Linear Algebra - Matrix Inverse Prove that if $A$ is invertible, and $B$ can be obtained from $A$ by applying elementary row operations, then $B$ is invertible. 
This is what i did:
Let $A$ be any nxn square matrix, 
Given: $E_{k}E_{(k-1)}...E_{2}E_{1}A=B$
We know that each Elementary Matrix is invertible because it can be obtained by applying one elementary row operation to identity matrix. 
For $k=1$, $$EA=B$$
Given that $A$ and $E$ are both invertible, 
$$(EA)^{-1}=A^{-1}E^{-1}$$
then, $$(B)^{-1}=A^{-1}E^{-1}$$
For $k=2$, $$E_{2}E_{1}A=B$$
Given that $A$ and $E_{1}$ and $E_{2}$ are invertible,
$$(E_{2}E_{1}A)^{-1}=(E_{2}(E_{1}A))^{-1}=(E_{1}A)^{-1}E_{2}^{-1}=A^{-1}E_{1}^{-1}E_{2}^{-1}$$
then, $$(B)^{-1}=A^{-1}E_{1}^{-1}E_{2}^{-1}$$
For $k>2$, inductively we can assume that, $$(E_{k}E_{(k-1)}...E_{2}E_{1}A)^{-1}=A^{-1}E_{1}^{-1}E_{2}^{-1}...E_{k-1}^{-1}E_{k}^{-1}$$
Then, $$B^{-1}=A^{-1}E_{1}^{-1}E_{2}^{-1}...E_{k-1}^{-1}E_{k}^{-1}$$
Therefore, $B$ is invertible.
Does this proof makes sense. 

There are 2 issues: there is no need for formal induction and, in each case, you're asserting that $B$ is invertible and then pointing out the correct inverse. To fix the second issue, you should propose a candidate and verify that it is the inverse of $B$.
So let $B=E_kE_{k-1}\cdots E_1A$ and consider $D\equiv A^{-1}E_1^{-1}\cdots E_k^{-1}$. Then:
$$
BD=\Big(E_kE_{k-1}\cdots E_1A\Big)\Big(A^{-1}E_1^{-1}\cdots E_k^{-1}\Big)=I_n
$$
where the last equality uses the associativity of matrix multiplication. Similarly,
$$
DB=\Big(A^{-1}E_1^{-1}\cdots E_k^{-1}\Big)\Big(E_kE_{k-1}\cdots E_1A\Big)=I_n.
$$
So $B$ is indeed invertible with $D$ being its inverse.
