# How to solve a system with more unknowns than equations?

1. May 18, 2014

### stonecoldgen

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

2. May 18, 2014

### micromass

Staff Emeritus
Here is the LaTeX code:

$$\left(\begin{array}{cccc|c} 1 & -1 & 3 & 2 & 1\\ 0 & 1 & 1 & 3 & -1\\ 0 & 0 & 1 & 1 & -1 \end{array}\right)$$

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:

$$\left\{ \begin{array}{l} x - y + 3z + 2t = 1\\ y + z + 3t = -1\\ z + t = -1 \end{array} \right.$$

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 $t$.