A Markov chain $\{X_n,n\geq0\}$ with states $0, 1, 2$, has the transition probability matrix $$\begin{bmatrix} \frac12& \frac13 &\frac16\\ 0&\frac13&\frac23\\ \frac12&0&\frac12 \end{bmatrix}$$ If $P(X_0=0)=P(X_0=1)=\frac14$, find $E[X_3]$.
I reasoned as follows,
Given the above transition probability matrix $\mathbf{P}$, I then found $$\mathbf{P}^3= \begin{bmatrix} \frac{13}{36} & \frac{11}{54} & \frac{47}{108} \\ \frac{4}{9} & \frac{4}{27} & \frac{11}{27} \\ \frac{5}{12} & \frac{2}{9} & \frac{13}{36} \\ \end{bmatrix} $$ and $P(X_0=2)=1-\left[P(X_0=0)+P(X_0=1)\right]=\frac12.$ I then found the expectation by conditioning, $$\begin{aligned} E[X_3]&=E_{X_0}[E[X_3\vert X_0]]=E\left[\sum_{i=0}^2 iP(X_3=i\vert X_0)\right]\\&=E[P(X_3=1\vert X_0)]+2E[P(X_3=2\vert X_0)]\\ &=\sum_{j=0}^2P(X_3=1\vert X_0=j)P(X_0=j)+2\sum_{j=0}^2P(X_3=2\vert X_0=j)P(X_0=j) \end{aligned}$$ so $$\begin{aligned}E[X_3]&=P(X_0=0)\mathbf{P}^3_{01}+P(X_0=1)\mathbf{P}^3_{11}+P(X_0=2)\mathbf{P}^3_{21}+\\&+2\left[P(X_0=0)\mathbf{P}^3_{02}+P(X_0=1)\mathbf{P}^3_{12}+P(X_0=2)\mathbf{P}^3_{22}\right]\\\\&=\frac14\cdot\frac{11}{54}+\frac{1}{4}\cdot\frac{4}{27}+\frac12\cdot\frac29+2\left(\frac14\cdot\frac{47}{108}+\frac14\cdot\frac{11}{27}+\frac12\cdot\frac{13}{36}\right)\\&=\frac{53}{54}.\end{aligned}$$ Would this be correct?
In terms of matrices, it suffices to compute $$\pi=[1/4\quad1/4\quad1/2]P^3[0\quad1\quad2]^\top$$