How exactly do we do Gauss elimination? 
This is a matrix:
  $$\begin{bmatrix}
1 & 1 & 1\\
1 & 2 & 3\\
1 & 3 & k
\end{bmatrix}\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=
\begin{bmatrix}
3\\
6\\
4+k
\end{bmatrix}$$
  Find $k$ so that it has no unique solution. Solve the equations for this value of $k$.

I found $k = 5$ by using $\text{determinant}=0$.
Then I tried Gauss elimination, from what I understand from my lecturer, I use the 3 rules of Gauss elimination to solve, I just randomly use whatever I need to achieve echelon form. I have come to this: 
$$\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 3\\
0 & 0 & 1 & 0\\
\end{bmatrix}$$
The answer is 
$$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\3\\0\end{bmatrix}+t\cdot\begin{bmatrix}1\\-2\\1\end{bmatrix}$$
Where does the $t$ come from? Does it involve eigenvalues? 
 A: You did not row reduce correctly. Since we are told that there are infinitely many solutions, we know that the final result should have free variables, so there should be at most two pivots. We should get:
$$
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
1 & 2 & 3 & 6 \\
1 & 3 & 5 & 9 \\
\end{array}\right]
\sim
\left[\begin{array}{ccc|c}
1 & 1 & 1 & 3 \\
0 & 1 & 2 & 3 \\
0 & 2 & 4 & 6 \\
\end{array}\right]
\sim
\left[\begin{array}{ccc|c}
1 & 0 & -1 & 0 \\
0 & 1 & 2 & 3 \\
0 & 0 & 0 & 0 \\
\end{array}\right]
$$
Translating back to a system, we have:
\begin{cases}
x - z = 0 \\
y + 2z = 3
\end{cases}
Taking $z = t$ and solving for $x$ and $y$ in terms of $t$, we conclude that the general solution is:
$$
\begin{bmatrix}x \\ y \\ z\end{bmatrix}
=
\begin{bmatrix}t \\ 3 - 2t \\ t\end{bmatrix}
= 
\begin{bmatrix}0 \\ 3 \\ 0\end{bmatrix}
+
t\begin{bmatrix}1 \\ -2 \\ 1\end{bmatrix}
$$
A: By Gauss-Jordan elimination I get the following:
$$\left(\begin{array}{ccc|c}
  1 & 1 & 1 & 3 \\
  1 & 2 & 3 & 6 \\
  1 & 3 & k & 4+k
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 1 & 1 & 3 \\
  0 & 1 & 2 & 3 \\
  0 & 2 &k-1& 1+k
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 1 & 1 & 3 \\
  0 & 1 & 2 & 3 \\
  0 & 0 &k-5& k-5
\end{array}\right)$$
In the next step I would like to divide the last row by $k-5$. Notice that this is valid only for $k-5\ne0$, i.e., $k\ne 5$. 

If $\boxed{k\ne5}$ I get
$$\left(\begin{array}{ccc|c}
  1 & 1 & 1 & 3 \\
  0 & 1 & 2 & 3 \\
  0 & 0 &k-5& k-5
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 1 & 1 & 3 \\
  0 & 1 & 2 & 3 \\
  0 & 0 & 1 & 1
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 1 & 0 & 2 \\
  0 & 1 & 0 & 1 \\
  0 & 0 & 1 & 1
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 0 & 0 & 1 \\
  0 & 1 & 0 & 1 \\
  0 & 0 & 1 & 1
\end{array}\right)$$
I see that for $k\ne5$ I have the unique solution $(1,1,1)$. 
I can check that this is indeed a solution by pluggin these values into the original system:
\begin{align*}
1+1+1&=3\\
1+2+3&=6\\
1+3+k&=4+k
\end{align*}

For $\boxed{k=5}$ I get
$$
\left(\begin{array}{ccc|c}
  1 & 1 & 1 & 3 \\
  0 & 1 & 2 & 3 \\
  0 & 0 & 0 & 0
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 0 &-1 & 0 \\
  0 & 1 & 2 & 3 \\
  0 & 0 & 0 & 0
\end{array}\right)
$$
This equation has infinitely many solutions $(t,3-2t,t)$, $t\in\mathbb R$, as explained in Adriano's answer.
