If A is invertible and A is row equivalent to B, prove B is also invertible I go

Question

If A is invertible and A is row equivalent to B, prove B is also invertible

I got..

since A is invertible, A^-1 exists and since A is row equivalent to B, there exists invertible elementary matrices E1,E2,...,Ek such that B = E1E2E3...EkA

this implies that (E1^-1)(E2^-1)(E3^-1)..(Ek^-1) exists because every elementary matrix has an inverse.. This is as far as I got..Please continue..Also, I see that one of the next steps is showing    B^-1 = (E1E2..EkA)^-1..If you do not know, at least tell me why this last step is so, as in what do you mulitply of both sides of the above equation to get that..Thank You!!

Explanation / Answer

Since A is invertible, A^(-1) exists.

Since B is row-equivalent to A, there exist invertible elementary matrices E_1, E_2, ...., E_n such that B = E_1*E_2*...*E_n*A. Note that E_1^(-1), (E_2)^(-1), ... E_n^(-1) exist.

B^(-1) = (E_1*E_2*...*E_n*A)^(-1)
= A^(-1) * (E_n)^(-1) * ... * (E_2)^(-1) * E_1^(-1).

Since A^(-1), E_1^(-1), (E_2)^(-1), ... , E_n^(-1) exist, B^(-1) exists and so B is invertible.

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