Answer:
Step-by-step explanation:
Given that:
[tex]P = \left[\begin{array}{ccc}0.22&0.20&0.65\\0.62&0.60&0.15\\0.16&0.20&0.20\end{array}\right][/tex]
For a steady-state of a given matrix [tex]\bar X[/tex]
[tex]\bar X = \left[\begin{array}{c}a\\b\\c\end{array}\right][/tex]
As a result P[tex]\bar X[/tex] = [tex]\bar X[/tex] and a+b+c must be equal to 1
So, if P[tex]\bar X[/tex] = [tex]\bar X[/tex]
Then;
[tex]P = \left[\begin{array}{ccc}0.22&0.20&0.65\\0.62&0.60&0.15\\0.16&0.20&0.20\end{array}\right]\left[\begin{array}{c}a\\b\\c\end{array}\right] =\left[\begin{array}{c}a\\b\\c\end{array}\right][/tex]
[tex]\implies \left\begin{array}{ccc}0.22a+&0.20b+&0.65c\\0.62a+&0.60b+&0.15c\\0.16a+&0.20b+&0.20c\end{array} \right = \left \begin{array}{c}a ---(1)\\b---(2)\\c---(3)\end{array}\right[/tex]
Equating both equation (1) and (3)
(0.22a+ 0.2b + 0.65c) - (0.16a + 0.2b + 0.2c) = a - c
0.06a + 0.45c = a - c
collect like terms
0.06a - a = -c - 0.45c
-0.94 a = -1.45 c
0.94 a = 1.45 c
[tex]c =\dfrac{ 0.94}{1.45}a[/tex]
[tex]c =\dfrac{ 94}{145}a --- (4)[/tex]
Using equation (2)
0.62a + 0.60b + 0.15c = b
where;
c = 94/145 a
[tex]0.62a + 0.60b + 0.15(\dfrac{94}{145}) a= b[/tex]
[tex]0.62a + 0.15(\dfrac{94}{145}) a= -0.60b+b[/tex]
[tex]0.62a + (\dfrac{141}{1450}) a= 0.40b[/tex]
[tex](0.62+\dfrac{141}{1450}) a= 0.40b[/tex]
[tex](\dfrac{62}{100}+\dfrac{141}{1450}) a= 0.40b[/tex]
[tex](\dfrac{1043}{1450})a= 0.40b[/tex]
[tex](\dfrac{1043}{1450})a= \dfrac{4}{10} b[/tex]
[tex](\dfrac{1043 \times 10}{1450 \times 4})a = \dfrac{4}{10} \times \dfrac{10}{4}[/tex]
[tex]b = (\dfrac{1043}{580}) a --- (5)[/tex]
From a + b + c = 1
[tex]a + \dfrac{1043}{580}a + \dfrac{94}{145} a = 1[/tex]
[tex]a + \dfrac{1043}{580}a + \dfrac{94*4}{145*4} a = 1[/tex]
[tex]a + \dfrac{1043}{580}a + \dfrac{376}{580} a = 1[/tex]
[tex]\dfrac{580+ 1043+376 }{580} a= 1[/tex]
[tex]\dfrac{1999}{580} a= 1[/tex]
[tex]a = \dfrac{580}{1999}[/tex]
∴
[tex]b = \dfrac{1043}{580} \times \dfrac{580}{1999}[/tex]
[tex]b = \dfrac{1043}{1999}[/tex]
[tex]c = \dfrac{94}{145} \times \dfrac{580}{1999}[/tex]
[tex]c= \dfrac{376}{1999}[/tex]
∴
The steady matrix of [tex]\bar X[/tex] is:
[tex]\bar X = \left[\begin{array}{c}\dfrac{580}{1999} \\ \\ \dfrac{1043}{1999}\\ \\ \dfrac{376}{1999}\end{array}\right][/tex]
