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

AI Thread Summary
The discussion focuses on solving a system of equations using Gaussian elimination, specifically the equations 3x - y + z = 1, 2x + 2y - 5z = 0, and 5x + y - 4z = 7. The user initially attempts the elimination process but makes arithmetic errors, leading to confusion about the solution. Ultimately, it is determined that the system has no solutions, as indicated by the determinant of the matrix being zero, which signifies that the matrix is non-invertible. The user acknowledges the mistakes and expresses a desire to improve their mathematical skills. The conversation highlights the importance of careful calculations in solving linear equations.
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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