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Hi I'm stuck on this homework question:

"For each $s ∈ R$, determine whether the vector $y$ is in the subspace of $R^ 4$ spanned by the columns of $A$ where:

A=$\begin{pmatrix}1 & 4 & -5\\3 & 13 &-13\\2 & 11 & -7\\-2 & -7 & -9\end{pmatrix}$ and $y=$ $\begin{pmatrix}1\\13\\26\\s\end{pmatrix}$."

What I initially thought is that I form an augmented matrix and see if it has a unique solution. If it has a unique solution then $y$ can be formed as a linear combination of the column vectors of $A$ so it is in the subspace. If it doesn't have a solution then $y$ can't be formed from the column vectors of $A$ so it's not in the subspace. Am I along the right lines? Thanks for your help.

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Almost correct.

If the system has one or more solutions then $y$ can be formed as a linear combination of the column vectors of $A$ so it is in the subspace.

It doesn't matter whether the solution is unique or not.

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Yes, you can make an augmented matrix and see what column vectors span the column space. If you succeed to write $y$ as a linear combination of those columns, $y$ is in the column space. Note that you were wrong when you talked about unique solution. This does not matter.

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