I have the following linear system: <mtable columnalign="right left right left right left right

Eliaszowyr1

Eliaszowyr1

Answered question

2022-05-22

I have the following linear system:
x + y + 2 z + 2 = 0 3 x y + 14 z 6 = 0 x + 2 y + 5 = 0
I immediately noticed that there was no z term in the last equation and thus determined that I will end with 0 z = some number 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.
L 1 : x + y + 2 z + 2 = 0 L 2 : 3 x y + 14 z 6 = 0 L 3 : x + 2 y + 5 = 0
Then 3 L 1 + L 2 L 2
x + y + 2 z + 2 = 0 0 x + 3 y 42 z 18 = 0 x + 2 y + 0 z + 5 = 0
Then L 1 + L 3 L 3
x + y + 2 z + 2 = 0 0 x + 3 y 42 z 18 = 0 0 x + 1 y + 0 z + 3 = 0
Then L 2 / 3 + L 3 L 3
x + y + 2 z + 2 = 0 0 x + 3 y 42 z 18 = 0 0 x + 0 y + 0 z + 3 = 0
Firstly, is my answer correct?

Answer & Explanation

Ismael Blackwell

Ismael Blackwell

Beginner2022-05-23Added 7 answers

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:
x 1 = 0 y 1 = 0 z 1 = 0
From which we can "read off" the unique solution:
( x y z ) = ( 1 1 1 )
There would be a problem (and no solution would exist) if you had the following (say, reduced) linear system of equations:
x + y + 2 z + 2 = 0 3 x y + 14 z 6 = 0 0 x + 0 y + 0 z + 5 = 0
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": 0 x + 0 y + 0 z = 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.
In your last elementary row operation, note that you didn't operate on the 42 z 18 of L 2 :
We go from:
x + y + 2 z + 2 = 0 0 x + 3 y 42 z 18 = 0 0 x + 1 y + 0 z + 3 = 0
Then applying, correctly, L 2 / 3 + L 3 L 3
x + y + 2 z + 2 = 0 0 x + 3 y 42 z 18 = 0 0 x + 0 y + 14 z + 9 = 0
Now, you'll see that a unique solution exists: 14 z = 9 To solve the system at this point, you might want to write your equations as follows:
x + y + 2 z = 2 0 x + 3 y 42 z = 18 0 x + 0 y + 14 z = 9
aniawnua

aniawnua

Beginner2022-05-24Added 5 answers

Your first idea is off, because, for example the system
x + y + z = 1 , x + 2 y = 1 , x + y + 2 z = 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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