We are given a hint, that for two such matrices $A$ and $B$:

$A$ and $B$ are row-equivalent $\iff \exists$ a product of elementary matrices, $\prod_{i = 1}^{m}{E_{i}}$ such that $B = \prod_{i = 1}^{m}{E_{i}}\cdot A$.

How would I go about showing this? I was thinking of using the hint to represent B in terms of A, and then showing that $\prod_{i = 1}^{m}{E_{i}}$ must be non-zero, but my argument looks like it's somewhat circular.


1 Answer 1


The product of determinant is non-zero iff determinant of each is non-zero iff both are invertible

You only need to prove that all invertible matrices are row equivalent to the identity matrix.

As you wrote, $B = \prod_{i = 1}^{m}{E_{i}}\cdot A$.

Now, since every elementary matrix is invertible, you can write $A = E_1^{-1}E_2^{-1} \cdots E_m^{-1}B$

As the product of invertible matrices is invertible, $A$ is invertible iff $B$ is invertible.

Since $B$ is row reduced echelon matrix (every row has non-zero entry and a leading $1$) and invertible, it has to be the identity matrix $I$

  • 1
    $\begingroup$ Thank you; that was neat, succinct and clear. $\endgroup$
    – SRSR333
    Commented Sep 25, 2018 at 0:42

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