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

Short Answer

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The steady-state vector for the given transition matrix is \(v = \begin{bmatrix}\frac{3}{4} \\ \frac{1}{4}\end{bmatrix}\).
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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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Most popular questions from this chapter

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