On solutions of matrix equation ${AX=B}$ over a Bezout domain

Let ${\mathrm R_{m,n}}$ be the set of $m\times n$ matrices over a Bezout domain $\mathrm R $ with identity $e\not= 0$ and let $0_{m,k}$ be the zero $m\times k$ matrix. Further, let $d_i(A)\in \mathrm{R}$ be an ideal generated by the $i$-th order minors of the matrix $A\in \mathrm{R}_{m,n},$ $i = 1, 2, \dots, \min\{m, n\}.$ In this article, we investigate a structure of solutions of a matrix equation $AX=B$, where $A\in {\mathrm R}_{m,n}$ and $B \in {\mathrm R}_{m,k}$ are known matrices and $X$ is unknown matrix over ${\mathrm R}$. It is known that matrix equation $AX=B$ is solvable over a Bezout domain $\mathrm{R}$ if and only if $ {\rm rank } A = {\rm rank }A_B=r$ and $d_i(A) = d_i(A_B)$ for all $i = 1, 2, \dots , r,$ where $A_B= \begin{bmatrix} A & B \end{bmatrix}.$ On the other hand, $AX=B$ is solvable over $\mathrm{R}$ if and only if matrices $\begin{bmatrix} A & 0_{m,k} \end{bmatrix}$ and $A_B$ are right-equivalent, that is, the Hermitian normal forms of these matrices coincide. In this article, we give alternative necessary and sufficient conditions for the solvability of equation $AX=B$ over a Bezout domain $\mathrm R .$ If a solution of this equation exists, we also give an algorithm for its construction. We prove also that the matrix equation $AX=B$ over $\mathrm{R}$ has a symmetric solution if and only if $AX=B$ has a solution over $\mathrm{R}$ and the matrix $AB^T$ is symmetric. If symmetric solution exists, we propose the method for its construction.