$ is also a solution. So the uniqueness would tell you that $\mathbf{s}=2\mathbf{s}$, and that can only hold if $\mathbf{s}=\mathbf{0}$ (assuming you are working over a field of characteristic $\neq2$... and if you don't know what that means, you are).
A: Hint
Case of a square matrix $A$
By the rank nullity theorem $A$ injective if and only if it’s kernel is reduced to the zero vector if and only if $A$ is surjective.
Case where $Ax=b$ has no solution
Take
$$A=\begin{pmatrix}
1\\
1\end{pmatrix} \text{ and }b=\begin{pmatrix}
1\\
2\end{pmatrix}$$
A: A useful result on linear systems is the following:

Let $K$ be a field,  $A$ the matrix of a linear map from $K^n$ to $K^p$, $\:b\in K^p$.
The linear equation $Ax=b$ ($x\in K^n$) has a solution if & only if the matrix $A$ and the augmented matrix $[A|b]$ have the same rank. Furthermore, this common rank is the codimension of the affine subspace of solutions.

A: 
My teacher told me that if $A\vec x =\vec0$ has one unique solution (the trivial) then $A\vec x =\vec b$  has only one unique solution. But I don't know how to prove this.

Simply, as hinted in comments by Arturo Magidin,
if $A\vec x=\vec b$ and $A\vec y=\vec b$, then $A\vec x - A\vec y=0$, so by linearity of $A$ then $A(\vec x-\vec y)=\vec0$,
so $\vec x-\vec y=\vec 0$, so $\vec x=\vec y$, so if there is a solution to $A\vec x=\vec b$, then it is unique.
