$g_{ij}$ calculation of randers metric Let $F=\alpha + \beta $ where $\alpha=\sqrt{a_{ij}(x)y^iy^j}$ is a riemannian metric and $\beta =b_i(x)y^i$ is a one form.that is F is Randers metric on a manifold $M$.  I want to calculate $g_{ij}$ where
$g_{ij}= \frac{1}{2}[F^2]_{y^iy^j}$



Clearly $g_{ij}=F_{y^i}F_{y^i}+FF_{y^iy^j}$.
Now we will calculate $F_{y^i}$.
$F_{y^i}=\alpha_{y^i}+\beta_{y^i}$.
Now $\alpha^2=a_{ij}(x)y^iy^j$ gives $\alpha_{y^k}=\frac{a_{ik}y^i}{\alpha}$
so$F_{y^i}=\frac{a_{ij}y^i}{\alpha}+b_i$

Again for $F_{y^iy^j}$ after a short calculation we get it as $0$
So $g_{ij}=F_{y^i}F{y^j}$.
so $g_{ij}=(\frac{a_{ij}y^j}{\alpha}+b_i)(\frac{a_{ij}y^i}{\alpha}+b_j)$  

Is it okay?  

How can I simplify it so that I can calculate the inverse of $g_{ij}$ using the following formula?

If $g_{ij}=h_{ij}+\delta C_iC_j$ where $C_i$ is a column vector , then det $g=(1+\delta C^2)\det(h_{ij})$ and $C^i=h^{ij}C_j$ and $g^{ij}=h^{ij}-\frac{\delta C^iC^j}{1+\delta C^2}$ where $C^2=h^{ij}C_iC_j$.

Can someone help me please.
First of all, the second derivative of F is not zero. Try that part again and you will get a nonzero part you need to add to what you already have.

Second, Once you have that part, you can collect all the terms with $a_{ij}$ in them, "factor" it out, and write it in a form that can be used in your formula.

edit:

apparently someone doesn't like my answer, so let me be more explicit.

you should get $g_{jk}=(1+b\cdot \hat{y})a_{jk}+(\hat{y}+b)_j(\hat{y}+b)_k-(1+b \cdot \hat{y})\hat{y}_j\hat{y}_k$.

Next apply your formula to the first two terms. Finally, apply the formula to the whole thing. Here, y-hat means $\frac{y}{\alpha}$. Also, in that formula, the vector index is raised and lowered by the inverse of the matrix. When you apply the formula to the second and third parts, you will need the inverse of the matrices (which are not just $a^{jk}$).
First of all, the second derivative of F is not zero. Try that part again and you will get a nonzero part you need to add to what you already have.

Second, Once you have that part, you can collect all the terms with $a_{ij}$ in them, "factor" it out, and write it in a form that can be used in your formula.

edit:

apparently someone doesn't like my answer, so let me be more explicit.

you should get $g_{jk}=(1+b\cdot \hat{y})a_{jk}+(\hat{y}+b)_j(\hat{y}+b)_k-(1+b \cdot \hat{y})\hat{y}_j\hat{y}_k$.

Next apply your formula to the first two terms. Finally, apply the formula to the whole thing. Here, y-hat means $\frac{y}{\alpha}$. Also, in that formula, the vector index is raised and lowered by the inverse of the matrix. When you apply the formula to the second and third parts, you will need the inverse of the matrices (which are not just $a^{jk}$).