For what values of a,b,c do there exist solutions to this system of equations? Let 
$$A=\begin{bmatrix} 1 & 1 & 0 & 3 & 4 \\ 1 & 2 & 1 & 5 & 7 \\ 1 & 3 & 2 & 7 & 10\end{bmatrix}$$
Consider the system 
$$A \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} a \\ b \\c \end{bmatrix}$$
Use row reduction to determine the general solution for this system. For what values of $a,b,c$ do there exist solutions to this system of equations? Find a basis for the null space $N(A)$.
My attempt: I row reduced all the way to 
$$\begin{bmatrix}
 1 & 0 & 1 & 1 & 1 & | & b-15 \\
 0 & 1 & -1 & 2 & 3 & | & a-b=15 \\
 0 & 0 & 0 & 0 & 0 & | & \frac{-c+a} 2+a-b
\end{bmatrix}$$
and got 
$$x_1=b-\alpha - \beta -y$$
$$x_2=a-b+\alpha -2\beta-3y$$
$$x_3=\alpha$$
$$x_4=\beta$$
$$x5=y$$
I don't know how to prove these answers, and whether I am right or wrong. Also, I need direction on how to solve for the basis of the null space.  Thank you for any help! :)
 A: Ok I think you might have some errors in your row reduction.  Step 1, you want to put a pivot at 1,1.  This means making a matrix that looks like:
$$\begin {bmatrix}
1 & ... \\
0 & ... \\
0 & ... \\
\end {bmatrix}$$
You can do this by subtracting the first row from the second and third:
$$\begin{bmatrix}
1 & 1 & 0 & 3 & 4 & | & a \\
0 & 1 & 1 & 2 & 3 & | & b - a \\
0 & 2 & 2 & 4 & 6 & | & c - a\\
\end{bmatrix}
$$
Now you want to add a second pivot at 2,2.  Do this by subtracting two times the second row from the third.
$$\begin{bmatrix}
1 & 1 & 0 & 3 & 4 & | & a \\
0 & 1 & 1 & 2 & 3 & | & b - a \\
0 & 0 & 0 & 0 & 0 & | & c - 2b + a\\
\end{bmatrix}
$$
To finish the RRE you want to subtract the 2nd row from the first so that your pivot columns only have a single nonzero element:
$$\begin{bmatrix}
1 & 0 & -1 & 1 & 1 & | & 2a-b \\
0 & 1 & 1 & 2 & 3 & | & b - a \\
0 & 0 & 0 & 0 & 0 & | & c - 2b + a\\
\end{bmatrix}
$$
So the question is, when does this system have a solution?  That's given by the nonpivot rows, the third row.  It is the equation:
  $$0 = c - 2b + a$$
which is a necessary condition for you to have a solution.  In your question you noticed that you can choose any $x_3$, $x_4$, and $x_5$ and those will give you the remaining $x_1$, and $x_2$.
Finally remains, given a matrix in reduced form, how do you find the null space.  Here's how:
1) If it is a square, leave it alone
2) If it is short and fat, add zero rows to make it square (edit: and it has to place the pivots along the diagonal)
3) If it is tall and skinny, remove the bottom rows to make it square (if you did RRE correctly the bottom rows will all be zeros anyway)
*) Finally, subtract the identity
This matrix is short and fat, so add zeros:
$$\begin{bmatrix}
1 & 0 & -1 & 1 & 1\\
0 & 1 & 1 & 2 & 3 \\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
\end{bmatrix}
$$
Subtract the identity:
$$\begin{bmatrix}
0 & 0 & -1 & 1 & 1\\
0 & 0 & 1 & 2 & 3 \\
0 & 0 & -1 & 0 & 0\\
0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & -1 \\
\end{bmatrix}
$$
And there is your null space basis, $\begin{bmatrix} -1 \\ 1 \\ -1 \\ 0 \\ 0 \end{bmatrix}$, $\begin{bmatrix} 1 \\ 2 \\ 0 \\ -1 \\ 0 \end{bmatrix}$, and $\begin{bmatrix} 1 \\ 3 \\ 0 \\ 0 \\ -1 \end{bmatrix}$.
