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

[sara_87]

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

 

[Dick]

For one thing in step 3, 34/2=17 not 16. You are just making arithmetic errors.

 

[sara_87]

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!

 

[Dick]

The same thing happens to me. You're welcome.

 

[mjsd]

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.

 

[sara_87]

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)

 

[gahasitha]

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