Writing the solution set(s) of the equation $Ax = 0$ 
Consider the following matrix $A = \begin{bmatrix}1&-4&0&0&1\\0&0&1&0&5\\0&0&0&1&1\end{bmatrix}$
a) Write the solution set of the equation $Ax = 0$
b) Write the solution set of the equation $Ax = \begin{bmatrix}1\\0\\0\end{bmatrix}$ (Hint: $\begin{bmatrix}1\\0\\0\\0\\0\end{bmatrix}$ is one solution to this equation)

I'm confused on both of these.
First for part a) I was thinking something like:
$x_1 = 4x_2 - x_5$
$x_3 = -5x_5$
$x_4 = -x_5$
I can't remember the method to solving this, if I could get one solution set then I could figure out how to obtain the rest I'm sure.
 A: For part $a)$, you got the right idea. The next step is just to solve for $x_1,x_2,x_3,x_4,x_5$
Or you can also write the system in the following form and reduce it into r:
$$Ax = \left[\begin{array}{ccccc|c}1&-4&0&0&1&0\\0&0&1&0&5&0\\0&0&0&1&1&0\end{array}\right]$$
It is easy to see that $x_2$ and $x_5$ are the free variables (because no leading $1$'s in the column), so let $x_2=r,x_5=s$, then solve for $x_1,x_3,x_4$ correspondingly using back substitution.

A: The algorithmic way:
The general strategy is to permute columns if necessary (which amounts to change the order of the unknowns), then operate full row reduction so as to write the system of linear equations in the form:
$$\begin{bmatrix}I_r&A\\ 0&0\end{bmatrix}=0,\quad\text{where $r$ is the rank of the matrix}.$$
Let's see what happens in the present case:
\begin{align*}
\begin{bmatrix}
1&-4&0&0&1 \\ 0&0&1&0&5 \\0&0&0&1&1
\end{bmatrix}
\rightsquigarrow
\begin{bmatrix}
1&0&0&\color{red}{-4}&\color{blue}1\\0&1&0&\color{red}0&\color{blue}5\\0&0&1&\color{red}0&\color{blue}1
\end{bmatrix}\qquad \scriptstyle(C_1C_2C_3C_4C_5\:\rightsquigarrow \:C_1C_3C_4C_2C_5)
\end{align*}
Thus the system has rank $3$, the main unknowns are $x_1,x_3, x_4$ and the solutions are
$$\begin{cases}
x_1=4x_2-x_5,\\x_3=-5x_5,\\x_4=-x_5.
\end{cases}$$
Note the last two columns of the transformed matrix give us a basis of the two dimensional space of solutions, after they have correctly been completed:
$$\begin{bmatrix}\color{red}4\\\mathbf{\color{brown}1}\\\color{red}0\\\color{red}0\\\mathbf{\color{brown}0}\end{bmatrix}\quad\text{and}\quad
\begin{bmatrix}\color{blue}{-1}\\\mathbf{\color{cyan}0}\\\color{blue}{-5}\\\color{blue}{-1}\\\mathbf{\color{cyan}1}\end{bmatrix}.$$
A: Your part (a) is correct.
For part (b), you need to add the general solution you got from (a) to any one particular solution to the nonhomogeneous linear system.
A: $A$ is a $3\times5$ matrix and its rank is 3.
That means that the Kernel as dimension 2.
$A = \begin{bmatrix}1&-4&0&0&1\\0&0&1&0&5\\0&0&0&1&1\end{bmatrix}\begin{bmatrix} x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix} = \mathbf 0$
I like the "take a guess" approach,$\mathbf v_1 = \begin{bmatrix} 4\\1\\0\\0\\0\end{bmatrix}$ Jumps out.  For the  next solution, we know that $x_4 = -x_5$ in order for the lasts line to be zero.  So, set them to $1,-1$ respectively.  Now we can solve for $x_3 = - 5x_5 = 5$.  And we can choose $x_1 = 1$ or $x_2 = -\frac 14,$
$\mathbf v_2 = \begin{bmatrix} 1\\0\\5\\1\\-1\end{bmatrix}$
a)
$a_1 \mathbf v_1 + a_2 \mathbf v_2$
b)
$\begin{bmatrix} 1\\0\\0\\0\\0\end{bmatrix} + a_1 \mathbf v_1 + a_2 \mathbf v_2$
