## Doing back sunstitution for Gauss Elimination

### Nicholas (view profile)

on 30 Sep 2013
Latest activity Edited by Jan Simon

### Jan Simon (view profile)

on 30 Sep 2013

I have Anew=[2 1 -1;0 2.5 -1.5;0 0 2.4] and bnew=[0;-3;4.8]

using back substitution only i need to find my x1, x2 and x3 values. I cant use inv(Anew)*bnew or rref() i need my program to use back projection. I think i need a for loop or something but i'm stuck and out of ideas at the moment please help. answers should be...

x1=1

x2=0

x3=2

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### Jan Simon (view profile)

on 30 Sep 2013
Edited by Jan Simon

### Jan Simon (view profile)

on 30 Sep 2013

Write down the equation on paper at first (I omit the "new" here):

```A * x = b
```
```[2 1   -1;       [x1;     [0;
0 2.5 -1.5;  *   x2;  =   -3;
0 0   2.4]       x3]      4.8]
```

You have a triangle matrix on the left, such the back-substitution can be applied directly. Start to solve the last row to get x3:

```0*x1 + 0*x2 + 2.4*x3 = 4.8
```

Use the result to solve the 2nd last row to get x2 and so on. A short look into WikiPedia might be useful also.