1. ## Gaussian Elimination

Determine the values of s, v and a using:

(A) Gaussian elimination
(B) Inverse matrices

3s + v – 2a = -5
2s – 2v + 3a = 22
s + 3v – a = -11

I may have to change the order, Ive given it ago but get stuck, I know the three methods. And I need to get the bottom left three zero.

Any help would be great! Thank you

2. Why don't you write out as many steps as you can so we can see what the difficulty is that you're having.

3. ok i will do this for you thank you

Originally Posted by size
Determine the values of s, v and a using:

(A) Gaussian elimination
(B) Inverse matrices

3s + v – 2a = -5
2s – 2v + 3a = 22
s + 3v – a = -11

I may have to change the order, Ive given it ago but get stuck, I know the three methods. And I need to get the bottom left three zero.

Any help would be great! Thank you
The basic idea is to first eliminate one of the variables using two of the equations, then eliminate another variable again using two of the equations, then use the derived equations to finally figure out the value of one of the two remaining values and work backwards to get s, v and a. To get the left corner numbers zero, you can use the first equation to eliminate s from the next two equations, then use the derived second equation to eliminate v from the third equation.

Does this make sense to you? Can you take it from here? (with respect to inverse matrices, it's been awhile since I've done linear algebra which is where you should have posted your question and I would refer you to your text and class notes).

5. Originally Posted by wonderboy1953
The basic idea is to first eliminate one of the variables using two of the equations, then eliminate another variable again using two of the equations, then use the derived equations to finally figure out the value of one of the two remaining values and work backwards to get s, v and a. To get the left corner numbers zero, you can use the first equation to eliminate s from the next two equations, then use the derived second equation to eliminate v from the third equation.
You might want to interchange the first and third equation first (or equivalently interchange the first and third row in the augmented matrix). This will simplify the first few computations.