the system of linear equations If you have a matrix M that looks like  
\begin{bmatrix}1&0&1&1\\-1&2&0&4 \\ 0&2&1&5\end{bmatrix}
Does Mx=b (x and b are vectors) have a unique solution for all possible values of b? 
And also how can I know the basis for the range space or the null space? 
I have done Gaussian elimination on this matrix and I got
\begin{bmatrix}1&0&1&1\\0&1&1/2&5/2 \\ 0&0&0&0\end{bmatrix}
so I was guessing that \begin{pmatrix}1\\0\\0\end{pmatrix} and \begin{pmatrix}0\\1\\0\end{pmatrix} are the basis for the range space of this matrix.
 A: Your proposed basis doesn’t work. I suspect that you’re confusing the rule about how to read a basis for the row space of a matrix from its RREF with that of how to read a basis for its column space—its range.  
Think about how you might compute the product of a matrix and vector. If you expand the product of $M$ with an arbitrary vector by columns, you have $$\begin{bmatrix}1&0&1&1\\-1&2&0&4 \\ 0&2&1&5\end{bmatrix} \begin{bmatrix}a\\b\\c\\d\end{bmatrix} = a\begin{bmatrix}1\\-1\\0\end{bmatrix} + b\begin{bmatrix}0\\2\\2\end{bmatrix} + c\begin{bmatrix}1\\0\\1\end{bmatrix} + d\begin{bmatrix}1\\4\\5\end{bmatrix},$$ so every such product is a linear combination of the columns of $M$, i.e., the range of $M$ is the span of its columns. In particular, every column of $M$ is an element of its range. Looking at your proposed basis for the range, we see that the last component of every linear combination of those vectors is $0$, so there’s no way to combine them to produce any of the last three columns of $M$. On the other hand, the original columns of $M$ that correspond to pivot columns in its RREF do form a basis for its range.  
Basically, elementary row operations leave the row space of a matrix unchanged but change its column space. Consider, for example, the matrix $\tiny{\begin{bmatrix}1&1\\1&1\end{bmatrix}}$. Its column space is obviously spanned by $(1,1)^T$, but its RREF is $\tiny{\begin{bmatrix}1&1\\0&0\end{bmatrix}}$, whose column space is spanned by $(1,0)^T$. More generally, performing an elementary row operation is equivalent to left-multiplying by a certain invertible matrix $E$. The row space of a matrix $M$ is the orthogonal complement of its null space $\mathscr N(M)$. Since $E$ has full rank, $\mathscr N(EM)=\mathscr N(M)$ ($E$ can’t grow the null space because the only vector that it maps to $0$ is $0$), therefore the row spaces of $M$ and $EM$ are also identical. On the other hand, the range of $EM$ is the image under $E$ of the range of $M$. In general these are different subspaces of the codomain, although they’ll have the same dimension since $E$ has full rank.  
Now, there’s a theorem that says that if $M\mathbf v$ and $M\mathbf w$ are linearly independent, then so are $\mathbf v$ and $\mathbf w$. (Proving this theorem is a useful exercise.) The sequence of elementary row operations applied to $M$ to obtain its RREF can be combined into a single invertible matrix $S$. This matrix preserves linear independence, so linearly-independent columns of $SM$ correspond to linearly-independent columns of $M$. The pivot columns of the RREF are obviously linearly independent, and the rank-nullity theorem tells us that the dimension of the range is equal to the number of pivot columns, so they form a basis for the range of $SM$. This means that the corresponding columns of $M$ are a basis for the range of $M$.
A: Hint: add first row to the second row

 The the second row is equal to the last row, so $\operatorname{rank}(M) = \dots$


Edit in response to OP's comment and edit:
OP has calculated the RREF of $M$
$$\begin{bmatrix}1&0&1&1\\0&1&1/2&5/2 \\ 0&0&0&0\end{bmatrix}.$$
$M$ can be viewed as a linear transformation from $\Bbb{R}^4$ to $\Bbb{R}^3$, so its range is its column space spanned by a basis $\{e_1,e_2\}$.  (OP has got it right.)
The nullspace of $M$ as the set of $x \in \Bbb{R}^4$ so that $Mx = 0$.  By the Rank-nullity Theorem, $\operatorname{nullity}(M) = 2$, so the nullspace of $M$ has dimension two.
From the third column of the RREF, we have basis vectors for the nullspace of $M$: $v_1 = (1, 1/2, -1, 0)^T$; from the fourth column, we have $v_2 = (1, 5/2, 0, -1)^T$.
Since $\operatorname{nullity}(M) > 0$, $Mx = b$ doesn't have unique solution.
