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To solve for the steady-state vector (or probability vector) of the matrix \( P \), we need to follow these steps:

1. **Steady-state equation**:
   The steady-state vector \( \mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \) satisfies the equation:
   \[
   P \mathbf{v} = \mathbf{v}
   \]
   where \( P = \begin{bmatrix} 0.4 & 0.5 \\ 0.6 & 0.5 \end{bmatrix} \).

   This expands to:
   \[
   \begin{bmatrix} 0.4 & 0.5 \\ 0.6 & 0.5 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}
   \]

2. **Setting up the system**:
   From this, we get two equations:
   \[
   0.4 v_1 + 0.5 v_2 = v_1
   \]
   \[
   0.6 v_1 + 0.5 v_2 = v_2
   \]

3. **Simplify the equations**:
   - First equation: \( 0.4 v_1 + 0.5 v_2 = v_1 \) becomes \( -0.6 v_1 + 0.5 v_2 = 0 \)
   - Second equation: \( 0.6 v_1 + 0.5 v_2 = v_2 \) becomes \( 0.6 v_1 - 0.5 v_2 = 0 \)

4. **Solving the system**:
   These two equations are:
   \[
   -0.6 v_1 + 0.5 v_2 = 0
   \]
   \[
   0.6 v_1 - 0.5 v_2 = 0
   \]
   These two equations are dependent, meaning they describe the same line in space. To find a unique solution, we add the condition that the sum of the components of \( \mathbf{v} \) should be 1, since it is a probability vector:
   \[
   v_1 + v_2 = 1
   \]

5. **Final solution**:
   From the first equation, solve for \( v_2 \) in terms of \( v_1 \):
   \[
   0.6 v_1 = 0.5 v_2 \quad \Rightarrow \quad v_2 = \frac{0.6}{0.5} v_1 = 1.2 v_1
   \]
   Now use the condition \( v_1 + v_2 = 1 \):
   \[
   v_1 + 1.2 v_1 = 1 \quad \Rightarrow \quad 2.2 v_1 = 1 \quad \Rightarrow \quad v_1 = \frac{1}{2.2} = \frac{5}{11}
   \]
   Thus, \( v_2 = 1.2 \times \frac{5}{11} = \frac{6}{11} \).

   Therefore, the steady-state vector is:
   \[
   \mathbf{v} = \begin{bmatrix} \frac{5}{11} \\ \frac{6}{11} \end{bmatrix}
   \]

Let me know if you'd like further details or have any questions.

Here are 5 related questions:
1. How does one check if a matrix is a valid stochastic matrix for steady-state calculations?
2. What is the significance of the steady-state vector in Markov chains?
3. Can a matrix have more than one steady-state vector?
4. How do eigenvalues relate to steady-state vectors?
5. What are the practical applications of steady-state vectors in probability theory?

**Tip**: The sum of the entries in a probability vector should always be 1, which helps in solving steady-state problems.

Let P = [[0.4, 0.5], [0.6, 0.5]], find its steady-state vector or the probability vector.

Learn how to determine the steady-state vector or probability vector for the matrix P = [[0.4, 0.5], [0.6, 0.5]]. This problem involves concepts from linear algebra, probability, and Markov chains, including solving systems of equations and using the stochastic matrix property. Suitable for high school students in Grades 10-12.

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