<s>[INST] 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...
 [/INST] 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):
$$\begin{cases}x=-2z+2u,\\y=z-5u,\\t=1-u.\end{cases}$$
</s>