 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

When $A$ is an <invertible matrix>, $B=A^{-1}(AB)$. Apply <1/a/i> first to $AB$ and then to $A^{-1}(AB)$:
$$
\operatorname{rowspace}(AB)\subseteq\operatorname{rowspace}(B)
=\operatorname{rowspace}(A^{-1}AB)\subseteq\operatorname{rowspace}(AB).
$$
Thus \b[the two <row spaces> are equal]. In particular, invertible left <matrix multiplication> preserves <row rank>.

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