Does the vector belong to the column space? I have to check whether the vector
$$b = \begin{vmatrix}
       -2\\
        4\\
        5\\
        6\\
        6\\
       \end{vmatrix}$$
belongs to the column space of matrix
$$A = \begin{bmatrix}
       1 & 1 & 1 & −2&  1\\
       1 & 0 & 2 &  4 & 2\\
       1 & 0 & 2 &  5 & 3\\
       1 & 0 & 2 &  6 & 4\\
       1 & 0 & 2 &  6 & 4\\
      \end{bmatrix}$$
I know that it is a question whether the matrix $A$ extended by vector $b$ has a solution.  Therefore we have
$$[A|b]= \left[
    \begin{array}{ccccc|c}
 1 &1& 1& −2& 1& -2\\
1 &0 &2 &4 &2&4\\
1 &0 &2 &5 &3&5\\
1 &0 &2 &6 &4&6\\
1 &0 &2 &6 &4&6\\
    \end{array}
\right] $$
After Gauss-Jordan elimination:
$$\left[
    \begin{array}{ccccc|c}
1&   0& 2& 0& -2& 0\\
0& 1& -1& 0& 5& 0\\
0& 0& 0& 1& 1& 1\\
0& 0& 0& 0& 0& 0\\
0& 0& 0& 0& 0& 0\\
   \end{array}
\right] $$
So it proves that the vector b belongs to the column space of matrix $A$? Because my prof. in his book gave the solution it doesn't. He make mistakes often but I'm not sure about it...

It obviously does, because the given matrix and the augmented matrix have the same rank. You even can read the solutions in the reduced row echelon form (I denote the unknowns $x, y,z,t,u$, in the order of the columns):
$$\begin{cases}x=-2z+2u,\\y=z-5u,\\t=1-u.\end{cases}$$
