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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").

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    $\begingroup$ Where did $x=1$ come from? $\endgroup$
    – ConMan
    Nov 21, 2021 at 22:34
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    $\begingroup$ "Setting x=1" you cannot do that because $x$ is not an independent variable (at least you have not proved that). $\endgroup$
    – markvs
    Nov 21, 2021 at 22:36
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    $\begingroup$ No: for “most” linear systems in three equations and two variables, there will be exactly one (ie theee minus two) free variable. And this is what happened here: you just forgot all about the second equation, so it’s not unexpected that the final result is one free variable for a solution of both equations (1) and (3). But there’s no reason why this solution should satisfy equation (2). $\endgroup$
    – Aphelli
    Nov 21, 2021 at 22:38
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    $\begingroup$ Combining the first and 3rd equations gives $x = 2z.$ And the algebra that follows is bad. Since you have 3 independent equations (and all equal to 0) the trivial solution is the only solution. $\endgroup$
    – Doug M
    Nov 21, 2021 at 22:44
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    $\begingroup$ Your friend found a solution to a system of two equations. Why is he surprised that it is not a solution to a third, unused and unrelated equation? $\endgroup$ Nov 22, 2021 at 7:27

5 Answers 5

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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.

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You wrote, correctly, that the second equation was not used at all. And the solutions of the system which consists of the other two equations are the triplets $(x,y,z)$ such that $x=-2y$ and that $z=-y$ indeed. Finally, there is not reason at all to pick $x=1$.

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  • $\begingroup$ worth pointing out, adding the first and third equations leads to $x-2z=0$ rather than $x-2y=0$ as the OP thinks. $\endgroup$
    – Will Jagy
    Nov 21, 2021 at 22:44
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    $\begingroup$ @WillJagy In fact, but adding $3$ times the first equation to the third one leads indeed to $x=-2y$. $\endgroup$ Nov 21, 2021 at 22:46
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Let me present an extreme version of your friend's argument:

By the first equation, we have

$$ z = x + y $$

Then, simply choose $ x = y = 1 $ and conclude $ z = 2 $.

Thus, $(1,1,2)$ is a solution to the system!

Or to a ridiculous degree:

By ignoring all equations, we can choose $x = y = z = 1$ and so $(1,1,1)$ is a solution!

I hope this illuminates the issue with your friend's solution. Each equation potentially restricts the possible solutions to the system, and so each equation not used potentially un-restricts those solutions to ones which are not actually valid.

One more example to demonstrate this fact:

Consider the system of two equations for one variable, $$ \begin{align*} x &= 1 \\ x &= 2 \end{align*} $$ By limiting our scope to only one equation, we may conclude either that $1$ is a solution or that $2$ is a solution, but in reality there are (obviously) no solutions, and so both are wrong.

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We can begin by substituting $(2)$ into $(1)$ or $(3)$. The problem is, using $(2)$, we find a contradiction between $(4)$ and $(6)$

\begin{align} z-x-y=0 \implies -x-y+z&=0 \tag{1}\\ z-2x=0 \implies z&=2x \tag{2}\\ 2x+y-3z=0\implies 2(x)+y-3z&=0 \tag{3} \end{align} Substitute $(2)$ into $(1)$ then $(4)$ into $(3)$

\begin{align} -x-y+(2x)=0\implies x&=y \tag{4}\\ 2(x)+y-3z=0\implies 3y-3z&= 0\tag{5}\\ \implies y=z\implies y&=2x \tag{6} \end{align}

or we can combine $(1)$ and $(3)$ as you did.

Sometimes rearranged things are earier to "see" \begin{align} z-x-y=0 \implies -x-y+z=0 \tag{1}\\ z-2x=0 \implies -2x+0y+z=0 \tag{2}\\ 2x+y-3z=0\implies 2(x)+y-3z=0 \tag{3} \end{align} Add $(1)$ to $(3)$, substitute $(4)$ into $(3)$ then $(5)$ into $(6)$, etc.

\begin{align} x-2z=0\implies x&=(2z) \tag{4} \\ \implies z&=\frac{x}{2} \tag{5}\\ 2(2z)+y-3z=0\implies y&=(-z) \tag{6}\\ \implies y&=\frac{-x}{2} \tag{7} \end{align}

Setting $x=1\quad \bigg(x,\dfrac{-x}{2},\dfrac{x}{2}\bigg) =\bigg(1,\dfrac{-1}{2},\dfrac{1}{2}\bigg) $

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It might be useful to see what you actually did solve: the set of three equations:

$$z-x-y=0$$

$$x=1$$

$$2x+y-3z=0$$

I've removed the equation that wasn't used, and replaced it with an equation that was used. <1,−1/2,1/2> is a solution to this particular problem. For this particular problem, I wouldn't start by combining 1 and 3. Instead, I'd first substitute 2 into 1 and then 2 into 3.

It's still a similar sort of problem: the intersection of three planes. But if should be fairly clear that by choosing different planes, you can get different solutions.

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