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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 xk=xk+1\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=[010.20.50.200.300.500.40.50.300.10], x0=[0.10.20.30.4]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

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First find Ax0Ax_{0},

Ax0=[010.20.50.200.300.500.40.50.300.10][0.10.20.30.4]=[0.460.110.370.06]x1=[0.460.110.370.06].\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 Ax1Ax_{1}

Ax1=[010.20.50.200.300.500.40.50.300.10][0.460.110.370.06]=[0.2140.2030.4080.175]x2=[0.2650.4050.1350.195].\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 Ax2Ax_{2}

Ax2=[010.20.50.200.300.500.40.50.300.10][0.2650.4050.1350.195]=[0.52950.09350.2840.093]x3=[0.52950.09350.2840.093].\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 Ax3Ax_{3}

Ax3=[010.20.50.200.300.500.40.50.300.10][0.52950.09350.2840.093]=[0.19680.19110.424850.18725]x4=[0.19680.19110.424850.18725].\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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