Matrix with 2 unknown variables, how to solve? Given that
$$Ax=b$$
where $$x = \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix},$$
$$A= \begin{bmatrix}
    1 & 1 & 3 \\
    1 & 2 & 4 \\
    k_1 & 3 & 5
\end{bmatrix},$$
and
$$ b= \begin{bmatrix}
    2\\
    3\\
    k_2
\end{bmatrix}$$
and $k_1 ,k_2\in\mathbb R$, I need to find the value of $k_1$ and $k_2$, so that $Ax=b$ works out.
when its


*

*inconsistent

*1 solution

*inf many solutions
and solve the system when its inf many solutions


I don't understand how to solve the system for both unknown variables. I tried to solve the matrix as I would if there was only one variable and I got
$$\begin{bmatrix}-k_1 & -2 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ k_2 - 5\end{bmatrix}$$
but this doesn't seem to make any sense to me because the R1 still has another unknown variable
if $k_2=5$ then the matrix would have inf many solutions, but still, there is $k_1$.
I would be grateful for any tip or solution.
 A: Row reduce the augmented matrix $\left[\begin{matrix}1&1&3&2\\1&2&4&3\\k_1&3&5&k_2\end{matrix}\right]$ to get $\left[ \begin{matrix}1&0 & 2 &1\\0&1&1&1\\ 0 & 0 & 2-2k_1 &k_2-k_1-3\end{matrix}\right]$. 
If $2-2k_1$ is $0$ and $k_2-k_1-3$ is nonzero, then the system is inconsistent. We now have to consider cases. If $2-2k_1 = 0$ and $k_2-k_1-3$ is nonzero, the system is inconsistent. This occurs when $k_1 =1$ and $k_2\neq 4$. 
The system is linearly dependent (i.e. it is consistent w/ infinitely many solutions) when $2-2k_1 = 0$ and $k_2 - k_1-3=0$. This occurs when $k_1 = 1$ and $k_2 = 4$. $k_1 \neq 1$, then the matrix can be further reduced to 
$\left[ \begin{matrix}1&0&0&\dfrac{4-k_2}{1-k_1}\\0&1&0&\dfrac{5-k_1-k_2}{2-2k_1}\\0&0&1&\dfrac{k_2-k_1-3}{2-2k_1}\end{matrix}\right]$. In this case, the system will have only one solution.
To summarize, if $k_1 =1$ and $k_2 =4$, the system is consistent and has infinitely many solutions, if $k_1 = 1$ and $k_2 \neq 4$, the system is inconsistent, and if $k_1\neq 1$, the system is consistent w/ one solution.
A: I'll call $k_1=x$, $k_2=y$ for simplicity. I don't think you did your row reduction right. you have 
$
\begin{bmatrix}
1&1&3&2\\
1&2&4&3\\
x&3&5&y
\end{bmatrix} 
\xrightarrow{R_2-R_1}
\begin{bmatrix}
1&1&3&2\\
0&1&1&1\\
x&3&5&y
\end{bmatrix} 
\xrightarrow{R_3-xR_1}
\begin{bmatrix}
1&1&3&2\\
0&1&1&1\\
0&3-x&5-3x&y-2x
\end{bmatrix} 
\xrightarrow{R_1-R_2}
\begin{bmatrix}
1&0&2&1\\
0&1&1&1\\
0&3-x&5-3x&y-2x
\end{bmatrix} 
\xrightarrow{R_3-(3-x)R_2}
\begin{bmatrix}
1&0&2&1\\
0&1&1&1\\
0&0&2-2x&y-x-3
\end{bmatrix} 
$
For the system to inconsistent, you need $2-2x=0$ and $y-x-3\ne0$. That is, $x=1$ and $y\ne4$. When $x\ne1$ the system is consistent, and it will have a unique solution. When $x=1$ and $y-x-3=0$, the last row will be zero and so there will be infinitely many solutions. In summary, 


*

*$x\ne1$: consistent, unique solution. 

*$x=1$, $y=4$: consistent, infinitely many solutions.

*$x=1$, $y\ne4$: inconsistent.  
In the case where $x=1$, $y=4$, the first two rows spell
$$
x_1+2x_3=1,\ \ \ x_2+x_3=1.
$$
So you can take $x_3=t$ as a parameter, and then the solution is 
$$
x_1=1-2t,\ \ \ x_2=1-t,\ \ \ x_3=t.
$$
