= Solution
For the <linear map> $\alpha$ represented by $A$, a solution to the <system of linear equations> exists precisely when $b\in\operatorname{im}\alpha$. If $x_0$ is one solution, then
$$
Ax=b\iff A(x-x_0)=0\iff x\in x_0+\ker\alpha.
$$
Thus the <affine solution space of a linear equation> completely describes the possibilities: \b[no solutions] when $b\notin\operatorname{im}\alpha$; \b[one solution] when $b\in\operatorname{im}\alpha$ and $\ker\alpha=\{0\}$; and \b[infinitely many solutions] when $b\in\operatorname{im}\alpha$ and $\ker\alpha\neq\{0\}$. In the last case, a nonzero $w\in\ker\alpha$ supplies the distinct solutions $x_0+tw$ for every real $t$. In dimension three, the <rank-nullity theorem> also says that a trivial <kernel> is equivalent to an invertible <matrix> $A$.
For the <matrix> equation, $AX=B$ means that $\beta=\alpha\circ\chi$, where $\chi$ is represented by $X$. Necessarily $\operatorname{im}\beta\subseteq\operatorname{im}\alpha$. Conversely, if that containment holds, choose an $\alpha$-preimage of each $\beta(e_j)$ for the three standard <basis vectors>, and use those preimages as the columns of $X$. Their linear extension gives $AX=B$. This proves the <image criterion for a matrix factorization>:
$$
\boxed{AX=B\text{ is solvable}\iff\operatorname{im}\beta\subseteq\operatorname{im}\alpha.}
$$
Whenever it is solvable, every solution is $X_0+N$ with $AN=0$. It is unique exactly when $\ker\alpha=\{0\}$. Indeed, a nonzero $w\in\ker\alpha$ produces a nonzero <matrix> $N$ with first column $w$ and other columns zero, so uniqueness fails for any compatible $B$ if the <kernel> is nontrivial.
For the printed numerical <matrix> $A$, its <determinant> is $-2$, so the solution is unique. To calculate it, solve $A(x,y,z)^T=(b_1,b_2,b_3)^T$. The last equation gives $z=b_3-3y$; the second then gives $x=b_2-b_3+y$. Substitution into the first yields
$$
y=\frac{b_1-4b_2+3b_3}{2},\quad
x=\frac{b_1-2b_2+b_3}{2},\quad
z=\frac{-3b_1+12b_2-7b_3}{2}.
$$
Applying these formulas to the three columns of the PDF's $B$ gives
$$
\boxed{X=\begin{pmatrix}2&0&3/2\\5&0&7/2\\-12&1&-17/2\end{pmatrix}.}
$$
Multiplication gives $AX=\begin{pmatrix}1&1&1\\0&1&0\\3&1&2\end{pmatrix}$, confirming the result against the printed <matrix>.
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