stonecoldgen said:
Because if you reduce to Row-Echelon form you don't end up with a triangulated form. Here is what I ended up with (How do I use Latex Reference to construct a Matrix? Can someone help me with that by the way please?)
1 -1 3 2 | 1
0 1 1 3 | -1
0 0 1 1 | -1
Here is the LaTeX code:
\left(\begin{array}{cccc|c}<br />
1 & -1 & 3 & 2 & 1\\<br />
0 & 1 & 1 & 3 & -1\\<br />
0 & 0 & 1 & 1 & -1<br />
\end{array}\right)
See our LaTeX tutorial:
https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
Anyway, to the math. In cases like this, you will not get a unique solution, you will get infinitely many solutions which depend on a free parameter. Let's first transfrom the system back to the usual form:
\left\{<br />
\begin{array}{l}<br />
x - y + 3z + 2t = 1\\<br />
y + z + 3t = -1\\<br />
z + t = -1<br />
\end{array}<br />
\right.
The idea is to write all variables in function of ##t## (or ##x## or whatever, it doesn't matter which one). We can certainly write ##z## in function of ##t## as follows:
z = -1 - t
Now, we know ##y + z + 3t = -1##, we can substitute the above version of ##z## into this equation and get
y + (-1 -t) + 3t = -1
Thus we can write ##y## in function of ##t##:
y = -2t
Finally, we can do the same for ##x##, we have the equation ##x - y + 3z + 2t = 1##. Substituting ##y## and ##z##, we get
x - (-2t) + 3(-1-t) + 2t = 1
Solving for ##x## gives us
x = 4-t
So we get the following solutions for the system:
x= 4-t,~y=-2t,~z=-1-t
This means that for every real number ##t## that we take, we get a solution for our system. For example, taking ##t=0##, we get ##(x,y,z,t) = (4,0,-1,0)##. On the other hand, taking ##t=1##, we get ##(x,y,z,t) = (3,-2,0,1)## as a solution. So we see that we have infinitely many solutions, one for each ##t##.
