Let $*$ denote the binary operation on a set $\mathbb{R}^3$ of ordered triples of real numbers defined such that: $(A_0,A_1,A_2) * (B_0,B_1,B_2)= (A_0B_0,A_0B_1 + A_1B_0 ,A_0B_2 + A_1B_1 + A_2B_0)$
for all $(A_0,A_1,A_2),(B_0,B_1,B_2)\in\mathbb{R}^3$.
I have proved that the set above is a monoid and I have worked out its identity element to be $(1,0,0)$.
I am trying to determine which of its elements are invertible and that's what I just can't figure out yet. This is what I have done
$(A_0,A_1,A_2) * (B_0,B_1,B_2)= (1,0,0)$
where $(B_0,B_1,B_2)$ is the inverse of $(A_0,A_1,A_2)$, so:
$(A_0B_0,A_0B_1 + A_1B_0 ,A_0B_2 + A_1B_1 + A_2B_0) = (1,0,0)$
Then:
$A_0B_0=1 \implies B_0=\frac{1}{A_0}$
$A_0B_1 + A_1B_0=0 \implies B_1 =-\frac{A_1B_0}{A_0}$
$A_0B_2 + A_1B_1 + A_2B_0=0 \implies B_2= \frac{-A_2B_0 -A_1B_1}{A_0}$
I do not know what to do next. I want to find which of the elements are invertible and what their inverses are.