System of linear equations - Resolution

Question

$$ \left( \begin{matrix} \pi_1 & \pi_2 & \pi_3 \end{matrix} \right) = \left( \begin{matrix} \pi_1 & \pi_2 & \pi_3 \end{matrix} \right) \begin{bmatrix} 0.6 & 0.3 & 0.1 \\ 0.2 & 0.7 & 0.1 \\ 0.1 & 0.3 & 0.6 \end{bmatrix} $$

I know also that $\pi_1$, $\pi_2$ and $\pi_3$ have the property $\pi_1 +\pi_2 + \pi_3 = 1 $ (*)

I am a old student of linear algebra. Is anyone could tell me how I can solve this system of linear equations? Is it sufficient to use the row reduction?

I would like to simplify the system of linear equations for using more easily (*) in such a way that I can find individually the values of $\pi_1$, $\pi_2$ and $\pi_3$

Answer 1

Let $M$ be your matrix and $\mathbf\pi$ the column vector of unknowns. The equation is then $\mathbf\pi^T=\mathbf\pi^T M$. Rearranged, this is $\mathbf\pi^T(I-M)=\mathbf\pi^T(M-I)=0$. Transposing, this becomes $(\mathbf\pi^T(M-I))^T=(M-I)^T\mathbf\pi=(M^T-I)\mathbf\pi=0$. You can then use standard row-reduction methods to solve it.

Note that the solutions will be eigenvectors of the eigenvalue $1$, which is indeed an eigenvalue of a stochastic matrix like this one.

Answer 2

This seems like a stationary distribution problem.

You should multiply the two matrices: $$\left( \begin{smallmatrix}\\ \pi_1 & \pi_2 & \pi_3 \end{smallmatrix} \right)\begin{bmatrix} 0.6 & 0.3 & 0.1 \\ 0.2 & 0.7 & 0.1 \\ 0.1 & 0.3 & 0.6 \end{bmatrix}=\left(\begin{matrix} 0.6\pi_1+0.2\pi_2+0.1\pi_3&0.3\pi_1+0.7\pi_2+0.3\pi_3&0.1\pi_1+0.1\pi_2+0.6\pi_3 \end{matrix}\right)=(\pi_1\quad\!\!\! \pi_2\quad\!\!\!\pi_3)$$ $$\begin{matrix} 0.4\pi_1-0.2\pi_2-0.1\pi_3=0\\ -0.3\pi_1+0.3\pi_2-0.3\pi_3=0\\ -0.1\pi_1-0.1\pi_2+0.4\pi_3=0 \end{matrix}$$ You would set this as an augmented matrix (if my matrix multiplication is correct) and do row reduction to find the solution, most likely with $\pi_1, \pi_2$ in terms of $\pi_3$, and you would find the stationary distribution by letting the sum of each $\pi_i=1$ and find each $\pi_i$ accordingly.