Why does Matlab have a different solution than mine? I tried other software, and they give me the same solution.

%matrix A, and vector of constants b
A = [2 1 0 0 0; 1 1 -1 0 -1; -1 0 1 0 1; 0 -1 0 1 1; 0 1 1 1 1]
b = [100; 0; 50; 120; 0]

%augmented matrix
Ab = [A b]

%row reduction
[rowreducedAb, pivotvarsAb] = rref(Ab)

%free and basic variables
[numqns, numvars] = size(A)
[numrows, numpivotvars] = size(pivotvarsAb)
numfreevars = numvars - numpivotvars

%LU decomposition of A
[L,U] = lu(A)
y = L\b

%inverse of U
invU = inv(U)

%original solution
x = invU*y

%cramer's rule
A1 = A
A1(:,1) = b
x1 = det(A1)/det(A)

My Solution:

Matrix A Vector of unknows Vector of constants

2 1 0 0 0 x1 100

1 1 -1 0 -1 x2 0

-1 0 1 0 1 x3 50

0 -1 0 1 1 x4 120

0 1 1 1 1 x5 0

Fourth Step: Gaussian Eliminations;

B + C -> B x2 = 50

A - B -> A x1 = 25

D + E -> D

D - C -> D

D) x1 + x5 = 70 x5 = 45

C) -x1 + x3 + x5 = 50 x3 = 25 - 45 + 50 x3 = 30

E) x2 + x3 - x4 + x5 = 0 50 + 30 + 45 = x4 x4 = 125

Answer 1

[Walter Roberson]

Back substitute on your proposed solution

>> A*[25;50;30;125;45]

ans =

   100
     0
    50
   120
   250

The first four entries match b, but the last entry of b is 0 not 250. Your solution is incorrect.

>> A\b
ans =

                        25
                        50
                      -220
                      -125
                       295

>> A*[25;50;-220;-125;295]
ans =
   100
     0
    50
   120
     0

MATLAB's solution works.

[Matteo Geraci]

x5 was 0 1 1 -1 1. I forgot the "-" sign. Thank you for your response.