Let $f : E \times F \to R$ be a bilinear form that is non-singular on the left, where $E, F$ are free $R$-modules of dimension $n$ (both of them).
Then if $X, Y$ are column vectors for general elements $x,y \in E, F$ respectively. Then the value $f(x,y)$ can be written as:
$$ f(x,y) = X^t G Y $$
Lang's Algebra on page 529 says that $X \mapsto X^t G$ is an isomoprhism (by non-singularity), and that isomorphic here implies that $G$ is invertible. How do we prove that $G$ is invertible, since things are unfamiliar from the usual way of matrix multiplying, i.e. $a(x) = AX$?
$G$ is an $n\times n$ matrix with entries in $R$.
You can revert back to a familiar way of matrix multiplication by exploiting the transpose operator as follows.
Define $\phi:R^{n \times 1} \rightarrow R^{1\times n}$ and $l:R^{n \times 1} \rightarrow R^{n \times 1}$ by $$\phi(X)=X^tG$$ $$l(X)=G^tX$$ Now $\phi(X)^t=l(X)$ for all $X\in R^{n\times 1}$ so $l$ is also an isomorphism. For each $1 \leq i \leq n$ find a unique $X_i\in R^{n\times 1}$ such that $$l(X_i)=G^tX_i=e_i$$ Such an $X_i$ exists by isomorphism. If $M$ is the matrix whose $i^{\text{th}}$ column is $X_i$ then its easy to see that $(G^t)^{-1}=M$ hence $G^t$ is invertible and so is $G$.