Simple Linear Algebra Problem I'm trying to solve for the full solution of $Ax=b $ where
$$ A = \pmatrix{20 &40 &60\\ 30 &60 &90}, b = \pmatrix{10\\ 15}$$
The answers I am getting are 
$$
\pmatrix{-2\\1\\ 0}x_1 + \pmatrix{-3\\0\\ 1}x_2 + \pmatrix{1/2\\0\\ 0}$$
I'm not exactly sure if this is correct or not as the answers given to me by my professor is different and I don't know why. Can someone offer some insight?
Thanks
 A: Your answer is correct, as you can check doing these calculations:
$$
A
\begin{pmatrix}
-2 \\
1  \\
0
\end{pmatrix}
=
A
\begin{pmatrix}
-3  \\
0   \\
1
\end{pmatrix}
=
\begin{pmatrix}
0  \\
0
\end{pmatrix}
$$
and
$$
A 
\begin{pmatrix}
1 / 2  \\
0  \\
0
\end{pmatrix}
=
\begin{pmatrix}
10  \\
15
\end{pmatrix}
\ .
$$
As for the answer given by your professor, you must take into account that those vectors $(-2 ,0 , 1), (-3,0, 1)$ and $(1/2, 0 ,0)$ are not unique. What is unique is the solution set
$$
(1/2, 0 ,0) + \ \mathrm{span}\left\{(-2 ,0 , 1), (-3,0, 1)  \right\}  \ .
$$
But there are infinitely many ways to write it. For instance, another way to write the same solution set could be the following:
$$
(0, 0, 1/6) +\ \mathrm{span}\left\{(-1 ,0 , 1/2), (-1,0, 1/3)  \right\} \ .
$$
EDIT. I forgot: in order to fully check that your solution is right, you should also verify that the rank of your matrix $A$ is one (as it is); so the solution set has indeed dimension = number of unknowns - $\mathrm{rank}\ A = 3 -1 = 2$.
