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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 proofs 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...

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    See that it is equal exactly to 4th column if everything is written right. So you can write it as $b= A^{(3)}$. So it obviously belong to column space of $A$. Just take $x^{T}=(0,0,0,1,0)$, and you'll have $Ax=b$. No need in further calcualtions.)). See that you did perhaps the same. – 2017-01-22
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    Thank you, also as I see that it belongs to Row Space of A? – 2017-01-22
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    I don't see that so obviously. Probably, if your calculations show that. – 2017-01-22

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You even can read the solutions in the reduced row echelon form (I denote the unknowns $s, y,z,t,u$, in this order): $$\begin{cases}x=-2z+2u,\\y=z-5u,\\t=1-y.\end{cases}$$

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    Oh! Yes. I misread the second row. Thanks for pointing it! – 2017-01-22