Finding A 1-form on $R^2 - {\{(0,0)}\}$ I want to find a 1-form on $R^2 - {\{(0,0)}\}$ such that $w(Y) = 0$ and $w(X) = 1$. Here, $$X = -y\frac{\partial }{\partial x}
      + x\frac{\partial}{\partial y}\ \text{and}\ Y = x\frac{\partial }{\partial x}
      + y\frac{\partial}{\partial y}.$$ 
I let $w = adx+bdy$ be a 1-form and then, using the given conditions compute: $$(adx+bdy)(-y\frac{\partial }{\partial x}
      + x\frac{\partial}{\partial y}) = 1,\ \text{and similarly},\ (adx+bdy)(x\frac{\partial }{\partial x}
      + y\frac{\partial}{\partial y}) = 0.$$ 
I then obtain the following equations: $-ay+bx=1$, $ax+by=0$. 
To solve for $a,b$, I then set it up as a matrix/vector system as 
$$\underset{\begin{array}{c}\\ A \end{array}}%
{%
\begin{pmatrix}
    x & y \\ 
    -y & x%
\end{pmatrix}%
}%
\underset{\begin{array}{c}\\ x \end{array}}%
{%
\begin{pmatrix}
    a \\ 
    b%
\end{pmatrix}%
}=\underset{\begin{array}{c}\\ b \end{array}}%
{%
\begin{pmatrix}
    0 \\ 
    1%
\end{pmatrix}%
          }$$
I then simply invert the matrix and obtain that $a$ = $\frac{-y}{x^2+y^2}$ and $b$ = $\frac{x}{x^2+y^2}$. 
However, this seems to be incorrect - the solution in the back of the book gives $$a = \frac{x}{x^2+y^2}\ \text{and}\ b = \frac{y}{x^2+y^2}.$$ Can someone kindly let me know what I did wrong?

Let $r = x^2 + y^2$. Note that if the values are $a = \frac{x}{r}$ and $b = \frac{y}{r}$, then we will have: $w = \frac{x}{r}$ $dx$ + $\frac{y}{r}$ $dy$. Then, the following follows: $ w(X) = \frac{1}{r} (-xy +xy)$ = $0$,$ w(Y) = \frac{1}{r} (x^2+y^2)$ = $\frac{1}{r} r = 1.$ This contradicts the conditions given ($w(X) =1$, $w(Y) =0$). Thus, the solution in the book can't be the 1-form on $R^2 -{\{(0,0)}\}$.
