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Problem 10

# Find the steady-state vector for the transition matrix. $$\left[\begin{array}{ll} \frac{4}{5} & \frac{3}{5} \\ \frac{1}{5} & \frac{2}{5} \end{array}\right]$$

Expert verified
The steady-state vector for the given transition matrix is $$v = \begin{bmatrix}\frac{3}{4} \\ \frac{1}{4}\end{bmatrix}$$.
See the step by step solution

## Step 1: 1. Verify that the matrix is a transition matrix.

A matrix is considered a transition matrix if it satisfies two conditions: a. Each element is non-negative. b. The sum of elements in each row is equal to 1. Given matrix: $$M = \left[\begin{array}{ll} \frac{4}{5} & \frac{3}{5} \\\ \frac{1}{5} & \frac{2}{5} \end{array}\right]$$ It is clear that all elements are non-negative. The sum of the elements of each row is: Row 1: $$\frac{4}{5}+\frac{3}{5}=1$$ Row 2: $$\frac{1}{5}+\frac{2}{5}=1$$ Since both conditions are satisfied, the given matrix is a transition matrix.

## Step 2: 2. Set up a system of equations to find the steady-state vector.

Let the steady-state vector $$v = \begin{bmatrix}v_1 \\ v_2 \end{bmatrix}$$. We need to solve the following equation: $$Mv = v$$ Replacing with the given matrix and vector, we get: $$\left[\begin{array}{ll} \frac{4}{5} & \frac{3}{5} \\\ \frac{1}{5} & \frac{2}{5} \end{array}\right] \begin{bmatrix}v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix}v_1 \\ v_2 \end{bmatrix}$$ Expanding the equation, we have the following system of equations: 1. $$\frac{4}{5}v_1 + \frac{3}{5}v_2 = v_1$$ 2. $$\frac{1}{5}v_1 + \frac{2}{5}v_2 = v_2$$ Additionally, the steady-state vector has the condition that its elements sum up to 1: 3. $$v_1 + v_2 = 1$$

## Step 3: 3. Solve the system of equations to obtain the steady-state vector.

To solve the system of equations, we can simplify and solve one of the equations first. Since equation 2 is simpler, let's start with it: $$\frac{1}{5}v_1 + \frac{2}{5}v_2 = v_2$$ Subtracting $$\frac{2}{5}v_2$$ from both sides, we get: $$\frac{1}{5}v_1 = \frac{3}{5}v_2$$ Now we can solve for $$v_1$$: $$v_1 = 3v_2$$ Using equation 3, we substitute the expression for $$v_1$$ and solve for $$v_2$$: $$3v_2 + v_2 = 1$$ $$4v_2 = 1$$ $$v_2 = \frac{1}{4}$$ Now, we can find $$v_1$$ using the expression we found earlier: $$v_1 = 3v_2 = 3\left(\frac{1}{4}\right) = \frac{3}{4}$$ Finally, the steady-state vector is: $$v = \begin{bmatrix} \frac{3}{4} \\ \frac{1}{4} \end{bmatrix}$$

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