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.
Suppose that $\langle a_0,a_1,a_2\rangle*\langle b_0,b_1,b_2\rangle=\langle 1,0,0\rangle$; then
$$\left\{\begin{align*} &a_0b_0=1\\ &a_0b_1+a_1b_0=0\\ &a_0b_2+a_1b_1+a_2b_0=0\;. \end{align*}\right.$$
This clearly implies that $a_0\ne 0$ and $b_0=\dfrac1{a_0}$, so the system becomes
$$\left\{\begin{align*} &a_0b_1+\frac{a_1}{a_0}=0\\ &a_0b_2+a_1b_1+\frac{a_2}{a_0}=0\;. \end{align*}\right.$$
The first of these equations implies that $b_1=-\dfrac{a_1}{a_0^2}$ and imposes no restriction on $a_1$. The second then reduces to
$$a_0b_2-\frac{a_1^2}{a_0^2}+\frac{a_2}{a_0}=0\;,$$
which can evidently be solved for $b_2$ without imposing any conditions on $a_2$. Thus, $\langle a_0,a_1,a_2\rangle$ is invertible precisely when $a_0\ne 0$, and I’ve computed most of the inverse above.