# Show that if the product of the determinants of two square matrices of order $n$ is non-zero, both matrices are row-equivalent.

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.

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 $$I$$

• Thank you; that was neat, succinct and clear. Commented Sep 25, 2018 at 0:42