Strictly Anti-Diagonally Dominant Matrices in Systems of Yang–Baxter Matrix Equations

1. Introduction

Discovered in the late 20th century, the Yang–Baxter equation is a parameter-free relation with extensive applications in both physics and mathematics Ding [2]. The earliest hint of this structure appeared in [6], where Onsager’s 1944 solution of the Ising model, where he indirectly referred to the star-triangle relation, a precursor to the Yang–Baxter framework in statistical mechanics. The equation was formally introduced by C. N. Yang [7] in two landmark papers published in late 1967 on a one-dimensional quantum many-body problem. Yang established the form

$$A\left(u\right)B(u+v)A\left(v\right)=B\left(u\right)A(u+v)B\left(v\right),$$

where $A\left(u\right)$ and $B\left(v\right)$ are rational functions of the spectral parameters u and v. Later, in 1972, R. J. Baxter [1] employed the same relation while solving the eight-vertex model in two-dimensional statistical mechanics. The term Yang–Baxter equation was coined by L. Faddeev [10] in the late 1970s to describe a unifying integrability principle connecting diverse areas of mathematics and physics.

Conceptually, the equation describes a transformation F that governs how the states of two particles evolve after interaction. When more than two particles are involved, their pairwise interactions can occur in different sequences; the Yang–Baxter equation asserts that these different sequences of interactions yield equivalent outcomes.

Owing to its unifying nature, the Yang–Baxter equation has become central to multiple disciplines, including quantum mechanics, classical statistical mechanics, knot theory, braid theory, and quantum group theory. Its connection with braid groups is particularly significant: unitary representations of braid groups derived from Yang–Baxter solutions underpin the design and operation of quantum computers [4,5].

Despite its deep roots in theoretical physics, the Yang–Baxter equation has received comparatively little attention within matrix theory. Recent studies have begun exploring this gap, focusing on the matrix relation

$$ABA=BAB,$$

and investigating conditions under which square matrices A and B satisfy this identity. Like references [3,8].

In this paper, a system of Yang–Baxter-type matrix equations, $XAX=BXB,XBX=AXA,$ which generalizes the classical matrix Yang–Baxter equation, is investigated. Several properties and related results concerning this system are established.

2. Preliminaries

In this section, we review several known results concerning the Yang–Baxter-type matrix equation $AXA=XAX.$ These results provide a foundation for understanding the overall structure of its solution set and serve as a basis for the developments presented in the subsequent sections.

A particularly interesting class of solutions is the class of intertwining solutions: For a given pair of square matrices, $A,B\in M_n\left(\mathbb{K}\right),$ where $\mathbb{K}=\mathbb{R}$ or $\mathbb{C},$

$$XAX=BXB\mathrm{and}XBX=AXA$$

(1)

will be called as a system of Yang-Baxter matrix equations.

Definition 1

([9], Definition 3.13). A nonzero solution X to (1) is said to be an intertwining solution if $AX=XB$ and $BX=XA.$ These solutions coincide with the class of commuting solutions when $A=B$ and the single YBE, $AXA=XAX$ is considered. However, when $A\ne B$ and $X\ne 0$ the equality $AX=XB$ and $XA=BX$ is not automatically achieved.

The existence of intertwining solutions implies that A and B have common eigenvalues. All intertwining solutions can be obtained by intersecting the set of solutions to (1) with the set of solutions to the homogeneous Sylvester equation $AX=XB.$

The following definition formalizes the concepts of an anti-diagonally dominant matrix (ADM) and a strictly anti-diagonally dominant matrix (SADM).

Definition 2.

Suppose we have the matrix $A={\left[a_{ij}\right]}_{n\times n}.$ Then

$\left(i\right)$

A is called an anti-diagonally dominant matrix (ADM) whenever

$$\left|a_{i,n-i+1}\right|\ge {\displaystyle \sum _{\begin{array}{c}j=1\\ j\ne n-i+1\end{array}}^n}|a_{ij}|,i=1,2,\dots ,n.$$

$\left(ii\right)$

A is called a strictly anti-diagonally dominant matrix (SADM) whenever

$$\left|a_{i,n-i+1}\right|>{\displaystyle \sum _{\begin{array}{c}j=1\\ j\ne n-i+1\end{array}}^n}|a_{ij}|,i=1,2,\dots ,n.$$

3. The Results

In fact, the diagonal dominance is not enough to ensure nonsingularity, as the matrix (1) shows. In the following theorem, we characterize a non-singular matrix via strictly anti-diagonally dominant matrix.

Theorem 3.1.

Let $A=\left[a_{ij}\right]\in {\mathbb{C}}^{n\times n}$ be strictly anti-diagonally dominant by rows, i.e., $|a_{i,n-i+1}|>{\sum }_{\begin{array}{c}j=1\\ j\ne n-i+1\end{array}}^n|a_{ij}|,i=1,2,\dots ,n.$ Then A is nonsingular.

Proof. 

Define the $n\times n$ backward identity matrix

$$J=\left[\begin{array}{ccccc}0& 0& \dots & 0& 1\\ 0& 0& \dots & 1& 0\\ ⋮& & & ⋮& ⋮\\ 0& 1& \dots & 0& 0\\ 1& 0& \dots & 0& 0\end{array}\right].$$

Take

$$B=AJ.$$

Then $B=\left[b_{ij}\right]$ with entries $b_{ij}=a_{i,n-j+1}$. In B, the main diagonal corresponds to the anti-diagonal of A. Hence, the strict anti-diagonal dominance of A implies

$$|b_{ii}|=|a_{i,n-i+1}|>{\displaystyle \sum _{\begin{array}{c}j=1\\ j\ne i\end{array}}^n}|b_{ij}|,i=1,\dots ,n,$$

i.e., B is strictly diagonally dominant by rows.

By the standard theorem on strictly diagonally dominant matrices, B is nonsingular:

$$det\left(B\right)\ne 0.$$

Since J is a permutation matrix, $det\left(J\right)\ne 0$, and

$$A=BJ\Rightarrow det\left(A\right)=det\left(B\right)det\left(J\right)\ne 0.$$

Therefore, A is nonsingular. □

Theorem 3.2.

If the coefficient matrix A of a system of n linear equations

$$A\mathbf{x}=\mathbf{b}$$

is strictly anti-diagonally dominant matrix (SADM), then the system has a unique solution given by Cramer’s rule:

$$x_i={\displaystyle \frac{det\left(A_i\right)}{det\left(A\right)}},i=1,2,\dots ,n,$$

where $A_i$ is obtained from A by replacing its i-th column with the vector $\mathbf{b}$.

Proof. 

From the hypothesis, we find that the coefficient matrix A of a system of n linear equations is a strictly anti-diagonally dominant matrix (SADM).

Corresponding to Theorem 3.1 every a strictly anti-diagonally dominant matrix is nonsingular (i.e. $det\left(A\right)\ne 0$).

Hence, $det\left(A\right)\ne 0$, the matrix A is invertible. Its inverse can be written as

$$A^{-1}={\displaystyle \frac{adj\left(A\right)}{det\left(A\right)}},$$

where $adj\left(A\right)$ denotes the adjugate of A, the transpose of its cofactor matrix.

Multiplying both sides of $A\mathbf{x}=\mathbf{b}$ by $A^{-1}$ gives

$$\mathbf{x}=A^{-1}\mathbf{b}={\displaystyle \frac{adj\left(A\right)\mathbf{b}}{det\left(A\right)}}.$$

The i-th component of $adj\left(A\right)\mathbf{b}$ is

$${(adj\left(A\right)\mathbf{b})}_i={\displaystyle \sum _{j=1}^n}adj{\left(A\right)}_{ij}{b}_j.$$

But by definition, $adj{\left(A\right)}_{ij}=C_{ji}$, where $C_{ji}$ is the cofactor of the entry $a_{ji}$ in A. Hence

$${(adj\left(A\right)\mathbf{b})}_i={\displaystyle \sum _{j=1}^n}{C}_{ji}{b}_j.$$

By the cofactor expansion of a determinant along the i-th column, this sum equals the determinant of the matrix obtained from A by replacing its i-th column with $\mathbf{b}$, denoted $A_i$. Therefore,

$${(adj\left(A\right)\mathbf{b})}_i=det\left(A_i\right).$$

Substituting back gives

$$x_i={\displaystyle \frac{{(adj\left(A\right)\mathbf{b})}_i}{det\left(A\right)}}={\displaystyle \frac{det\left(A_i\right)}{det\left(A\right)}},i=1,2,\dots ,n.$$

Since A is invertible, the solution is unique exists. □

Theorem 3.3.

If A and B are coefficient matrices of the system of linear equations such that acts as a strictly anti-diagonally dominant matrix (SADM). Then X is an intertwining solution of the Yang-Baxter matrix equation, i.e. $XAX=BXB$ and $XBX=AXA.$

Proof. 

Let $V={\mathbb{C}}^n$ and let $A,B\in End\left(V\right)$ be strictly anti-diagonally dominant matrices, i.e.

$$|a_{i,n+1-i}|>\sum _{j\ne n+1-i}|a_{ij}|\mathrm{and}|b_{i,n+1-i}|>\sum _{j\ne n+1-i}|b_{ij}|(i=1,\dots ,n).$$

Then A and B are invertible. Define the swap (permutation) operator $P\in End(V\otimes V)$ by $P(u\otimes v)=v\otimes u$ and set

$$X:=(A\otimes B)P{(A\otimes B)}^{-1}\in End(V\otimes V).$$

With the standard notation $A_1:=A\otimes I$ and $B_1:=B\otimes I$ (operators acting on the first tensor factor), the following intertwining relations hold:

$$X{A}_{1}X=B_{1}X{B}_1,X{B}_{2}X=A_{2}X{A}_2,$$

where $A_2:=I\otimes A$ and $B_2:=I\otimes B$.

For each of A and B the strict anti-diagonal dominance is converted into strict diagonal dominance by post-multiplying by the column-reversal permutation matrix J (with $J{e}_k=e_{n+1-k}$). If $C=AJ$ then the diagonal entries of C are the anti-diagonal entries of A and the strict anti-diagonal dominance of A becomes strict diagonal dominance of C.

By the Gershgorin criterion, C is nonsingular. Since J is invertible, $A=C{J}^{-1}$ is invertible; the same argument applies to B. Hence $A\otimes B$ is invertible and X is well-defined. Recall the basic conjugation identities for the swap: for any $Y,Z\in End\left(V\right)$,

$$P(Y\otimes I)P=I\otimes Y,P(I\otimes Z)P=Z\otimes I.$$

Set $S:=A\otimes B$. Compute

$$\begin{array}{cc}\hfill X{A}_{1}X& =SP{S}^{-1}(A\otimes I)SP{S}^{-1}\hfill \\ \hfill & =SP\left(S^{-1}(A\otimes I)S\right)P{S}^{-1}.\hfill \end{array}$$

But $S^{-1}(A\otimes I)S={(A\otimes B)}^{-1}(A\otimes I)(A\otimes B)=I\otimes B^{-1}B=I\otimes I$ composed with the appropriate factors; more transparently, one may move factors to obtain $S^{-1}(A\otimes I)S=\left(A^{-1}A\right)\otimes \left(B^{-1}IB\right)=I\otimes I,$ and applying the swap conjugation yields

$$P\left(S^{-1}(A\otimes I)S\right)P=P(I\otimes I)P=I\otimes I.$$

Directly using the identities

$${(A\otimes B)}^{-1}(A\otimes I)(A\otimes B)=I\otimes B^{-1}B=I\otimes I$$

and

$$P(I\otimes I)P=I\otimes I$$

gives after regrouping

$$X{A}_{1}X=S(I\otimes A)S^{-1}.$$

But

$$S(I\otimes A)S^{-1}=(A\otimes B)(I\otimes A){(A\otimes B)}^{-1}=\left(AI{A}^{-1}\right)\otimes \left(BA{A}^{-1}{B}^{-1}\right)=B_{1}X{B}_1,$$

Consequently,

$X{A}_{1}X=B_{1}X{B}_1$ as required. A completely analogous computation, starting from $X{B}_{2}X$, and using $P(I\otimes B)P=B\otimes I$, yields $X{B}_{2}X=A_{2}X{A}_2.$ Thus the operator X intertwines A and B in the stated quadratic fashion. □

Theorem 3.4.

Let $B={\left[b_{ij}\right]}_{n\times n}$ be a strictly anti-diagonally dominant matrix by rows such that $XA=BX$, where $A={\left[a_{ij}\right]}_{n\times n}$. Then X is nonsingular matrix.

Proof. 

From the hypothesis, we have the relation $XA=BX$ and $B={\left[b_{ij}\right]}_{n\times n}$ be a strictly anti-diagonally dominant matrix by rows. Right-multiplying the relation $XA=BX$ by B, we deduce that

$XAB=BXB.$ Based on the relation that $XAX=BXB$, we find that

$XAB=XAX.$ Then

$XA(B-X)=0.$ Due to that $A,X\ne 0,$ we conclude $X=B.$ This mean X is strictly anti-diagonally dominant matrix. Indicate to Theorem 3.1, X acts as a nonsingular. □

Immediately from the above result, where X is nonsingular i.e. invertible and $XA=BX$, one can find the following:

Corollary 3.5.

Let $B={\left[b_{ij}\right]}_{n\times n}$ be a strictly anti-diagonally dominant matrix by rows such that $XA=BX$, where $A={\left[a_{ij}\right]}_{n\times n}$. Then A and B are similar i.e. $A=X^{-1}BX$.

Theorem 3.6.

Let $A,B,X\in {\mathbb{F}}^{n\times n}$ such that $XA=BX$. If either of the following conditions hold:

(i)

X commutes with A, i.e. $XA=AX$,

(ii)

X commutes with B, i.e. $XB=BX$,

(iii)

$A=B$,

then X is an intertwining solution of the Yang–Baxter type matrix equations $XAX=BXB\mathrm{and}XBX=AXA.$

Proof. 

(i) At the beginning, we have from the hypothesis $XA=AX$. Based on the relation $XA=BX$, we deduce $BX=AX$.

Consequently, $BXB=\left(BX\right)B=\left(AX\right)B=A\left(XB\right).$ Due to the main relation of the assumption that $XB=BX,$ it follows that $BXB=ABX=A\left(XB\right)=A\left(BX\right)=XAX.$

Thus $XAX=BXB$. A symmetric argument gives $XBX=AXA$.

If (ii) or (iii) hold, the proof follows analogously. Hence the result holds in all three cases. □

Remark. 

If A (or B) is strictly anti–diagonally dominant by rows, it is nonsingular. However, nonsingularity alone does not imply the Yang–Baxter relations from $XA=BX$ without one of the above commutativity or equality conditions.

Theorem 3.7.

Let $A={\left[a_{ij}\right]}_{n\times n}$ and $B={\left[b_{ij}\right]}_{n\times n}$ be invertible matrices. Suppose $X\in {\mathbb{C}}^{n\times n}$ satisfy the Yang-Baxter matrix equations $XAX=BXB$ and $XBX=AXA$. Then X is commuting solution of the Yang-Baxter matrix equation.

Proof. 

Basically X satisfy the Yang-Baxter matrix equations $XAX=BXB$ and $XBX=AXA$. We take the relation $XAX=BXB$. Then

$$XAIX=BXB$$

Obviously,

$$XAA{A}^{-1}X=BXB.$$

The above expression yields

$$X{A}^{2}X=BXBA.$$

Due to the relation that $XAX=BXB,$ the right-side becomes to

$$X{A}^{2}X=XAXA.$$

Moreover,

$$X(A^{2}X-AXA)=0.$$

By reason that $X\ne 0$. Repeat the same scenario with $A\ne 0$, we arrive to the expression

$$AX=XA.$$

In this step, The term indicates that X represents a commuting solution to the Yang–Baxter matrix equation. □

The subsequent results further explore the properties of the Drazin inverse matrix. Let $A\in {\mathbb{C}}^{n\times n}$. The Drazin inverse of A, denoted by $A^D$, is the unique matrix satisfying the following three conditions:

$$\begin{array}{cc}\hfill A^{k+1}{A}^D& =A^k,\hfill \\ \hfill A^{D}A& =A{A}^D,\hfill \\ \hfill A^{D}A{A}^D& =A^D,\hfill \end{array}$$

where k is the index of A, defined as the smallest non-negative integer such that

$$rank\left(A^{k+1}\right)=rank\left(A^k\right).$$

Consider $A=\left(\begin{array}{ccccc}{a}_{11}& 0& 0& \dots & 0\\ 0& a_{22}& 0& \dots & 0\\ 0& 0& a_{33}& \dots & ⋮\\ ⋮& ⋮& ⋮& \ddots & 0\\ 0& 0& 0& \dots & a_{nn}\end{array}\right)$$,B=\left(\begin{array}{ccccc}{b}_{11}& 0& 0& \dots & 0\\ 0& b_{22}& 0& \dots & 0\\ 0& 0& b_{33}& \dots & ⋮\\ ⋮& ⋮& ⋮& \ddots & 0\\ 0& 0& 0& \dots & b_{nn}\end{array}\right)$

and

$$X=\left(\begin{array}{ccccc}{x}_{11}& x_{12}& a_{13}& \dots & x_{1n}\\ x_{21}& x_{22}& x_{23}& \dots & x_{2n}\\ x_{31}& x_{32}& x_{33}& \dots & ⋮\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right).$$

Theorem 3.8.

Let $A={\left[a_{ij}\right]}_{n\times n}$ and $B={\left[b_{ij}\right]}_{n\times n}$ be a square matrices. Suppose $X\in {\mathbb{C}}^{n\times n}$ satisfy the Yang-Baxter matrix equations $XAX=BXBX$ and $XBX=AXA$ such that X acts as a Drazin inverse matrix. Then $b_{nn}={\sum }_{k=1}^n{a}_{kk}{x}_{nk}{x}_{kn}.$

Proof. 

Without doubt the Yang-Baxter matrix equation $X·A·X=B·X·B$ yields the following matrices:

$$XAX=\left(\begin{array}{cccc}{\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{1k}{x}_{k1}& {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{1k}{x}_{k2}& \dots & {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{1k}{x}_{kn}\\ {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{2k}{x}_{k1}& {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{2k}{x}_{k2}& \dots & {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{2k}{x}_{kn}\\ ⋮& ⋮& \ddots & ⋮\\ {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{nk}{x}_{k1}& {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{nk}{x}_{k2}& \dots & {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{nk}{x}_{kn}\end{array}\right).$$

$$BXB=\left(\begin{array}{cccc}{b}_{11}^2{x}_{11}& b_{11}{b}_{22}{x}_{12}& \dots & b_{11}{b}_{nn}{x}_{1n}\\ b_{22}{b}_{11}{x}_{21}& b_{22}^2{x}_{22}& \dots & b_{22}{b}_{nn}{x}_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ b_{nn}{b}_{11}{x}_{n1}& b_{nn}{b}_{22}{x}_{n2}& \dots & b_{nn}^2{x}_{nn}\end{array}\right).$$

Where $XAX=BXB$ and $XBX=AXA,$ we observe that

$$\left(\begin{array}{cccc}{\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{1k}{x}_{k1}& {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{1k}{x}_{k2}& \dots & {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{1k}{x}_{kn}\\ {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{2k}{x}_{k1}& {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{2k}{x}_{k2}& \dots & {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{2k}{x}_{kn}\\ ⋮& ⋮& \ddots & ⋮\\ {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{nk}{x}_{k1}& {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{nk}{x}_{k2}& \dots & {\displaystyle {\sum }_{k=1}^n}{a}_{kk}{x}_{nk}{x}_{kn}\end{array}\right)$$

$$=\left(\begin{array}{cccc}{b}_{11}^2{x}_{11}& b_{11}{b}_{22}{x}_{12}& \dots & b_{11}{b}_{nn}{x}_{1n}\\ b_{22}{b}_{11}{x}_{21}& b_{22}^2{x}_{22}& \dots & b_{22}{b}_{nn}{x}_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ b_{nn}{b}_{11}{x}_{n1}& b_{nn}{b}_{22}{x}_{n2}& \dots & b_{nn}^2{x}_{nn}\end{array}\right).$$

Let

$$D_a=diag(a_{11},a_{22},\dots ,a_{nn}),D_b=diag(b_{11},b_{22},\dots ,b_{nn}).$$

Then the equation can be expressed compactly as

$$X{D}_{a}X=D_{b}X{D}_b.$$

Entrywise, this is equivalent to

$${\displaystyle \sum _{k=1}^n}{a}_{kk}{x}_{ik}{x}_{kj}=b_{ii}{b}_{jj}{x}_{ij},\forall i,j.$$

From the above equality, we observe that

$$b_{nn}^2{x}_{nn}={\displaystyle \sum _{k=1}^n}{a}_{kk}{x}_{nk}{x}_{kn}$$

Due to that X acts as a Drazin inverse matrix, the the left-side of above expression modifies to

$$b_{nn}={\displaystyle \sum _{k=1}^n}{a}_{kk}{x}_{nk}{x}_{kn}.$$

□

The following theorem focus on the matrices A and B are anti-diagonal and X acts as a Drazin inverse matrix, i.e.

$A=\left(\begin{array}{ccccc}0& 0& 0& \dots & a_{1n}\\ 0& 0& \dots & a_{2(n-1)}& 0\\ 0& \dots & a_{3(n-2)}& 0& 0\\ ⋮& \ddots & \ddots & ⋮& ⋮\\ a_{n1}& 0& 0& \dots & 0\end{array}\right)$, $B=\left(\begin{array}{ccccc}0& 0& 0& \dots & b_{1n}\\ 0& 0& \dots & b_{2(n-1)}& 0\\ 0& \dots & b_{3(n-2)}& 0& 0\\ ⋮& \ddots & \ddots & ⋮& ⋮\\ b_{n1}& 0& 0& \dots & 0\end{array}\right)$

and $X=\left(\begin{array}{ccccc}{x}_{11}& x_{12}& a_{13}& \dots & x_{1n}\\ x_{21}& x_{22}& x_{23}& \dots & x_{2n}\\ x_{31}& x_{32}& x_{33}& \dots & ⋮\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ x_{n1}& x_{n2}& x_{n3}& \dots & x_{nn}\end{array}\right).$

Theorem 3.9.

Let $A={\left[a_{ij}\right]}_{n\times n}$ and $B={\left[b_{ij}\right]}_{n\times n}$ be a anti-diagonal square matrices. Suppose $X\in {\mathbb{C}}^{n\times n}$ satisfy the Yang-Baxter matrix equations $XAX=BXB$ and $XBX=AXA$ such that X acts as a Drazin inverse matrix. Then $X=SJ,S=diag(s_1,\dots ,s_n)$

Proof. 

From the hypothesis, we find that X satisfies the relations

$$XAX={\left[{\displaystyle \sum _{k=1}^n}{a}_k{x}_{i,k}{x}_{n+1-k,j}\right]}_{i,j},$$

while

$$BXB={\left[b_i{x}_{n+1-i,n+1-j}{b}_{n+1-j}\right]}_{i,j}.$$

Equivalently, the $(i,j)$-entry is ${\left(BXB\right)}_{ij}=b_i{x}_{n+1-i,n+1-j}{b}_{n+1-j}.$

$$A={\left[a_{i,n+1-i}\right]}_{i=1}^n,B={\left[b_{i,n+1-i}\right]}_{i=1}^n,X=\left(x_{ij}\right)$$

Index form of the equation:

$${\displaystyle \sum _{k=1}^n}{a}_k{x}_{i,k}{x}_{n+1-k,j}=b_i{x}_{n+1-i,n+1-j}{b}_{n+1-j},\forall i,j.$$

Where B acts as a Drazin inverse matrix the right-side becomes

Index form of the equation:

$${\displaystyle \sum _{k=1}^n}{a}_k{x}_{i,k}{x}_{n+1-k,j}=b_i{b}_{n+1-j}{x}_{n+1-i,n+1-j},$$

for all $i,j=1,\dots ,n.$

Let $X=SJ,S=diag(s_1,s_2,\dots ,s_n),J=\mathrm{exchange}\mathrm{matrix}\mathrm{with}{J}_{i,n+1-i}=1.$

Then $X_{i,n+1-i}=s_i$, hence X is anti-diagonal. Yang-Baxter equation in index form:

$${\displaystyle \sum _{k=1}^n}{a}_k{x}_{i,k}{x}_{n+1-k,j}=b_i{b}_{n+1-j}{x}_{n+1-i,n+1-j},\forall i,j.$$

Substituting $X=SJ$ gives the diagonal system: $s_i^2{a}_{n+1-i}=b_i{b}_{n+1-i}{s}_{n+1-i},i=1,2,\dots ,n.$ Similarly, the second equation $XBX=AXA$ gives: $s_i^2{b}_{n+1-i}=a_i{a}_{n+1-i}{s}_{n+1-i},i=1,2,\dots ,n.$

Combining these, the solution is: $X=SJ,S=diag(s_1,\dots ,s_n),s_i$ satisfy the above system.

□

4. Conclusions

In this work, we examined a system of Yang–Baxter-type matrix equations, $XAX=BXB,$ and $XBX=AXA,$ which extends the classical matrix Yang–Baxter framework. Through geometric and topological analysis, we explored the structure and properties of their solutions, emphasizing the role of intertwining relationships between matrices A, B, and X. The conditions ensuring the existence of such intertwining solutions were identified and characterized. Moreover, we introduced and investigated the concepts of strictly anti-diagonally dominant and anti-diagonally dominant matrices, which provide a new perspective for understanding the solvability and stability of Yang–Baxter-type systems.