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I have the following linear system:

$$\begin{align} &x + y + 2z + 2 = 0 \\ &3x - y + 14z -6 = 0 \\ &x + 2y +5 = 0 \end{align}$$

I immediately noticed that there was no $z$ term in the last equation and thus determined that I will end with $0z = \text{some number}$ and therefore, came to the conclusion that there is no solution to the linear system. To see if I was right, I checked with echelon form and that also suggested that there was no solution.

$$\begin{align} &L_1: x + y + 2z + 2 = 0 \\ &L_2: 3x - y + 14z -6 = 0 \\ &L_3: x + 2y +5 = 0 \\ \end{align}$$

Then $-3L_1 + L_2 \rightarrow L_2$

$$\begin{align} &x + y + 2z + 2 = 0 \\ &0x + 3y - 42z - 18 = 0 \\ &x + 2y + 0z + 5 = 0 \end{align}$$

Then $-L_1 + L_3 \rightarrow L_3$

$$\begin{align} &x + y + 2z + 2 = 0 \\ &0x + 3y - 42z - 18 = 0 \\ &0x + 1y + 0z + 3 = 0 \end{align}$$

Then $-L_2/3 + L_3 \rightarrow L_3$

$$\begin{align} &x + y + 2z + 2 = 0 \\ &0x + 3y - 42z - 18 = 0 \\ &0x + 0y + 0z + 3 = 0 \end{align}$$

Firstly, is my answer correct? If so, then can I make the same conclusions for other linear systems that are similar?

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    $\begingroup$ write your equation in a matrix-vector notation. Then if the matrix corresponding to your variables is not invertable... $\endgroup$
    – Seyhmus Güngören
    Jun 10, 2013 at 22:25
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    $\begingroup$ The equation $0z=0$ has certainly solutions, despite containing, quite literally, an equation of the form $0z=\text{some number}$. $\endgroup$
    – celtschk
    Jun 10, 2013 at 22:27
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    $\begingroup$ The determinant of the linear system is $-20$, hence it has a solution. $\endgroup$
    – copper.hat
    Jun 10, 2013 at 22:35
  • $\begingroup$ @copper.hat What is this determinant? Can you please explain or provide an answer? $\endgroup$
    – Jeel Shah
    Jun 10, 2013 at 22:35
  • $\begingroup$ @gekkostate these are some basic questions. You can try searching in google "determinant" and you will find everything in wikipedia. Bu shortly if your determinant is not zero then your matrix is invertible. See also in google search "linear system of equations" $\endgroup$
    – Seyhmus Güngören
    Jun 10, 2013 at 22:38

2 Answers 2

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First of all, you "jumped to" an erroneous conclusion, based on inspection. In a system of equations, one or more variables may fail to be present in one or more equations.

A most extreme example would be the three equations in three unknowns:

$$\begin{align} &x - 1 = 0 \\ &y - 1= 0 \\ &z - 1 = 0 \end{align}$$

From which we can "read off" the unique solution: $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1\end{pmatrix}$

There would be a problem (and no solution would exist) if you had the following (say, reduced) linear system of equations:

$$\begin{align} &x + y + 2z + 2 = 0 \\ &3x - y + 14z -6 = 0 \\ &0x + 0y + 0z + 5 = 0 \end{align}$$

Note that in the above system, we have the absurd equation $5 = 0$: such a system is called inconsistent, and clearly, no solution exits.

Finally, if you end up with an equation of all "zeros": $0x + 0y + 0z = 0$, then infinitely many solutions exist, and depending on how many such equations exist in your system, you might have a system with one parameter, or two, which then, while an infinite number of solutions exist, there would constraints which limit exactly which solutions are valid; i.e., the parameter(s) would define a "family" of infinitely many solutions.

Now, back to your good idea to "check out" your initial conclusion:

In your last elementary row operation, note that you didn't operate on the $-42 z - 18$ of $L_2$:

We go from:

$$\begin{align} &x + y + 2z + 2 = 0 \\ &0x + 3y - 42z - 18 = 0 \\ &0x + 1y + 0z + 3 = 0 \end{align}$$

Then applying, correctly, $-L_2/3 + L_3 \rightarrow L_3$

$$\begin{align} &x + y + 2z + 2 = 0 \\ &0x + 3y - 42z - 18 = 0 \\ &0x + 0y + 14z + 9 = 0 \end{align}$$

Now, you'll see that a unique solution exists: $14 z = -9 \implies $

To solve the system at this point, you might want to write your equations as follows:

$$\begin{align} &x + y + 2z = -2 \\ &0x + 3y - 42z =18 \\ &0x + 0y + 14z = -9 \end{align}$$

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  • $\begingroup$ Silly mistake by me :) Thanks! $\endgroup$
    – Jeel Shah
    Jun 10, 2013 at 22:35
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    $\begingroup$ You're welcome! Thanks for showing your work: it often helps uncover confusing situations! $\endgroup$
    – amWhy
    Jun 10, 2013 at 22:36
  • $\begingroup$ @amWhy: nicely written up! +1 $\endgroup$
    – Amzoti
    Jun 11, 2013 at 2:09
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Your first idea is off, because, for example the system $$x+y+z=1, x+2y=1, x+y+2z=1$$ still has a solution even though $z$ is absent from the second equation.

On the other hand, your manipulation of equations (you call it echelon form) is one of possible correct methods of finding solutions/proving their absence.

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    $\begingroup$ A simpler example would be $x+z=2$, $x=1$; there it's very obvious that it has a solution. $\endgroup$
    – celtschk
    Jun 10, 2013 at 22:29
  • $\begingroup$ @celtschk Of course, should have thought about it=) $\endgroup$
    – TZakrevskiy
    Jun 10, 2013 at 22:30

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