Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

Markov chain algebraic problem?

equations:

p1+p2+p3=1

-6p1+2p2+9p3=0

3p1-12p2+p3=0

3p1+10p2-10p3=0

Update:

what is the value of p1 p2 and p3

1 Answer

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  • 2 months ago

    (1) : p₁ + p₂ + p₃ = 1

    (2) : 6p₁ + 2p₂ + 9p₃ = 0

    (3) : 3p₁ - 12p₂ + p₃ = 0

    You calculate (2) - [6 * (1)] and you obtain the equation (4)

    (4) : (6p₁ + 2p₂ + 9p₃) - 6.(p₁ + p₂ + p₃) = 0 - (6 * 1)

    (4) : 6p₁ + 2p₂ + 9p₃ - 6p₁ - 6p₂ - 6p₃ = - 6

    (4) : - 4p₂ + 3p₃ = - 6

    (4) : 4p₂ = 3p₃ + 6

    (4) : p₂ = (3p₃ + 6)/4

    You calculate (2) - [2 * (3)]

    (6p₁ + 2p₂ + 9p₃) - 2.(3p₁ - 12p₂ + p₃) = 0 - (2 * 0)

    6p₁ + 2p₂ + 9p₃ - 6p₁ + 24p₂ - 2p₃ = 0

    26p₂ + 7p₃ = 0 → recall (4): p₂ = (3p₃ + 6)/4

    26.[(3p₃ + 6)/4] + 7p₃ = 0

    13.[(3p₃ + 6)/2] + 7p₃ = 0

    13.(3p₃ + 6) + 14p₃ = 0

    39p₃ + 78 + 14p₃ = 0

    53p₃ = - 78

    → p₃ = - 78/53

    Recall (4): p₂ = (3p₃ + 6)/4

    p₂ = [3.(- 78/53) + 6]/4

    p₂ = [(- 234/53) + (318/53)]/4

    p₂ = [84/53]/4

    → p₂ = 21/53

    Recall (1): p₁ + p₂ + p₃ = 1

    p₁ = 1 - p₂ - p₃

    p₁ = 1 - (21/53) - (- 78/53)

    p₁ = (53 - 21 + 78)/53

    → p₁ = 110/53

    • Wayne DeguMan
      Lv 7
      2 months agoReport

      Equation 2 is -6p₁ + 2p₂ + 9p₃ = 0 

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