= Solution
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 subspace>
$$
\mathbf x_0+\ker A,
$$
because subtracting two solutions produces a vector in the <kernel>, and adding any <kernel> vector preserves the equation. Therefore the necessary and sufficient cases are:
* \b[No solution:] $\operatorname{rank}[A\mid\mathbf b]>\operatorname{rank}A$, equivalently $\mathbf b$ is outside the <column space>.
* \b[Exactly one solution:] $\operatorname{rank}A=3$, equivalently $\det A\ne0$; then $\mathbf x=A^{-1}\mathbf b$ for every $\mathbf b$.
* \b[Infinitely many solutions:] $\operatorname{rank}[A\mid\mathbf b]=\operatorname{rank}A<3$. The <rank-nullity theorem> gives a nonzero <kernel> vector $\mathbf v$, and $\mathbf x_0+t\mathbf v$ gives distinct solutions for all real $t$.
For the matrix equation, each column of $X$ solves the same <system of linear equations> with the corresponding column of $B$. Thus \b[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 $\mathbf x_2=\mathbf b_1-\mathbf b_2-\mathbf x_3$. Hence
$$
\boxed{X=\begin{pmatrix}1&1&-1\\1&-1&0\\1&0&1\end{pmatrix}.}
$$
Multiplying $AX$ gives the prescribed $B$, which both checks the arithmetic and confirms the unique answer.
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