Show that $\lambda =1$ is always an eigenvalue for M and find the other eigenvalue in terms of $a$ and $b$.

Find the steady state probability vector for M in terms of $a$ and $b$.

        M1A1

$\Rightarrow {\lambda }^2-\left(a+b\right)\lambda +a+b-1=0\Rightarrow \left(\lambda -1\right)\left(\lambda +\left(1-a-b\right)\right)=0$         A1

$\Rightarrow \lambda =1\text{or}\lambda =a+b-1$         AGA1

[4 marks]

       M1A1

$\Rightarrow 1-b=\left(2-a-b\right)p\Rightarrow p=\frac{1-b}{2-a-b}$         M1

So vector is $\left(\begin{array}{c}\frac{1-b}{2-a-b}\\ \frac{1-a}{2-a-b}\end{array}\right)$         A1A1

[5 marks]