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:

$$(A_0B_0,A_0B_1 + A_1B_0 ,A_0B_2 + A_1B_1 + A_2B_0) = (1,0,0)$$


$$A_0B_0=1 \implies B_0=\frac{1}{A_0}$$

$$A_0B_1 + A_1B_0=0 \implies B_1 =-\frac{A_1B_0}{A_0}$$

$$A_0B_2 + A_1B_1 + A_2B_0=0 \implies B_2= \frac{-A_2B_0 -A_1B_1}{A_0}$$

I do not know what to do next. I want to find which of the elements are invertible and what their inverses are.

  • $\begingroup$ You've actually already done all the real work that's needed. All that's left is to notice two things: (1) If $A_0=0$ then the $B_i$'s that you computed don't exist (because you can't divide by 0), so there is no inverse for $(A_0,A_1,A_2)$. (2) If $A_0\neq0$ then your formulas for the $B_i$'s give you an inverse for $(A_0,A_1,A_2)$. $\endgroup$ Dec 30, 2012 at 1:26

2 Answers 2


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


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.

  • $\begingroup$ thank you ijust cant seem to make the connection which implies that a0!=0 $\endgroup$
    – Jack welch
    Dec 29, 2012 at 23:47
  • $\begingroup$ @Jack: You have to have $a_0b_0=1$. If $a_0=0$, this is impossible. If $a_0\ne 0$, it simply requires that we set $b_0=\dfrac1{a_0}$. $\endgroup$ Dec 29, 2012 at 23:54

As you can see, $(A_0,A_1,A_2)$ has an inverse iff $A_0\not=0$. (Otherwise you wont be able to solve for an inverse)

Actually this monoid is ismorphic to $R[x]/<x^3>$ under multiplication

  • $\begingroup$ how do i list all the inverses for all invertible elements for this set $\endgroup$
    – Jack welch
    Dec 29, 2012 at 22:20
  • $\begingroup$ You want to find real numbers $(B_0,B_1,B_2)$ such that it is an inverse for $(A_0,A_1,A_2)$. The system of equations that you wrote will have a solution iff $A_0\not=0$ $\endgroup$
    – Amr
    Dec 29, 2012 at 22:25
  • $\begingroup$ This would mean, of course, that $B_0 \neq 0$ as well, yes? $\endgroup$
    – amWhy
    Dec 29, 2012 at 22:44
  • $\begingroup$ Sure. This is because $\frac{1}{A_0}\not=0$ $\endgroup$
    – Amr
    Dec 29, 2012 at 22:45
  • $\begingroup$ So it would seem any element $X = \langle X_0, X_1, X_2\rangle$ in this "monoid/ring" is invertible provided $X_0 \ne 0$; that is, those elements $Y = \langle 0, Y_1, Y_2\rangle$ are not invertible? $\endgroup$
    – amWhy
    Dec 29, 2012 at 22:50

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