# Markov chain algebraic problem?

equations:

p1+p2+p3=1

-6p1+2p2+9p3=0

3p1-12p2+p3=0

3p1+10p2-10p3=0

what is the value of p1 p2 and p3

### 1 Answer

- la consoleLv 72 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

Equation 2 is -6p₁ + 2p₂ + 9p₃ = 0