To find the equilibrium vector for a transition matrix, we need to find a vector v such that:

$P \cdot \mathbf{v} = \mathbf{v}$

where $P$ is the transition matrix and $\mathbf{v}$ is the equilibrium vector. This equation implies that v is an eigenvector of P with eigenvalue 1. Additionally, the elements of v must sum to 1, because v represents a probability distribution in a Markov chain.

Given Matrix

The matrix $P$ is: $P = \begin{bmatrix} 0.65 & 0.10 & 0.25 \\ 0.10 & 0.80 & 0.10 \\ 0.10 & 0.40 & 0.50 \end{bmatrix}$

Steps to Solve

1. Set up the equations for the equilibrium vector: $P \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ which translates to:

   0.65x + 0.10y + 0.25z = x \\ 0.10x + 0.80y + 0.10z = y \\ 0.10x + 0.40y + 0.50z = z \\ \end{cases}$$

2. Rewrite each equation by moving terms to one side to set up a system of linear equations:

   -0.35x + 0.10y + 0.25z = 0 \\ 0.10x - 0.20y + 0.10z = 0 \\ 0.10x + 0.40y - 0.50z = 0 \\ \end{cases}$$

3. Add the constraint $x + y + z = 1$ (since it’s a probability vector).

4. Solve this system of equations to find the values of $x$, $y$, and $z$.

I'll proceed with the calculations.The equilibrium vector for the given transition matrix is:

[ \mathbf{v} = \begin{bmatrix} \frac{1}{3} \ \frac{1}{3} \ \frac{1}{3} \end{bmatrix} = \begin{bmatrix}