Find the steady-state vector for the matrix below. (Enter exact values for components as an ordered n-tuple.)

$\begin{bmatrix} 0.82 & 0.27 \\ 0.18 & 0.73 \end{bmatrix}$

Neetesh Kumar | October 21, 2024

Linear Algebra Homework Help

This is the solution to Math2B Course: Linear Algebra
Final Exam Question Number 17
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Step-by-step solution:

To find the steady-state vector $\mathbf{v}$ for the given matrix, we need to solve the equation:

$A \mathbf{v} = \mathbf{v}$

Where $A$ is the given matrix:

$A = \begin{bmatrix} 0.82 & 0.27 \\ 0.18 & 0.73 \end{bmatrix}$

The steady-state vector $\mathbf{v}$ is of the form $\mathbf{v} = \begin{bmatrix} x \\ y \end{bmatrix}$, and it must satisfy the equation $A \mathbf{v} = \mathbf{v}$.
This can be rewritten as a system of equations:

$\begin{aligned} 0.82x + 0.27y &= x \\ 0.18x + 0.73y &= y \end{aligned}$

Step 1: Solve the system of equations

Start by simplifying each equation:

1. For the first equation: $0.82x + 0.27y = x$ Subtract $x$ from both sides: $0.82x + 0.27y - x = 0$
   This simplifies to: $-0.18x + 0.27y = 0$ So, $0.18x = 0.27y \quad \Rightarrow \quad \frac{x}{y} = \frac{0.27}{0.18} = \frac{3}{2}$ Thus, $x = \frac{3}{2}y$

2. For the second equation: $0.18x + 0.73y = y$ Subtract $y$ from both sides: $0.18x + 0.73y - y = 0$
   Simplifying gives: $0.18x - 0.27y = 0$ Which is the same equation we obtained from the first part.
   Thus, we have confirmed that $x = \frac{3}{2}y$.

Step 2: Normalize the solution

The steady-state vector $\mathbf{v}$ is a probability vector, so the sum of its components must be 1.

We know that $x = \frac{3}{2}y$. Therefore:

$x + y = 1$

Substitute $x = \frac{3}{2}y$ into this equation:

$\frac{3}{2}y + y = 1$

This simplifies to:

$\frac{5}{2}y = 1 \quad \Rightarrow \quad y = \frac{2}{5}$

Now substitute $y = \frac{2}{5}$ into $x = \frac{3}{2}y$:

$x = \frac{3}{2} \times \frac{2}{5} = \frac{6}{10} = \frac{3}{5}$

Thus, the steady-state vector is:

$\mathbf{v} = \left( \frac{3}{5}, \frac{2}{5} \right)$

Final Answer:

The steady-state vector is $\left( \frac{3}{5}, \frac{2}{5} \right)$.

$\begin{bmatrix} \frac{3}{5} & \frac{2}{5} \\ \frac{3}{5} & \frac{2}{5} \end{bmatrix}$

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