# Gaussian Elimination

1. Jan 23, 2007

### 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

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: Jan 23, 2007
2. Jan 23, 2007

### Dick

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

3. Jan 23, 2007

### 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

4. Jan 23, 2007

### Dick

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

5. Jan 23, 2007

### mjsd

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.

6. Jan 23, 2007

### sara_87

yes you're right! thanx 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)

7. Jul 26, 2009

### 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