Dual basis with non-degenerate bilinear form Let $V$ be $n$-dimensional vector space and $\{x_1,\cdots,x_n\}$ a basis. Let $\beta:V\times V\rightarrow F$ be a non-degenerate (symmetric) bilinear form. This implies that there exists a dual basis $\{y_1,\cdots, y_n\}$ of $V$ w.r.t. $\beta$, i.e. 
$$\beta(x_i,y_j)=\delta_{ij}.$$
Can we write expression for basis elements $y_i$'s in terms of $x_i$'s and matrix of $\beta$ w.r.t. basis $\{x_1,\cdots, x_n\}$?

This is an obvious computational question and may be very trivial, but I didn't find in any book mentioning about computation of dual basis. For example, as a concrete example, let $V$ be the space of column vectors of length $n$ over field $\mathbb{R}$. Consider dot product on this column space.  Then given any invertible $n\times n$ matrix $A$ over $\mathbb{R}$, its $n$ columns define a basis of $V$; what is dual basis w.r.t. dot product? Perhaps it is columns of $(A^{-1})^t$, am I right? Then next question comes more general, which I put above.

Let $B$ be the matrix with entries $b_{i,j}=\beta(x_i,x_j)$. It's symmetric
and invertible. Write $C=B^{-1}=(c_{i,j})$. Then $y_j=\sum_kc_{j,k}x_k$.
One checks
$$\beta(x_i,y_j)=\sum_k c_{j,k}b_{i,k}=\delta_{i,j}$$
using the fact that $CB^t=CB=I$.
