If the $m\times n$ matrix $A$ has rank $n$ ($n$ pivot columns), then $A$ has a left inverse. If $L$ is such a left inverse and $v_1$ and $v_2$ are solutions of $Ax=b$, then $Av_1=b=Av_2$, so
$$
v_1=I_nv_1=LAv_1=LAv_2=I_nv_2=v_2
$$
Hence the system $Ax=b$ has at most one solution.
In another way: the system $Ax=b$ can have infinitely many solutions only if the matrix $A$ has some nonpivot column (in this case the system has either infinitely many solutions or none). Indeed, when you perform Gaussian elimination, pivot columns correspond to unknowns whose value is determined by the vector $b$, whereas nonpivot columns correspond to “free variables” which can be given arbitrary values.
Conversely, if every system $Ax=b$ has a solution, then the matrix $A$ must have $m$ pivot columns, or some system would have no solution. If both conditions hold (that is, $n$ pivot columns and every system has a solution), then $m=n$ and the matrix $A$ is invertible.
$x+y+w=0$
$x-z+2w=0$
The first equation leads to $x = -y-w$ and the second leads to $x = z-2w$. Then we have $-y-w = z-2w$, so $y = -z+w$
Then $x = z-2w$ and $y = -z+w$. We have two variables $z,w$ that are free, as expected.
So you have a vector $(x,y,z,w)$ and want to write it as a combination of basis vectors. As the equations showed: $(x,y,z,w) = (z-2w,-z+w,z,w) = (1,-1,1,0)z + (-2,1,0,1)w$, hence a basis would be $\{(1,-1,1,0),(-2,1,0,1)\}$
Best Answer
Sure: $$ \begin{cases} x+y=0\\ x+y=1 \end{cases} $$ The rank of the incomplete matrix is $1$, so one variable is free, but the system is inconsistent.
But this might depend on the definition of free variable to begin with.