2.Use MATLAB/Octave to find the solution(s) toAx=b where(a)A=111111−1−11−100001−1b=1111,(b)A=1032−421650−11−3−11,b=47−5.(a)> A=[1 1 1 1; 1 1 -1 -1 ; 1 -1 0 0; 0 0 1 -1];> b=[1 1 1 1]’;> C=[A b];> rref(C)ans =1 0 0 0 10 1 0 0 00 0 1 0 0.50 0 0 1 -0.5and so the answer isx1x2x3x4=100.5−0.5We could have instead done> A=[1 1 1 1; 1 1 -1 -1 ; 1 -1 0 0; 0 0 1 -1];> b=[1 1 1 1]’;> A\bans =100.5-0.5and found the same answer — be careful though (see next part).(b)> A=[1 0 3 2 -4; 2 1 6 5 0; -1 1 -3 -1 1];> b = [4 7 -5]’;> C=[A b];> rref(C)ans =1 0 3 2 0 40 1 0 1 0 -10 0 0 0 1 0The solution isx1x2x3x4x5=4−1000+s−2−1010+t−301002