Show that $(T_{ij})=\begin{pmatrix} \alpha &\omega &0\\-\omega & \alpha &0\\ 0 &0& \beta \end{pmatrix}$ 
A tensor has components $T_{ij}$ with respect to Cartesian coordinates
   $x$. If the tensor is invariant under arbitrary rotations around the
   $x_3$-axis, show that it must have the form $$(T_{ij})=\begin{pmatrix}
 \alpha &\omega &0\\-\omega & \alpha &0\\ 0 &0& \beta \end{pmatrix}$$

Surely we want $\begin{pmatrix} \cos\theta &\sin\theta &0\\-\sin\theta & \cos\theta&0\\ 0 &0& 1\end{pmatrix}\begin{pmatrix} T_{11} &T_{12} &T_{13}\\T_{21}& T_{22}&T_{23}\\ T_{31}&T_{32}&T_{33}\end{pmatrix}=\begin{pmatrix} T_{11} &T_{12} &T_{13}\\T_{21}& T_{22}&T_{23}\\ T_{31}&T_{32}&T_{33}\end{pmatrix}$ but we then have $(\cos\theta )T_{11}+(\sin\theta) T_{21}=T_{11}$ and $(\cos\theta) T_{21}-(\sin\theta) T_{11}=T_{21}$ which eventually leads to $1-\cos\theta=\cos\theta-1$ which must hold for all $\theta$ which is clearly wrong, unless $T_{11}/T_{21}=0$ which again is wrong. 
What am I doing wrong? Is my approach fundamentally wrong? I imagine I am underestimating the question quite a but (if this method were to work that would make the question far easier than expected), I suspect there's a lot more to the question I'm not considering or my approach is wrong altogether.
Any help would be much appreciated
Thank you

I think your calculation is right.
Let $R_\theta$ be a $2\times 2$ rotational matrix, observe that
\begin{align}
R_\theta M = M \ \ \Rightarrow \ \ (I-R_\theta)M= 0
\end{align}
for all $\theta \in [0, 2\pi)$ implies $M = 0$. 
I think you might be confusing something here.
