$$ \boldsymbol{\omega}:= \left( \omega_{0},\omega_{1},\omega_{2} \right) ~~ \leftarrow~~ \text{each element of the vector is a real number} $$
$$ \omega_{0}+ \omega_{1}+ \omega_{2} =1 $$
$$ \Pi :=\text{3*3 real matrix} $$
$$ \boldsymbol{\omega}= \boldsymbol{\omega} \Pi $$
I want to determine each element of $~ \boldsymbol{\omega} ~$
What I tried are as below.
$$ \boldsymbol{\omega} -\boldsymbol{\omega} \Pi =O $$
$$ \boldsymbol{\omega} \left( E-\Pi \right) =O $$
$$ A:=\left( E-\Pi \right) $$
$$ \boldsymbol{\omega} A=O $$
$$ \boldsymbol{\omega} A A ^{-1} =O \cdot A ^{-1} $$
$$ \boldsymbol{\omega} E = O $$
$$ \boldsymbol{\omega} =O $$
But actually $~ \boldsymbol{\omega} ~$ is not a zero vector.
Any other more good way exists?
$\Pi=$$ \begin{bmatrix} 0.6 & 0.4 &0\\ 0.3&0 & 0.7\\ 0.2&0 & 0.8\\ \end{bmatrix} $
$$ \boldsymbol{\omega}= \boldsymbol{\omega} \Pi $$
Do the computations of RHS to gain the one matrix.
And you can obtain $\boldsymbol{\omega}= \alpha \left(s,t,u\right) \leftarrow\text{symbol alpha is a variable}$
Using $$\omega_{0}+\omega_{1}+\omega_{2}=1~~,$$
You can determine the value of $\alpha~~~$ .