Solving Gaussian Elimination Problem: 3x - y + z = 1

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Homework Help Overview

The problem involves solving a system of linear equations using Gaussian elimination. The equations presented are: 3x - y + z = 1, 2x + 2y – 5z = 0, and 5x + y – 4z = 7.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply Gaussian elimination step-by-step but encounters inconsistencies in their calculations, leading to a conclusion of no solutions. Some participants question the arithmetic accuracy in the steps taken. Others suggest checking the determinant of the coefficient matrix as a method to verify the solvability of the system.

Discussion Status

Participants are actively discussing the arithmetic errors made during the elimination process. There is recognition of the determinant method as a valid approach to assess the system's solvability, and some express a desire to apply this method to other problems. Multiple interpretations of the equations' relationships are being explored, particularly regarding the implications of similar equations leading to contradictions.

Contextual Notes

There is mention of previous instruction on using determinants to determine the solvability of systems of equations, which may influence participants' approaches to the problem.

sara_87
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Question:

solve by using gaussian elimination:

3x - y + z = 1
2x + 2y – 5z = 0
5x + y – 4z = 7

what i did:

step 1: new row 1 = old row 1 – row 2, I got:

x – 3y + 6z = 1
2x + 2y – 5z = 0
5x + y – 4z = 7

step 2: new row 2 = old row 2 – (2 * row1) and new row 3 = old row 3 – (5 * row 1), I got:

x – 3y + 6z = 1
0 + 8y – 17z = -2
0 + 16y – 34z = 2

step 3: new row 3 = old row 3 * (1/2) I got:

x – 3y + 6z = 1
0 + 8y – 17z = -2
0 + 8y – 16z = 1

step 4: new row 3 = old row 3 – row 2, I got:

x – 3y + 6z = 1
0 + 8y – 17z = -2
0 + 0 + z = 3

the Real Answer:

Gaussian elimination gives 0z = 6, ie, 0 = 6 which is clearly impossible.
NO solutions.



i obviously didn’t get that, and i would really appreciate it if someone could check whether i am going wrong somewhere or if my teacher is wrong

thanx
 
Last edited:
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For one thing in step 3, 34/2=17 not 16. You are just making arithmetic errors.
 
you know i did that question 3 times and i made that mistake three times how stupid of me lol!
i was never meant to do maths
thanx for your help!
 
The same thing happens to me. You're welcome.
 
sara_87 said:
you know i did that question 3 times and i made that mistake three times how stupid of me lol!
i was never meant to do maths
thanx for your help!

by the way, the ability to do maths is somewhat different from whether you are meticulous.

the quick way to check whether a system of equations are solvable, find the determinant of the matrix, in this case the determinant of
\begin{pmatrix}3&-1&1\\ 2&2&-5\\ 5&1&-4\end{pmatrix}
is actually 0 , ie. the matrix is non-invertible, so no solutions.
 
yes you're right! thanks i'll find the determinant for the rest of the questions (we were actually taught that method - but I've just been winding my self up)
 
3x - y + z = 1
2x + 2y – 5z = 0
5x + y – 4z = 7

look by subtracting 2nd from 3rd it gives
3x-y+z= 7

so there a two similar equations
3x - y + z = 1
3x-y+z= 7

so this question don't have an answer
 

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