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.

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