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Am I right?

If there are 3 operations will call them E1,E2,E3

then A would be A = E1E2E3, or A = E3E2E1 A

for inverse if there are 3 operations to get the inverse then

$A^{-1} = E1^{-1}E2^{-1}E3^{-1} or A^{-1} = E3^{-1}E2^{-1}E1^{-1}A$

im a bit confused and my textbook is very very brief.

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    A row operation to $A$ amounts to multiplication $EA$ of some elementary matrix $E$ to the right of $A$. To say $A$ can be turned into the identity matrix after finitely many row operation is the same as saying there are elementary matrices $E_1,...,E_k$ such that $E_1 \cdots E_kA = \mathrm{Id}$. It will follow that $E_1 \cdots E_k$ is the multiplicative inverse of $A$ – 2017-02-16

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Order matters for matrix multiplication: $AB \neq BA$, in general.

$(AB)^{-1} = B^{-1}A^{-1}$

Check: $AB (AB)^{-1} = AB B^{-1}A^{-1}=I$

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    I'm just trying to understand all the elementary operation to get identity / inverse. I relooked at would this be right: $$E_4 E_3 E_2 E_ 1 A = I, \text{ then } E_4 E_3 E_2 E_1 = A^{-1}$$ How do do I get A? – 2017-02-16
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    @user349557 If you leftmultiply your original equation by $E_4^{-1}$ you have $E_4^{-1}E_4E_3E_2E_1A=E_4^{-1}I$. Does this simplify nicely? What more can you do to continue moving things to the other side? Assuming you know all of $E_1,E_2,\dots$ can you see how to use this information to find $A$? – 2017-02-16
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    Okay I understand so $A = E4^{-1}E3^{-1}E2^{-1}E1^{-1}I$ – 2017-02-16