Commutative Monoid - matrix set

Question

Let $M$={$\begin{bmatrix} a & b & c \\ c & a & b \\ b & c & a \end{bmatrix}|a,b,c\in \mathbb{R}, a+b+c=0$}. The matrices in $M$ are a special kind of Toeplitz matrices (circulant). It is easy to show that $M$ is closed under the regular operation "$\cdot$" of multiplication. Associativity and commutativity follow in the same manner. When trying to prove that $(M,\cdot)$ is a commutative monoid, a problem arose. So we are searching for an element $I\in M$ such that $AI=IA=A,\forall A\in M$. Denoting $A=\begin{bmatrix} a_1 & b_1 & c_1 \\ c_1 & a_1 & b_1\\ b_1 & c_1 & a_1 \end{bmatrix}$ and B=$\begin{bmatrix} x & y & z\\ z & x & y\\ y & z & x \end{bmatrix}$, after calculations I obtained the following system: $$\left\{\begin{matrix} (a_1-b_1)x-(a_1+2b_1)y=a_1 \\ (2a_1+b_1)x+(a_1-b_1)y=a_1+b_1 \end{matrix}\right.\;,$$ where $a_1,b_1$are real numbers (varying) and $x, y$ are the unknowns (we need to find $x,y$ such that this solution(s) work for all $a,b$). It seems that $x=\frac{2}{3}, y=\frac{-1}{3}$ are good and then, from $x+y+z=0$ we get $z=\frac{-1}{3}$. My question is: how can we prove that this is the only good solution (so that $$I=-\begin{bmatrix} \frac{2}{3}&\frac{1}{3} &\frac{1}{3} \\ \frac{1}{3}& \frac{2}{3} &\frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} & \frac{2}{3} \end{bmatrix}$$

Answer 1

This follows directly from the system of equations that you derived. Since the system must be satisfied for arbitrary $a_1$, $b_1$, the coefficients of $a_1$ and $b_1$ must vanish separately. (Equivalently, the equations you get when substituting $a_1=1$, $b_1=0$ and when substituting $a_1=0$, $b_1=1$ must be satisfied.) This gives a system of equations which can easily be shown to have only that one solution.