Give an example of a regular stochastic $2\times2$ matrix with steady state vector $\begin{bmatrix}\frac{1}{3}\\\frac{2}{3}\end{bmatrix}$.
Solution: If our answer is $A = \begin{bmatrix}1 − a &b \\ a & 1 − b\end{bmatrix}$, then we need $\begin{bmatrix}−a & b \\ a & −b\end{bmatrix} * \begin{bmatrix}\frac{1}{3}\\\frac{2}{3}\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$. This becomes $a = 2b$ so $\begin{bmatrix}1 − 2b & b \\2b& 1 − b\end{bmatrix}$ works for any $b ∈ [0, 1/2]$.
My reasoning: Let q be the steady state vector.
$Aq=q$
$Aq-q=0$
$(A-I)q=0$
$A = \begin{bmatrix}a -1 &b \\ a & b-1\end{bmatrix}$
I don't understand how they have it the other way around with $A = \begin{bmatrix}1 − a &b \\ a & 1 − b\end{bmatrix}$, and I don't see how they proceeded to $\begin{bmatrix}−a & b \\ a & −b\end{bmatrix} * \begin{bmatrix}\frac{1}{3}\\\frac{2}{3}\end{bmatrix}$
It looks like you might be a bit confused about what’s going on in the answer. You’re looking for a matrix of the form $$A=\begin{bmatrix}1-a&b\\a&1-b\end{bmatrix}.$$ The steady-state distribution $q$ is a right eigenvector of $1$, so satisfies the equation $Aq=q$, which can be written as $(A-I)q=0$, which is what you have, too. Somehow from that, you jump to an incorrect expression for $A$. You haven’t explained how you got there, so I can’t really help you, but it kind of looks like you went backwards through the solution to the equation.
The correct approach is to expand $$(A-I)q=\left[\begin{array}{rr}-a&b\\a&-b\end{array}\right]\begin{bmatrix}\frac13\\\frac23\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}.$$ This yields the condition $a=2b$, from which the rest follows.