I have the information that: distribution of probabilities is
$$ P_{X} =\{\frac{11}{24},\frac{7}{24},\frac{1}{8},\frac{1}{8}\}$$ And the matrix has the form; $$A= \left[ \begin{array}{cccc} p_{1}&p_{2}&p_{2}&p_{4}\\ p_{2}&p_{1}&p_{2}&p_{5}\\ p_{3}&p_{3}&p_{2}&p_{5}\\ p_{3}&p_{3}&p_{2}&p_{5}\\ \end{array} \right] $$
I have resolve:
$$A\cdot P_{X}=P_{X}$$
I can see that the sums of the columns is $1$ $$4p_{2}=1 \rightarrow p_{2}=\frac{1}{4}$$
Also we can see row 4 and row 5 are equal
so $$ \left[ \begin{array}{cccc} p_{1}&\frac{1}{4}&\frac{1}{4}&p_{4}\\ \frac{1}{4}&p_{1}&\frac{1}{4}&p_{5}\\ p_{3}&p_{3}&\frac{1}{4}&p_{5}\\ p_{3}&p_{3}&\frac{1}{4}&p_{5}\\ \end{array} \right] \cdot \left[ \begin{array}{c} \frac{11}{24}\\ \frac{7}{24}\\ \frac{1}{8}\\ \frac{1}{8}\\ \end{array} \right]=\left[ \begin{array}{c} \frac{11}{24}\\ \frac{7}{24}\\ \frac{1}{8}\\ \frac{1}{8}\\ \end{array} \right] $$ multiply except the last row(and add the last equation $=1$): $$\frac{11}{24}p_{1}+\frac{1}{8}p_{4}=\frac{17}{48}\\ \frac{7}{24}p_{1}+\frac{1}{8}p_{5}=\frac{7}{48}\\ \frac{3}{4}p_{3}+\frac{1}{8}p_{5}=\frac{3}{32}\\ p_{1}+\frac{1}{4}+p_{3}+p_{4}+p_{5}=1$$
How can I continue and solve this?
Thank you very much
Hints: write the system as a matrix equation, say $Ax=b$, where $x=(p_1,\cdots,p_5)^t$ is a column vector of variables, $b$ is a constant column vector, consider the argumented matrix $B=[A \ b].$\begin{bmatrix} 1&0&1&1&1&\frac{3}{4}\\ \frac{11}{24}&0&0&\frac{1}{8}&0&\frac{17}{48}\\ \frac{7}{24}&0&0&0&\frac{1}{8}&\frac{3}{32}\\ 0&0&\frac{3}{4}&0&\frac{1}{8}&\frac{3}{4} \end{bmatrix} For above matrix $B$, by a successive elementary transformation of rows , simplifying $B$ to get an upper-triangular matrix, then solving the last equation to get solutions and iterating step by step