$Ax = b$ has a unique solution then $Ax =c$ also has a unique solution? Let $n \in \mathbb{N},A \in M_{n\times n}(\mathbb{R}),b,c \in \mathbb{R}^n$. If $Ax = b$ has a unique solution, then $Ax = c$ also has a unique solution?
I can show that if $Ax = c$ has a solution then it must has a unique solution:
Assume the contrary, that $Ax_1 = Ax_2 = c$ with $x_1 \not = x_2.$
$\ $
Now we have $A(x_1 - x_2) = 0$. Suppose $Ax = b$, then $A(x + x_1- x_2) = b $, since $x$ is unique, we get a contradiction.
But is it possible that $Ax = c$ has no solutions? 
Hope someone can help me, thanks!
 A: $Ax = b$ has unique solution that means $A$ has full rank. Hence $Ax = c$ has unique solution. (Here $A$ is square also)
A: I like to think this kind of problems in terms of transformations. Remember that a matrix $A$ can be thought as encoding a transformation which maps a given vector $x$ to another, $y=Ax$. Both $y$ and $x$ are in the same vector space if A is square. 
For a system $Ax=b$, having a unique solution in this context means that there is only one vector $x$ which multiplication by A transforms to $b$. 
The rank of $A$ is the dimension of its image. If its not singular, it is full rank, meaning that, as a transformation, it maps the whole space back to the whole space.
This means that for every vector in the image of the transformation, ($b$ or $c$ or any other, since it's the whole space if $A$ is not singular, as we said) you can find one and only one vector $x$ which $A$ maps to.  
I have a small, prototype app I use to show this to students. Disclaimer: chart legend in spanish. 
A: My way of looking at it:
If, for some $b$, the equation
$Ax = b \tag{1}$
had a unique solution, then
$\ker A = \{v \in \Bbb R^n | Av = 0\} =\{0\};  \tag{2}$
for if there exists a non-zero $w \in \ker A$, then
$A(x + w) = Ax + Aw = Ax = b, \tag{3}$
since $Aw = 0$; but then $x$ is not unique,
for $w \ne 0$ implies $x + w \ne x$; this contradiction shows $\ker A = {0}$.  Now it
is a well-known general fact that an $n \times n$ matrix is invertible if and only if its kernel consists solely of the $0$ vector; thus $A^{-1}$ exists.  So we 
may take
$y = A^{-1}c,  \tag{4}$
from which
$Ay = A(A^{-1}c) = (AA^{-1})c = Ic = c, \tag{5}$
showing such a $y$ solves $Ax = c$.  Furthermore, since $y$ is given by the formula
(4), it is necessarily unique; there is only one value
for $A^{-1}c$; indeed, $Ay = c$ yields
$y = Iy = (A^{-1}A)y = A^{-1}(Ay) = A^{-1}c; \tag{6}$
only one way to do it.
If one is unfamiliar with, or otherwise wishes to circumvent direct appeal to the theorem that $\ker A = {0}$ if and only if $A^{-1}$ exists, an argument based upon linear independence may be used.
If $v_1, v_2, . . . , v_m \in \Bbb R^n$ are linearly independent, the so are the $Av_j$, $1 \le j \le m$, since a relation of the form $\sum_j \beta_j Av_j = 0$ implies $A(\sum_j \beta_j v_j)= 0$ implies $\sum_j \beta_j v_j = 0$ since $\ker A = {0}$, contradicting the linear independence of the $v_j$.  Thus the $Av_j$ are linearly independent.  Now if $m = n$, so that the $v_j$ and hence the $Av_j$ are bases, we may write $c = \sum c_j Av_j = A(\sum c_j v_j)$, whence $y = \sum c_j v_j$ solves $Ay = c$.  Uniqueness is then argued as in the usual manner:  if $Ay_1 = Ay_2 = b$, then $A(y_1 -y_2) = 0$, whence $\ker A = \{0\}$ forces $y_1 = y_2$.
