I was recently solving a system of linear equations, 3 equations and 3 unknowns. I first solved via Row Reduction of the matrix and got a valid answer, but my friend attempted to solve the system using informal algebra methods and got the wrong answer. I know his answer is wrong, but I am struggling to explain what mathematical rule he broke.
Here is the system:
$z-x-y=0$
$z-2x=0$
$2x+y-3z=0$
Combining the first and third equation, one gets $x=-2y$. Plugging this back into equation one, one gets $z=-y$. Setting $x=1$, one gets the vector $<1,-1/2,1/2>$. This vector is valid for equations 1 and 3, but not for equation 2.
Now I know that this is not the proper technique for solving a system of three variables and that equation 2 was not used so how should one expect it to be satisfied. I know that this solution is wrong, but I am unsure how to explain what is wrong about it other than saying "that's not the way it's done." I personally made this mistake when first learning linear algebra and "that's not the way it's done" is all my teacher could say. If anyone has a better explanation for what exactly is wrong about this, I would greatly appreciate. Also, since equation 2 is not being used, it is 2 equations, 3 unknowns so there should be 2 free variables, not one (again I think this will occur if elimination is "done correctly").
Best Answer
Your friend used just two equations, which (since they were linearly independent) is enough to limit the solution space to one dimension. The correct conclusion is not that $\langle x,y,z\rangle = \langle 1,-\frac12,\frac12\rangle$ is the solution, but that the solution has the form $$\langle x,y,z\rangle = \left\langle t,-\frac12t,\frac12t\right\rangle$$ for some real number $t.$
This is true. The correct solution does have that form.
Now compare this with your solution and see if you can tell what value(s) of $t$ give a solution to all three equations.