Given a quartet $(s,p,m, n)$ and a law
$$(s,p,m,n).(q,r,t,u)=(\frac{2}{3}sq, pr,m+(1-t), nu)$$
where $s,p,m,n,q,r,t$ and $u$ are in $\mathbb{R} \setminus \{0\}$.
The left and right identities have been found to be different which is ok. I believe these to be $(3/2,1,1,1)$ for the right identity and $(3/2,1,-1,1)$ for the left identity. Please confirm this to be true.
How does one prove that every element obeying this law has an inverse or perhaps even a unique inverse (perhaps on the left and on the right)?
I want to reason that we start with a declaration (theorem?) that
$$(s,p,m,n)(q,r,t,u) = (3/2,1,1,1)$$
An example:
$$(s,p,m,n )(1,2,3,4) =(3/2,1,1,1)$$
Then $(s,p,m,n) =(9/4,1/2,3,1/4)$
$$(s,p,m,n)(q,r,t,u) = (3/2,1,-1,1)$$
$$(9/4,1/2,3,1/4)(q,r,t,u) = (3/2,1,-1,1)$$
then $(q,r,t,u) = (1,2,5,4)$
Please check if these are even true. From checking it definitely looks like for every element there is a unique left and right inverse. What is a good way of proving this?
Here $(3/2, 1, -1,1)$ is not a left identity, since
$$(3/2, 1, -1, 1)\cdot(q,r,t,u)=(q, r, \color{red}{-t}, u).$$