Let r(A)=r 1,? r(B)=r2，

Then there are n-r 1 solution vectors in the basic solution system of Ax=0, and n-r2 solution vectors in the basic solution system of Bx=0. Because all solutions of Ax=0 are solutions of Bx=0, the n-r 1 solution vectors in the basic solution system of Ax=0 can be expressed linearly by the n-r2 solution vectors in the basic solution system of Bx=0. =n-r2, so r 1≥r2. That is, rank (A)≥ rank (b). So ①? Is correct.

If Ax=0 and Bx=0 have the same solution, then the n-r 1 solution vector in the basic solution system of Ax=0 is equivalent to the n-r2 solution vector in the basic solution system of Bx=0, so n-r 1=n-r2,? So r 1=r2. Namely rank (A)= rank (b).

②? ④? None of them are correct. You can cite counterexamples. For example: