Can someone more explicitly describe Theorem 4.11 in Algebra?
Let $E$ be a module over a commutative ring $R$, and let $v_1,\dots,v_n$ be elements of $E$. Let $A=(a_{ij})$ be a matrix in $R$, and let $A(v_1,\dots,v_n)^T=(w_1,\dots,w_n)^T$. Let $\Delta$ be an $n$-multilinear alternating map on $E$. Then $$ \Delta(w_1,\dots,w_n)=D(A)\Delta(v_1,\dots,v_n).$$
Lang's proof: Use the fact that $D(A)=D(A^T)$ and expand $$\Delta(a_{11}v_1+\cdots+a_{1n}v_n,\dots, a_{n1}v_1+\cdots+a_{nn}v_n)$$ and find precisely what you want. $\square$
What explicitly are you looking for in the expansion? Thanks.
Expanding $\Delta(a_{11}v_1+\cdots a_{1n}v_n,\cdots,a_{n1}v_1+\cdots+a_{nn}v_n)$ using linearity, we obtain
$$\sum a_{1i_1}a_{2i_2}\cdots a_{ni_n}\Delta(v_{i_1},v_{i_2},\cdots,v_{i_n}),$$
where the sum is taken over all available tuples $(i_1,\cdots,i_n)$. As $\Delta$ is an alternating map, it is zero if any two of its arguments are the same, so we can drop out all of these cases and we are left with those summands when $i_1,\cdots,i_n$ are all distinct, which means they must be given by some permutation $i_k=\pi(k)$ of $k\in\{1,2,\cdots,n\}$. We thus obtain
$$\sum_{\pi\in S_n}a_{1\pi(1)}\cdots a_{n\pi(n)}\Delta(v_{\pi(1)},\cdots,v_{\pi(n)})=\left(\sum_{\pi\in S_n}a_{1\pi(1)}\cdots a_{n\pi(n)}{\rm sgn}(\pi)\right)\Delta(v_1,\cdots,v_n)$$
which is equal to $D(A)\Delta(v_1,\cdots,v_n)$ (or at least I assume this is the working definition of the determinant at hand).