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...
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.