Linear system - number of solutions depending on the parameter k Determine for what value of $k$ the following system has 


*

*unique solution,

*no solution and

*infinitely many solutions.


\begin{cases}
x+2y+z=3\\
2x-y-3x=5\\
4x+3y-z=k
\end{cases}
I did till 
$$
\left[\begin{array}{ccc|c} 
1 & 2 & 1 &  3\\
1 & 1 & 1 &  1/5\\
0 & 0 & 1 &  -k+42/5
\end{array}\right]
$$
I am not sure how to continue.
 A: You're not arrived at the end of the elimination:
\begin{align}
\left[\begin{array}{ccc|c} 
1 & 2 & 1 &  3\\
2 & -1 & -3 &  5\\
4 & 3 & -1 &  k
\end{array}\right]
&\to
\left[\begin{array}{ccc|c} 
1 & 2 & 1 &  3\\
0 & -5 & -5 &  -1\\
4 & 3 & -1 &  k
\end{array}\right]
&&R_2\gets R_2-2R_1
\\&\to
\left[\begin{array}{ccc|c} 
1 & 2 & 1 &  3\\
0 & -5 & -5 &  -1\\
0 & -5 & -5 &  k-12
\end{array}\right]
&&R_3\gets R_3-4R_1
\\&\to
\left[\begin{array}{ccc|c} 
1 & 2 & 1 &  3\\
0 & 1 & 1 &  1/5\\
0 & -5 & -5 &  k-12
\end{array}\right]
&&R_2\gets -\frac{1}{5}R_2
\\&\to
\left[\begin{array}{ccc|c} 
1 & 2 & 1 &  3\\
0 & 1 & 1 &  1/5\\
0 & 0 & 0 &  k-11
\end{array}\right]
&&R_3\gets R_3+5R_2
\end{align}
Now you should be able to end.
A: 1) First step:
$$\begin{cases}
x+2y+z=3\\
2x-y-3x=5\\
4x+3y-z=k
\end{cases}\Longleftrightarrow$$

$x+2y+z=3 \Longleftrightarrow x=3-2y-z$

$$\begin{cases}
-3+y+z=5\\
12-5y-5z=k
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
-15+5y+5z=25\\
12-5y-5z=k
\end{cases}\Longleftrightarrow$$
$$-3=25+k \Longleftrightarrow$$
$$k=-28 $$
2) Second step:
$$\begin{cases}
x+2y+z=3\\
2x-y-3x=5\\
4x+3y-z=-28
\end{cases}\Longleftrightarrow$$

$2x-y-3x=5 \Longleftrightarrow y=-5-x$


$x+2y+z=3 \Longleftrightarrow z=28+4x+3y \Longleftrightarrow z=13+x$


$x+2y+z=3 \Longleftrightarrow x=3-2y-z \Longleftrightarrow x=z-13$

So we got:
$$k=-28,y=-x-5,x=z-13,z=13+x$$
