Let given$M =$ 1     0     1 0     1     1 1     1     1and$b =$ 1 0 1How to find the solution $x_3$ where $x=${$x_1,x_2,x_3$}? Solution: Based on Wiedemann algorithm we has $u_i=[0; 0; 1]$. I am getting confused in the Step2. How to find $M_s$? Is it just the $3^th$ columns of $M_t$ as $M_s=[1 1 1]$ or it is created from Krylov sequence? Please help me find M_s and minimal polynomial? Matlab code:%% Input Generation Matrix M, b and index ith%% Output single symbol xM=[1 0 1;0 1 1;1 1 1];b=[1; 0; 1];index=2;[K N]=size(M);Mt=M';Imatrix=eye(K,K);x=[];for index=1:Ku_i=Imatrix(:,index);se_Krylov=[];se_Krylov(:,1)=u_i;    for i=2:(2*K) % due to i=1 is u_i       se_Krylov(:,i)=mod((Mt^i)*u_i,2);    end%% M_s is the operator Mt retricted to S    M_s=se_Krylov(index,:);%% Compute the polynominal using Berlekmap Massey[f, LCP] = Berlekamp_Massey2(M_s);d=size(f,2)-1; %deg of f: x^d+x^(d-1)+....x_comma=zeros(K,1);    for i=d:-1:1        x_comma=mod(x_comma+f(d).*(Mt)^(i-1)*u_i,2);    endx_single=mod(x_comma'*b,2) ;%Inner product x(index)=x_single;end
Find the solution of linear equation using Wiedemann/ Krylov method
matlab;linear algebra