Solve the primal solution:
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min z = -3x_1 + x_2 + x_3 + 0x_4 + 0x_5
subjec to   (AA)x = bb
                x nonnegative
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By hand calculation, one need to use artifial variables 
and the two phase method.

c = [0 0 0 0 0 1 1]
A =[]; b=[];
AA = [1 -2 1 1  0 0 0
-4  1 2 0 -1 1 0
-2  0 1 0  0 0 1];
bb = [11 3 1]';
LB=[0 0 0 0 0 0 0];
UB = [];
[x,fval] = linprog(c,A,b,AA,bb,LB,UB)

Ans: x = [0 1 1 12 0 0 0]', z = 0.
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Then we use the second, third, fourth column of A to form

B = [-2 1 1
1 2 0
0 1 0];
AA = B^(-1)*AA;
AA = AA([1,2,3],[1,2,3,4,5]);
bb = B^(-1)*bb;
c = [-3, 1, 1, 0, 0];
LB =[0 0 0 0 0]
UB = []
[x,fval] = linprog(c,A,b,AA,bb,LB,UB)

Ans: x = [4,1,9,0,0]', z = -2.
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In Matlab, we can directly use:

c = [-3 1 1 0 0];
A =[]; b=[];
AA =  [1 -2 1 1  0 
-4  1 2 0 -1
-2  0 1 0  0];
bb = [11 3 1]';
LB=[0 0 0 0 0];
UB = [];
[x,fval] = linprog(c,A,b,AA,bb,LB,UB)

Ans: x = [4,1,9,0,0]', z = -2

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Now, we use the first, second, third column to form

B = [1 -2 1
-4 1 2
-2 0 1];

B^(-1)*[AA b],
y = [-3,1,1]*B^(-1),
c - c_B*AA

ans =

    1.0000    0.0000         0    0.3333   -0.6667
         0    1.0000         0         0   -1.0000
         0    0.0000    1.0000    0.6667   -1.3333

y =

   -0.3333    0.3333    0.6667

ans =

    0.0000   -0.0000         0    0.3333    0.3333

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Set y^t = c_B B^(-1).
Then  c^t - y^tA is nonnegative.
So,  c - A^t y is nonnegative, i.e., y is dual feasible.
One can check  y^t b = c_B  B^(-1)b = z = the primal optimal.
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