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enter image description here I am having a hard time trying to figure out a nonzero vector in $\text{im}(A)$ that is not a column of $A$. I know that first two columns have pivots so $\text{col}(A)=\text{span}\{(1 \ 1 \ 1) , (-2 \ -2 \ 1)\}$ and that the solution of the Matrix is

$x_1=5t-s+2r$
$x_2=-4s-4r$
$x_3=t$
$x_4=s$
$x_5=r$

I was thinking about adding the vectors that make up the $\text{col}(A)$ in order to get $\text{im}(A)$ so I would get $\text{im}(A)= [-1 \ -1 \ \ \ 0]^T$. I am not sure if I am solving this correctly.

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    What about $(2,2,2)^T$ or $(3,3,3)^T$ or ...? – 2017-01-25
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    @copper.hat how did you get those? – 2017-01-25
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    $A(2,0,0,0,0)^T, A(3,0,0,0,0)^T,...$. Choosing random numbers for $x$ and then computing $Ax$ will (almost certainly) find you a point that is not a column of $A$. There are only 5 columns of $A$ and an entire continuum of points in the range/image. – 2017-01-25
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    For b), note that the top two components of each column are the same, so the same will be true of $Ax$, from which you can easily find a point satisfying b). – 2017-01-25
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    @copper.hat for b) and c) I have gotten the correct answers. I just having a hard time understanding how to find im(A) that is not a column of A. – 2017-01-25
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    The image is the set of all points $Ax$. The columns are $Ae_1,...,Ae_5$, where the $e_k$ are unit vectors along each axis. Choose an $x$ such that $Ax \notin \{ Ae_1,...,A e_5 \}$. If $Ae_k \neq 0$ you could just try $A (n e_k)$ until you find some $n$ such that $A(ne_k) \notin \{ Ae_1,...,A e_5 \}$. – 2017-01-25
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    @copper.hat Thank you very much! I understand it now and I got the correct answer. Once again thank you for your help! – 2017-01-25
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    You are very welcome. Just to clarify; there are an infinite number of correct answers, but only 5 incorrect ones here. – 2017-01-25

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The Image of A, would be the set spanned by the independent column vectors of A. Any linear combination of those vectors is in the image of A.