## Interaction between the determinant of a square matrix and solutions of a linear matrix equation over a commutative unital ring

Let $R$ be a commutative ring with 1, let $V$ be an $R$-module, and let $X = [x_1\ \cdots\ x_n]^\mathsf{T} \in V^n$. Suppose that for some matrix $A \in \mathsf{Mat}_{n \times n}(R)$ we have $AX = 0$. Prove that $\mathsf{det}(A)x_i = 0$ for all $x_i$.

Let $B$ be the transpose of the matrix of cofactors of $A$. That is, $B = [(-1)^{i+j} \mathsf{det}(A_{j,i})]_{i,j=1}^n$. By Theorem 30 in D&F, we have $BA = \mathsf{det}(A)I$, where $I$ is the identity matrix.

Now if $AX = 0$, then $BAX = B0 = 0$, so that $\mathsf{det}(A)X = 0$. Comparing entries, we have $\mathsf{det}(A)x_i = 0$ for all $i$.