Problem Help - Matrix/Markov

a) We have $\begin{bmatrix}\\ \pi & 1-\pi  \end{bmatrix} \begin{bmatrix}1-\alpha & \alpha\\ \lambda & 1-\lambda \end{bmatrix} = \begin{bmatrix}\\ \pi & 1-\pi \end{bmatrix}$. So 
$$ \begin{cases} \pi(1-\alpha)+(1-\pi)\lambda=\pi\\ \pi\alpha+(1-\pi)(1-\lambda)=1-\pi \end{cases}\implies\pi=\frac{\lambda}{\alpha+\lambda},\;1-\pi=\frac{\alpha}{\alpha+\lambda}$$
The long run probability of being employed is $\pi =\frac{\lambda}{\alpha+\lambda}$

b) The long run probability of being unemployed is $1-\pi = {\alpha \over \alpha +\lambda }$.

c) $ \begin{bmatrix}\\ E_{t+1} & U_{t+1} \end{bmatrix} = \begin{bmatrix}\\ E_t & U_t \end{bmatrix} \begin{bmatrix}1-\alpha & \alpha\\ \lambda & 1-\lambda \end{bmatrix} $ So $$ U_{t+1}=\alpha E_t +(1-\lambda )U_t = \alpha (1-U_t) +(1-\lambda )U_t = \alpha + (1-\alpha -\lambda )U_t$$

d) If $U_{t+1}=U_t$ then $U_t = \alpha + (1-\alpha -\lambda )U_t$ So $$ U_t = \frac{\alpha}{\alpha+\lambda}=1-\pi$$So the steady state unemployment level is the same as long run probability of being unemployed (as it should be since we have Markov process).