Asked by Nicholas
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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Answer by Jan Simon
on 30 Sep 2013

Edited by Jan Simon
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.

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