Common Annihilator Through Matrices (11.4.3)

Dummit and Foote Abstract Algebra, section 11.4, exercise 3:

Let $R$ be a commutative ring with $1$ and let $V$ be an $R$-module with $x_1,...,x_n∈V$. Letting $W$ be the column matrix of these elements, assume that for some $A∈M_{n × n}(R)$, $$AW=0$$ Prove $(\text{det }A)x_i=0$ for $i∈\{1,...,n\}$.

Proof: This implies a system $$α_{11}x_1+α_{12}x_2+...+α_{1n}x_n=0$$ $$α_{21}x_1+α_{22}x_2+...+α_{2n}x_n=0$$ $$...$$ $$α_{n1}x_1+α_{n2}x_2+...+α_{nn}x_n=0$$ In fashion of constructing the cofactor formula for the determinant along the first column, multiply the first row by $\text{det }A_{11}$ and for $k∈\{2,...,n\}$ add $(-1)^{k+1}\text{det }A_{k1}$ times the $k^{th}$ row to the first row to obtain $$(\text{det }A)x_1+\sum_{j=2}^n (α_{1j}\text{det }A_{11}+\sum_{k=2}^n(-1)^{k+1}α_{kj}\text{det }A_{k1})x_j=0$$ $$(\text{det }A)x_1+\sum_{j=2}^n (\sum_{k=1}^n(-1)^{k+1}α_{kj}\text{det }A_{k1})x_j=0~~~~~(*)$$ For each $j$, let $B_j$ be the matrix $A$ with the first column replaced by the $j^{th}$ column. We see $$0 = \text{det }B_j = \sum_{k=1}^n(-1)^{k+1}β_{k1}\text{det }B_{k1} = \sum_{k=1}^n(-1)^{k+1}α_{kj}\text{det }A_{k1}$$ so that $(*)$ collapses to $$(\text{det }A)x_1 = 0$$ By interchanging arbitrary $x_i$ with $x_1$ and letting $A_i$ be the matrix $A$ with its first and $i^{th}$ column interchanged, since this operation negates the determinant and by the argument above we have $$-(\text{det }A)x_i = 0 = (\text{det }A)x_i~~~\square$$