If $A_1=\begin{pmatrix}1&0\\1&1\end{pmatrix}$ and $A_2=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, and the measure on $\{A_1,A_2\}^{\mathbb Z}$ is the $(\frac 12,\frac 12)$ Bernoulli measure, then there is a unique stationary measure:
given a sequence of 1's and 2's indexed by $\mathbb Z^-$, $\ldots,\omega_{-3},\omega_{-2},\omega_{-1}$, define $x_\omega$ to be the point of $\mathbb{PR}^2$ in the direction $\lim_{n\to\infty}A_{\omega_{-1}}\ldots A_{\omega_{-n}}\begin{pmatrix}1\\1\end{pmatrix}$. (The limit exists since the $A$'s are contracting on the positive cone). The unique stationary measure is $\int \delta_{x_\omega}\,d\mathbb P(\omega)$, where $\mathbb P$ is the Bernoulli measure on $\{1,2\}^{\mathbb Z^-}$. It's not hard to check from the definition that it is stationary. It's also quite easy to check from the above that any measure on the non-negative cone is attracted to this measure. A bit more work is needed to deal with measures with support off the non-negative cone, but one can convince oneself that the mass is gradually attracted to the non-negative cone.
The second example is the easy one: if $B_1$ and $B_2$ are any invertible diagonal matrices, then it is immediate to check that $\delta_X$ and $\delta_Y$, the measures supported on the $X$ and $Y$ direction are stationary.