I consider the variety of quasigroups to be one of the main examples of reversible varieties since the quasigroups are simply the algebraic structures $(X,\cdot,\backslash,/)$ such that the mappings $(x,y)\mapsto(x\cdot y,y),(x,y)\mapsto(x/y,y)$ are inverses and the mappings $(x,y)\mapsto(x,x\cdot y),(x,y)\mapsto(x,x\backslash y)$ are also inverses. There are ways to generalize the variety of quasigroups to more complicated reversible varieties such as varieties of mutually orthogonal Latin squares.

The variety of quasigroups also generalizes to reversible varieties with operations of higher arity, and the finite algebras in some of these varieties are up-to-isomorphism in a many-to-one correspondence with the square matrices $A$ such that every entry in $A$ is a $0$ or a $1$ and each entry in $A^{k}$ is $1$. I don't think that anyone has studied these varieties before, but they seem to be interesting and maybe even useful in cryptography for block ciphers where the block size is $k$ times as long as the round key size.

**$k+1$-ary quasigroups**

Suppose that $X$ is a set and $$S_{1},\dots,S_{k},T_{1},\dots,T_{k}:X^{k+1}\rightarrow X,\Gamma_{1},\dots,\Gamma_{k}:X^{2k}\rightarrow X$$ are operations.
For each $i\in X$, define $f_{i}:X^{k}\rightarrow X^{k}$ by letting
$$f_{i}(u_{1},\dots,u_{k})=(S_{1}(i,u_{1},\dots,u_{k}),\dots,S_{k}(i,u_{1},\dots,u_{k})),$$ and define
$$g_{i}(u_{1},\dots,u_{k})=(T_{1}(i,u_{1},\dots,u_{k}),\dots,T_{k}(i,u_{1},\dots,u_{k})).$$ Define binary operations $\cdot,\backslash,/$ on $X^{k}$ by letting $(u_{1},\dots,u_{k})\cdot(v_{1},\dots,v_{k})=f_{u_{1}}\circ\dots\circ f_{u_{k}}(v_{1},\dots,v_{k})$ and
$$(u_{1},\dots,u_{k})\backslash(v_{1},\dots,v_{k})=g_{u_{k}}\circ\dots\circ g_{u_{1}}(v_{1},\dots,v_{k})$$ and
$$(u_{1},\dots,u_{k})/(v_{1},\dots,v_{k})=(w_{1},\dots,w_{k})$$ where we define
$w_{i}=\Gamma_{i}(u_{1},\dots,u_{k},v_{1},\dots,v_{k})$ for $1\leq i\leq k$. Then we say that $$(X,S_{1},\dots,S_{k},T_{1},\dots,T_{k},\Gamma_{1},\dots,\Gamma_{k})$$ is a $k+1$-ary quasigroup if $(X^{k},\cdot,\backslash,/)$ is a quasigroup and
$f_{i},g_{i}$ are inverses for each $i\in X$. The class of all $k+1$-ary quasigroups is a reversible variety. If $k=1$, then the $k+1$-ary quasigroups are precisely the quasigroups.

**Obtaining $k+1$-ary quasigroups from matrices**

We shall now discuss a way to obtain $k+1$-ary quasigroups from certain matrices. Let us begin with a couple lemmas.

Lemma: Suppose that $R,S$ are both $n\times n$-matrices. Suppose that the sum of each row in $R$ is $r$, the sum of each column in $R$ is $r$, the sum of each row in $S$ is $s$, and the sum of each column in $S$ is $s$. Then the sum of each row in $RS$ is $rs$, and the sum of each column in $RS$ is $rs$.

Lemma: Suppose that $A$ is an $n\times n$-matrix where each entry in $A$ is a $0$ or a $1$ and each entry in $A^{k}$ is $1$. Then $n=m^{k}$ for some natural number $m$ where the sum of each row of $A$ is $m$ and the sum of each column of $A$ is $m$ as well.

Proof: Suppose that $A=(a_{i,j})_{i,j}$. Then
$$A^{k+1}=A\cdot A^{k}=(a_{i,j})_{i,j}\cdot(1)_{i,j}=(\sum_{\ell}a_{i,\ell})_{i,j}.$$ Similarly, $$A^{k+1}=A^{k}\cdot A=(1)_{i,j}(a_{i,j})=(\sum_{\ell}a_{\ell,j})_{i,j}.$$ Therefore, we have $\sum_{\ell}a_{i,\ell}=\sum_{\ell}a_{\ell,j}$ for each $i,j$. Therefore, there exists some $m$ such that the sum of each row in $A$ is $m$ and the sum of each column in $A$ is $m$ as well. By the above lemma, we conclude that the sum of each row and column in $A^{k}$ is $m^{k}$. Therefore, $n=m^{k}$. Q.E.D.

Suppose now that $A$ is a matrix where every entry in $A$ is a $0$ or a $1$ and every entry in $A^{k}$ is $1$. If $f$ is a permutation, then let $\phi(f)$ denote its corresponding permutation matrix. Then there are permutations $f_{1},\dots,f_{m}\in S_{n}$ such that $A=\phi(f_{1})+\dots+\phi(f_{m})$ and such permutations $f_{1},\dots,f_{m}$ are easy to compute (you obtain such a decomposition using the Birkhoff-von-Neumann algorithm)).

By associating $[n]$ with $[m]^{k}$, we can assume that $f_{1},\dots,f_{m}$ are bijections from $[m]^{k}$ to $[m]^{k}$ such that whenever $u_{1},\dots,u_{k},v_{1},\dots,v_{k}\in[m]$ there is a unique tuple $(w_{1},\dots,w_{k})$ such that $$f_{w_{1}}\circ\dots\circ f_{w_{k}}(u_{1},\dots,u_{k})=(v_{1},\dots,v_{k}).$$ From the functions $f_{1},\dots,f_{m}$, we obtain our $k+1$-ary quasigroup $$([m],S_{1},\dots,S_{k},T_{1},\dots,T_{k},\Gamma_{1},\dots,\Gamma_{k})$$
by letting $S_{i}(j,u_{1},\dots,u_{k})=\pi_{i}f_{j}(u_{1},\dots,u_{k})$ and
$T_{i}(j,u_{1},\dots,u_{k})=\pi_{i}f_{j}^{-1}(u_{1},\dots,u_{k})$. Define operations $\Gamma_{1},\dots,\Gamma_{k}:[m]^{2k}\rightarrow[m]$ by letting
$\Gamma_{i}(u_{1},\dots,u_{k},v_{1},\dots,v_{k})=w_{i}$ for $1\leq i\leq k$ such that $$(f_{w_{1}}\circ\dots\circ f_{w_{k}})(v_{1},\dots,v_{k})=(u_{1},\dots,u_{k}).$$

Likewise, every $k+1$-ary quasi-group with underlying set $[m]$ produces a matrix $A$ where each entry in $A$ is a $0$ or a $1$ and each entry in $A^{k}$ is $1$.

There are plenty of non-trivial examples of matrices $A$ where each entry in $A$ is a $0$ or a $1$ and each entry in $A^{k}$ is $1$.