Is there NO solution to this linear system of 3 equations, $3$ unknowns?

Question

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?

Answer 1

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.

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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:

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

Answer 2

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.