Try by contradiction: if then ...
Can I claim that suppose one of the entries in both R and S are not the same, then using the same solution, claim that there is a contradiction and actually the entry(that I had supposed not the same) is the same?
That is suppose both R and S have the same second row but different first row with the 1,1 and 1,2 entries the same but 1,3 entry differs. Let the 1,3 entries of R and S be r and s respectively. Then if we have a solution and , we will have . But I am stucked here as if is zero, then my argument does not create contradiction.
Hi FernandoRevilla,
The R and S that you had given are in row-reduced form and the X that is given by you is also a solution for RX=0 and SX=0. Does this mean that what I want to prove(the initial question) is not correct?