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The system of linear equations is consistent precisely when $\mathbf{b}$ belongs to the column space of $A$. If ${\mathbf{x}}_0$ is one solution, all solutions form the affine subspacebecause subtracting two solutions produces a vector in the kernel, and adding any kernel vector preserves the equation. Therefore the necessary and sufficient cases are:

For the matrix equation, each column of $X$ solves the same system of linear equations with the corresponding column of $B$. Thus a unique $X$ exists if and only if $A$ is invertible, and then $X=A^{-1}B$. Necessity does not depend on $B$: if $A$ is singular and one $X$ exists, put any nonzero kernel vector in one column of a matrix $Y$ and zeros in its other columns. Then $AY=0$ and every $X+tY$ is another solution.

For the specified data, $\det A=3$. If ${\mathbf{x}}_j$ denotes row $j$ of $X$ and ${\mathbf{b}}_j$ row $j$ of $B$, row elimination yields ${\mathbf{x}}_3=(2{\mathbf{b}}_1-{\mathbf{b}}_2-{\mathbf{b}}_3)/3$, ${\mathbf{x}}_1={\mathbf{b}}_2-{\mathbf{x}}_3$, and ${\mathbf{x}}_2={\mathbf{b}}_1-{\mathbf{b}}_2-{\mathbf{x}}_3$. HenceMultiplying $AX$ gives the prescribed $B$, which both checks the arithmetic and confirms the unique answer.