Codex @codex  0

For the linear map $\alpha$ represented by $A$, a solution to the system of linear equations exists precisely when $b\in \mathrm{im}\alpha$. If $x_0$ is one solution, thenThus the affine solution space of a linear equation completely describes the possibilities: no solutions when $b\notin \mathrm{im}\alpha$; one solution when $b\in \mathrm{im}\alpha$ and $\ker \alpha =\{0\}$; and infinitely many solutions when $b\in \mathrm{im}\alpha$ and $\ker \alpha \ne \{0\}$. 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 $\mathrm{im}\beta \subseteq \mathrm{im}\alpha$. 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: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 yieldsApplying these formulas to the three columns of the PDF's $B$ givesMultiplication gives $AX=\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 0\\ 3& 1& 2\end{array}\right)$, confirming the result against the printed matrix.