<s>[INST] Solving linear system equation How would I solve this linear system equation?
$$\begin{cases} 
2w+x-y=4\\
3z-x=6\\
-2y-x+9z+4w=7 
\end{cases}$$
First I arranged them so I could make the metric and then I was stuck and I don't realy know how to continue, please help me.
 [/INST] Write your system into normal form:
\begin{cases} 
x-y+2w=4\\
x-3z=-6\\
x+2y-9z-4w=-7 
\end{cases}
Now, depending on the tools you have available, there are several possibilities. The most efficient is Gaussian elimination: the matrix of the system is
\begin{align}
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
1 & 0 & -3 & 0 & -6 \\
1 & 2 & -9 & -4 & -7
\end{array}\right]
&\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
0 & 1 & -3 & -2 & -10 \\
1 & 2 & -9 & -4 & -7
\end{array}\right] && R_2\gets R_2-R_1 \\[6px]
&\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
0 & 1 & -3 & -2 & -10 \\
0 & 3 & -9 & -6 & -11
\end{array}\right] && R_3\gets R_3-R_1 \\[6px]
&\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
0 & 1 & -3 & -2 & -10 \\
0 & 0 & 0 & 0 & 19
\end{array}\right] && R_3\gets R_3-3R_2 \\[6px]
\end{align}
which shows the system has no solution.
Another possibility is to get $x$ from the second equation: $x=3z-6$. Substitute in the third equation to get
$$
3z-6+2y-9z-4w=-7
$$
or
$$
2y-6z-4w=-1
$$
Substitute also in the first equation to find $3z-6-y+2w=4$, so you get
$$
y=3z+2w-10
$$
and now back in the other equation
$$
6z+4w-6z-4w-20=-1
$$
or
$$
-20=-1
$$
that is false.
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