reserve x,y for set,
  i,j,k,l,m,n for Nat,
  K for Field,
  N for without_zero finite Subset of NAT,
  a,b for Element of K,
  A,B,B1,B2,X,X1,X2 for (Matrix of K),
  A9 for (Matrix of m,n,K),
  B9 for (Matrix of m,k,K);
reserve D for non empty set,
  bD for FinSequence of D,
  b,f,g for FinSequence of K,
  MD for Matrix of D;

theorem Th52:
  for B be Matrix of K st Solutions_of(0.(K,n,k),B) is non empty
  holds B = 0.(K,n,width B)
proof
  let B be Matrix of K;
  set A=0.(K,n,k);
  set ZERO=0.(K,n,width B);
  assume Solutions_of(0.(K,n,k),B) is non empty;
  then consider x being object such that
A1: x in Solutions_of(0.(K,n,k),B);
A2: len A=n by MATRIX_0:def 2;
A3: dom A=Seg n;
A4: len ZERO=n by MATRIX_0:def 2;
  then
A5: len B=len ZERO by A1,A2,Th33;
  then reconsider B9=B as Matrix of n,width B,K by A4,MATRIX_0:51;
A6: ex X st X=x & len X = width A & width X = width B & A * X = B by A1;
  now
    let i such that
A7: 1<=i & i<=n;
A8: width A=k by A7,MATRIX_0:23;
A9: i in Seg n by A7;
    then Line(A,i) = A.i by MATRIX_0:52
      .= width A|->0.K by A9,A8,FINSEQ_2:57;
    then width B|-> 0.K = Line(B,i) by A1,A6,A3,A9,Th41
      .= B9.i by A9,MATRIX_0:52;
    hence B.i=ZERO.i by A9,FINSEQ_2:57;
  end;
  hence thesis by A4,A5;
end;
