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Question

# Determine to six decimal places the steady state vector corresponding to the given initial state vector. Also find the smallest integer k such that $\mathbf{x}_{k}=\mathbf{x}_{k+1}$ to 6 decimal places for all entries. (NOTE: Even if it is less computationally efficient, it may be easier to compute state vectors using powers of A instead of the recursive formula.)$A=\left[\begin{array}{rrrr} 0 & 1 & 0.2 & 0.5 \\ 0.2 & 0 & 0.3 & 0 \\ 0.5 & 0 & 0.4 & 0.5 \\ 0.3 & 0 & 0.1 & 0 \end{array}\right], \ \mathbf{x}_{0}=\left[\begin{array}{c} 0.1 \\ 0.2 \\ 0.3 \\ 0.4 \end{array}\right]$

Solution

Verified
Step 1
1 of 4

First find $Ax_{0}$,

\begin{align*} Ax_{0}&=\begin{bmatrix} 0&1&0.2&0.5\\ 0.2&0&0.3&0\\ 0.5&0&0.4&0.5\\ 0.3&0&0.1&0 \end{bmatrix}\begin{bmatrix} 0.1\\ 0.2\\ 0.3\\ 0.4 \end{bmatrix}=\begin{bmatrix} 0.46\\ 0.11\\ 0.37\\ 0.06 \end{bmatrix}\\ &\Rightarrow x_{1}=\begin{bmatrix} 0.46\\ 0.11\\ 0.37\\ 0.06 \end{bmatrix}. \end{align*}

Then find $Ax_{1}$

\begin{align*} Ax_{1}&=\begin{bmatrix} 0&1&0.2&0.5\\ 0.2&0&0.3&0\\ 0.5&0&0.4&0.5\\ 0.3&0&0.1&0 \end{bmatrix}\begin{bmatrix} 0.46\\ 0.11\\ 0.37\\ 0.06 \end{bmatrix}=\begin{bmatrix} 0.214\\ 0.203\\ 0.408\\ 0.175 \end{bmatrix}\\ &\Rightarrow x_{2}=\begin{bmatrix} 0.265\\ 0.405\\ 0.135\\ 0.195 \end{bmatrix}. \end{align*}

Find $Ax_{2}$

\begin{align*} Ax_{2}&=\begin{bmatrix} 0&1&0.2&0.5\\ 0.2&0&0.3&0\\ 0.5&0&0.4&0.5\\ 0.3&0&0.1&0 \end{bmatrix}\begin{bmatrix} 0.265\\ 0.405\\ 0.135\\ 0.195 \end{bmatrix}=\begin{bmatrix} 0.5295\\ 0.0935\\ 0.284\\ 0.093 \end{bmatrix}\\&\Rightarrow x_{3}=\begin{bmatrix} 0.5295\\ 0.0935\\ 0.284\\ 0.093 \end{bmatrix}. \end{align*}

Find $Ax_{3}$

\begin{align*} Ax_{3}&=\begin{bmatrix} 0&1&0.2&0.5\\ 0.2&0&0.3&0\\ 0.5&0&0.4&0.5\\ 0.3&0&0.1&0 \end{bmatrix}\begin{bmatrix} 0.5295\\ 0.0935\\ 0.284\\ 0.093 \end{bmatrix}=\begin{bmatrix} 0.1968\\ 0.1911\\ 0.42485\\ 0.18725 \end{bmatrix}\\&\Rightarrow x_{4}=\begin{bmatrix} 0.1968\\ 0.1911\\ 0.42485\\ 0.18725 \end{bmatrix}. \end{align*}

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