Find the steady-state vector for the matrix below.

$\begin{bmatrix} 0.6 & 0.1 \\ 0.4 & 0.9 \end{bmatrix}$

The steady-state vector is:

(Type an integer or decimal for each matrix element. Round to the nearest thousandth as needed.)

Neetesh Kumar | October 18, 2024

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This is the solution to Math2B Course: Linear Algebra
Assignment: Ch5 Section 9 Question Number 3
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Step-by-step solution:

The steady-state vector for a matrix $P$ satisfies the equation:

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

$P$ is the given transition matrix, and $\mathbf{v}$ is the steady-state vector. For the given matrix:

$P = \begin{bmatrix} 0.6 & 0.1 \\ 0.4 & 0.9 \end{bmatrix}$

Let $\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$ be the steady-state vector. Then, we have the following system of equations from $P \mathbf{v} = \mathbf{v}$:

$0.6 v_1 + 0.1 v_2 = v_1$

$0.4 v_1 + 0.9 v_2 = v_2$

Simplifying each equation:

For the first equation:

$0.6 v_1 + 0.1 v_2 = v_1$

$0.6 v_1 + 0.1 v_2 - v_1 = 0$

$-0.4 v_1 + 0.1 v_2 = 0$
(Equation 1)

For the second equation:

$0.4 v_1 + 0.9 v_2 = v_2$

$0.4 v_1 + 0.9 v_2 - v_2 = 0$

$0.4 v_1 - 0.1 v_2 = 0$
(Equation 2)

Step 1: Solve the system of equations

We now have the system of equations:

$-0.4 v_1 + 0.1 v_2 = 0$
$0.4 v_1 - 0.1 v_2 = 0$

From Equation 1:

$v_2 = 4 v_1$

Step 2: Use the fact that the steady-state vector sums to 1

Since $v_1 + v_2 = 1$, substitute $v_2 = 4 v_1$:

$v_1 + 4 v_1 = 1$

$5 v_1 = 1$

$v_1 = \frac{1}{5} = 0.2$

Substitute $v_1 = 0.2$ into $v_2 = 4 v_1$:

$v_2 = 4(0.2) = 0.8$

Final Answer

The steady-state vector is:

$\begin{bmatrix} 0.2 \\ 0.8 \end{bmatrix}$

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