We have that $A$ and $B$ are $m\times n$ matrices, which are row equivalent. If we also have that $CA=B$, how to prove that $C$ is invertible?

  • 1
    $\begingroup$ It's not true in general: take $A$ and $B$ to be the zero matrix - then $C$ can be anything. In general, it depends upon the rank of $A$ and the dimensions $m,n$. $\endgroup$ Mar 26, 2014 at 7:42
  • $\begingroup$ Maybe you mean that you need to prove that there EXISTS such an invertible $C$? If so, please see my answer. $\endgroup$
    – DKal
    Mar 26, 2014 at 8:11

2 Answers 2


Not true. Take $ A = \pmatrix{a & a &a \cr b&b&b\cr 0 &0&0\cr}\quad B= \pmatrix{b&b&b\cr a & a &a \cr 0 &0&0\cr}.$ The (non-singular) permutation matrix $P$ can interchange first two rows and so they are row equivalent. Now define $C$ to be the matrix obtained from $P$ changing the last row to be the zero row. Then $CA=B$ is still true but $C$ is not invertible.


Hint: You probably want to show that the matrix $C$ is a product of elementary matrices (that is, it is a matrix that does the elementary row operations on $A$ to get to $B$).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .