Prove that a matrix is symplectic?

Question

Let $\omega$ be the skew-symmetric bilinear form on $R^{2n}$ given by $$\omega(x,y)=\sum_{j=1}^{n}(x_{j}y_{n+j}-x_{n+j}y_{j})$$ Let $\Omega$ be the $2n \times 2n$ matrix \begin{pmatrix} 0 & I_{n}\\ -I_{n} & 0 \end{pmatrix} show that for all $x,y\in R^{2n}$, we have $\omega(x,y)=\langle x,\Omega y\rangle $ and a $2n \times 2n $ matrix A belong to Sp(n;R) iff $$-\Omega A^{tr}\Omega=A^{-1}.$$ First Part: \begin{align} \langle x,\Omega y\rangle &=\langle \{x_{1},x_{2},...,x_{2n}\}, \Omega \{y_{1},y_{2},...y_{2n}\}\rangle \\ &=\langle \{x_{1},x_{2},...,x_{2n}\},\{y_{n+1},...y_{2n},-y_{1}...-y_{n}\}\rangle \\ &=\sum_{j=1}^{n}(x_{j}y_{n+j}-x_{n+j}y_{j})\\ &=\omega(x,y). \end{align} second part: A matrix will be in real symplectic group iff $\omega(Ax,Ay)=\omega(x,y)$. so $$\langle Ax,\Omega Ay\rangle =\langle x,\Omega y\rangle \Rightarrow \langle A^{tr}\Omega^{tr}Ax,y\rangle =\langle \Omega^{tr}x,y\rangle \Rightarrow\Omega^{-tr}A^{tr}\Omega^{tr}=A^{-1}.$$ But $\Omega^{tr}\neq\Omega.$ Where did I do wrong?