# Gaussian elimination system of Eqs, the case of no solution

• late347
In summary: Two equations are dependent if a number can be found such that you get the other when multiplying the first with that number.Like ##2x + 2y = 4## and ##x + y = 2##: basically only one equation instead of 2.
late347

## Homework Statement

system of equations is as follows
4x +2y -2z = 0
2x + y -z= 1
3x +y -2z = 1

## The Attempt at a Solution

[/B]

Using gaussian elimination

we can multiply mid-eq, by (-2) [[[actually... it is simply a basic equation procedure]]]
2x+y-z=1 |*(-2)
=
-4x -2y +2z = -2

further with using gaussian elimination (?) we sum together
4x+2y-2z=0
-4x-2y+2z=-2
result
0=-2´my teacher said something to the effect that, it is concluded that the system of equations does not have solutions... because a non-sensical result came out of the procedure.

Now, my teacher gave me some kind of proof for the Gaussian elimination method, but I'm still little bit uncertain why the "untrue equation midresult" causes the entire system of eqiuation to not have solution...
Our course ended today also, so I can't ask my teacher except by email.

perhaps it's a dumb question but anyway...

1.) By what reasoning is it arrived to this conclusion that when a "non-sensical equation result" comes out from the procedure, that this "mid-result" if you allow me to call it that, causes the system of equations to not have solution (Therefore, you don't have to calculate any further using Gaussian elimination?)

sorry if I failed to think about the problem rigorously enough as required by homework forum rules.

If you look at the first two equations: an ##\ (x,y,z) \ ## that satisfies equation two would satisfy ##\ 4x +2y-2z=2\ ## ( a simple multiplication by 2) which makes it impossible to satisfy the first equation.

The wording is: equations 1 and 2 are not independent.

BvU said:
If you look at the first two equations: an ##\ (x,y,z) \ ## that satisfies equation two would satisfy ##\ 4x +2y-2z=2\ ## ( a simple multiplication by 2) which makes it impossible to satisfy the first equation.

The wording is: equations 1 and 2 are not independent.

I can see how when you compare the equations...

One of the equations equals zero
And the same equation equals two... seems difficult to find an (x, y, z)

What does that even mean equations are not independent?

late347 said:
I can see how when you compare the equations...

One of the equations equals zero
And the same equation equals two... seems difficult to find an (x, y, z)

What does that even mean equations are not independent?

If you set ##2x+y-z = w## the first two equations say ##2w = 0## and ##w = 1##. There is very obviously NO possible solution.

late347 said:
I can see how when you compare the equations...
What does that even mean equations are not independent?

Two equations are dependent if a number can be found such that you get the other when multiplying the first with that number.
Like ##2x + 2y = 4## and ##x + y = 2##: basically only one equation instead of 2.
For more equations: if you can manipulate a group of one or more equations to get one of the remaining equations.
(manipulate: mutiply all terms with a number, add, etc.)​

Maybe my classification wasn't correct: when you multiply the first by 1/2 you don't exactly get the second equation, only the same coefficients for ##x,y## and ##z##.
Perhaps the term for ##2w = 0## and ##w = 1## is 'conflicting' or 'contradictory' . Can a native english purist mathemagician help me out?

## 1. What is Gaussian elimination system of equations?

Gaussian elimination is a method used in linear algebra to solve a system of linear equations. It involves transforming the system into an equivalent triangular system, making it easier to solve for the unknown variables.

## 2. How does Gaussian elimination work?

Gaussian elimination works by using elementary row operations to transform the system of equations into an equivalent triangular system. These operations include multiplying a row by a constant, adding a multiple of one row to another, and swapping rows. This process continues until the system is in its triangular form, at which point the unknown variables can be easily solved for.

## 3. What is the case of no solution in Gaussian elimination?

The case of no solution in Gaussian elimination occurs when the system of equations is inconsistent, meaning there is no set of values for the variables that will satisfy all of the equations simultaneously. This can happen if the equations contradict each other or if there are more unknown variables than equations.

## 4. How can you tell if a system of equations has no solution?

If the triangular form of the system contains a row of all zeros except for the last column, then the system has no solution. This indicates that the equations are contradictory and cannot be satisfied.

## 5. Can Gaussian elimination be used to solve systems of non-linear equations?

No, Gaussian elimination can only be used to solve systems of linear equations. Non-linear equations involve variables with exponents or other non-linear terms, which cannot be solved using this method. Other techniques, such as substitution or elimination, must be used to solve non-linear systems of equations.

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