2026-07-072026-07-07http://salesiana.dossiersoluciones.com/handle/123456789/225966Grinshpon has proved that if $S$ is a commutative subring of a ring $R$ and $A\in M_n(S)$ is invertible in $M_n(R)$, then $det(A)$ is invertible in $R$. We give a very short proof of the result.1 pageRings and Algebrastext