Determining whether or not a vector is a linear combination of a given matrix $$
A=
\begin{bmatrix}
1 & 0 & 5\\
-2 & 1 & -6\\
0 & 2 & 8
\end{bmatrix}
,b= \begin{bmatrix}
2\\
-1\\
6
\end{bmatrix}
$$
The problem asks to determine whether or not vector $b$ is a linear combination of the vectors formed by matrix $A$. I have used the elementary row operations to reduce augmented matrix and got
$$\begin{bmatrix}
1 & 0 & 5 & 2\\
0 & 1 & 4 & 3\\
0 & 0 & 0 & 0
\end{bmatrix}$$
I am not sure what to do after this. According to the matrix in reduced row echelon form, the linear system is consistent so there are solutions. The answer in the back of the book states that $b$ is not a linear combination of matrix $A$. I thought that if a matrix in reduced row echelon form was consistent then there existed a linear combination?
Why doesn't $1a_1+1a_2+0a_3=b$ work?
 A: There are three spaces commonly associated with a matrix: column space, row space and null space.
One solution to the system of linear equations $Ax=b$ is $x=\begin{bmatrix}\color{blue}{2} \\ \color{red}{3} \\ \color{green}{0} \end{bmatrix}$, as can be seen from the given row echelon form.  This implies the solution
$$\require{cancel}\underbrace{\begin{bmatrix}2 \\ -1 \\ 6\end{bmatrix}}_b=\color{blue}{2}\begin{bmatrix}1 \\ -2 \\ 0\end{bmatrix}+\color{red}{3}\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}+\cancel{\color{green}{0}\begin{bmatrix} 5 \\ -6 \\ 8 \end{bmatrix}}$$ which could alternatively be found by inspection.  This implies $b$ is in the column space of $A$.
We can also see that $(2,-1,6)$ is the negative of the second row of $A$.  This implies $b$ is in the row space of $A$ (or, perhaps more formally, $b^T$ is not in the row space of $A$).
Converting $A$ to row echelon form $R$ gives
$$
R=\begin{bmatrix}
1 & 0 & 5 \\
0 & 1 & 4 \\
0 & 0 & 0 \\
\end{bmatrix}
$$
and $Rx=\mathbf{0}$ has the solution space $\{(-5t,-4t,t):t \in \mathbb{R}\}$ -- this is the null space of $A$.  Since $b$ is not of the form $(-5t,-4t,t)$, we see that $b$ is not in the null space of $A$.
