Sets with algebraic operations

In item 2, as an example on $\mathbb {Z}$, the operation $*: n * m=-n-m$ was introduced, which is commutative but not associative. In the algebraic structure $(\mathbb {Z}, *)$ the relations $(n * m) * m=n, m *(m * n)=n$ hold. Suppose now we are given an arbitrary algebraic structure $(X, *)$, in which $(x * y) * y=x, y *(y * x)=x$ for all $x, y \in X$. Prove that $x * y=y * x$, i.e., that the operation * is commutative.

Show that the set

with the usual operation of matrix multiplication is a semigroup. Is $\left(M_{n}^{0}(\mathbb {R}), \cdot \right)$ a monoid?

In a multiplicative monoid $M$, an arbitrary element $t$ is chosen and a new operation $*: x * y=x t y$ is introduced. Show that $(M, *)$ is a semigroup and that the invertibility of the element $t$ in $M$ is a necessary and sufficient condition for $(M, *)$ to be a monoid with neutral (identity) element $t^{-1}$.

Show that the set $\mathbb {Z}$ with the operation $\circ : n \circ m=n+m+n m=(1+n) \times (1+m)-1$ is a commutative monoid. What serves as the neutral element in $(\mathbb {Z}, \circ )$? Find all invertible elements in $(\mathbb {Z}, \circ )$.