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Siong Thye Goh
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We can solve linear system of equations.

Equation $(a)$ can be converted to $$(0.2-1)\pi_1 + 0.1\pi_2 + 0.55\pi_3=0\tag{a'}$$

Similarly for $b$ and $c$.

Also, with the constraint $\pi_1+\pi_2+\pi_3=1$

We have $3$ variables and $4$ constraints.

$$\pi=P^T\pi$$ $$e^T\pi=1$$

$$\begin{bmatrix} P^T-I \\ e^T\end{bmatrix}\pi =\begin{bmatrix} 0_3 \\ 1\end{bmatrix}$$

You can perform Gaussian Elimination to get the solution.

Here is the Octave solution:

octave:1> A = [-0.8, 0.1, 0.55, 0; 0.3, -1, 0, 0; 0.5,  0.9, -0.55,  0; 1, 1, 1, 1]
A =

  -0.80000   0.10000   0.55000   0.00000
   0.30000  -1.00000   0.00000   0.00000
   0.50000   0.90000  -0.55000   0.00000
   1.00000   1.00000   1.00000   1.00000

octave:2> rref(A)
ans =

   1.00000   0.00000   0.00000   0.37037
   0.00000   1.00000   0.00000   0.11111
   0.00000   0.00000   1.00000   0.51852
   0.00000   0.00000   0.00000   0.00000

We can solve linear system of equations.

Equation $(a)$ can be converted to $$(0.2-1)\pi_1 + 0.1\pi_2 + 0.55\pi_3=0\tag{a'}$$

Similarly for $b$ and $c$.

Also, with the constraint $\pi_1+\pi_2+\pi_3=1$

We have $3$ variables and $4$ constraints.

You can perform Gaussian Elimination to get the solution.

Here is the Octave solution:

octave:1> A = [-0.8, 0.1, 0.55, 0; 0.3, -1, 0, 0; 0.5,  0.9, -0.55,  0; 1, 1, 1, 1]
A =

  -0.80000   0.10000   0.55000   0.00000
   0.30000  -1.00000   0.00000   0.00000
   0.50000   0.90000  -0.55000   0.00000
   1.00000   1.00000   1.00000   1.00000

octave:2> rref(A)
ans =

   1.00000   0.00000   0.00000   0.37037
   0.00000   1.00000   0.00000   0.11111
   0.00000   0.00000   1.00000   0.51852
   0.00000   0.00000   0.00000   0.00000

We can solve linear system of equations.

Equation $(a)$ can be converted to $$(0.2-1)\pi_1 + 0.1\pi_2 + 0.55\pi_3=0\tag{a'}$$

Similarly for $b$ and $c$.

Also, with the constraint $\pi_1+\pi_2+\pi_3=1$

We have $3$ variables and $4$ constraints.

$$\pi=P^T\pi$$ $$e^T\pi=1$$

$$\begin{bmatrix} P^T-I \\ e^T\end{bmatrix}\pi =\begin{bmatrix} 0_3 \\ 1\end{bmatrix}$$

You can perform Gaussian Elimination to get the solution.

Here is the Octave solution:

octave:1> A = [-0.8, 0.1, 0.55, 0; 0.3, -1, 0, 0; 0.5,  0.9, -0.55,  0; 1, 1, 1, 1]
A =

  -0.80000   0.10000   0.55000   0.00000
   0.30000  -1.00000   0.00000   0.00000
   0.50000   0.90000  -0.55000   0.00000
   1.00000   1.00000   1.00000   1.00000

octave:2> rref(A)
ans =

   1.00000   0.00000   0.00000   0.37037
   0.00000   1.00000   0.00000   0.11111
   0.00000   0.00000   1.00000   0.51852
   0.00000   0.00000   0.00000   0.00000
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Source Link
Siong Thye Goh
  • 7k
  • 3
  • 21
  • 31

We can solve linear system of equations.

Equation $(a)$ can be converted to $$(0.2-1)\pi_1 + 0.1\pi_2 + 0.55\pi_3=0\tag{a'}$$

Similarly for $b$ and $c$.

Also, with the constraint $\pi_1+\pi_2+\pi_3=1$

We have $3$ variables and $4$ constraints.

You can perform Gaussian Elimination to get the solution.

Here is the Octave solution:

octave:1> A = [-0.8, 0.1, 0.55, 0; 0.3, -1, 0, 0; 0.5,  0.9, -0.55,  0; 1, 1, 1, 1]
A =

  -0.80000   0.10000   0.55000   0.00000
   0.30000  -1.00000   0.00000   0.00000
   0.50000   0.90000  -0.55000   0.00000
   1.00000   1.00000   1.00000   1.00000

octave:2> rref(A)
ans =

   1.00000   0.00000   0.00000   0.37037
   0.00000   1.00000   0.00000   0.11111
   0.00000   0.00000   1.00000   0.51852
   0.00000   0.00000   0.00000   0.00000