Solving a 4 variable linear system equation with known coefficients

Question

The knowledge is trivial and I'm kinda ashamed to ask, but I am not able to solve the following system using substitution or elimination.

$ λP_0 = μP_{1a} + μP_{1b} \\ μP_{1a} + λP_{1a} = λP_0 + μP_2 \\ 2μP_2 = λP_{1a} + λP_{1b} \\ P_0 + P_{1a} + P_{1b} + P_2 = 1 $

where λ,μ are known.

Do you have any recommendations?

P.S: It needs to be solved... by hand

Answer 1

We can exploit the structure of the system to simplify a little and temporarily merge $P_{1a}+P_{1b}$.

$$\begin{cases}λP_0 &= μP_1 \\ 2μP_2 &= λP_1\\ P_0 + P_1+ P_2 &= 1 \end{cases}.$$

Then

$$\frac\mu\lambda P_1+P_1+\frac{\lambda}{2\mu}P_1=1$$ gives you $P_1=\dfrac{2\lambda\mu}{2\mu^2+2\lambda\mu+\lambda^2}$ and from there $P_0,P_2.$

Reintegrating the second equation of the initial system, you now have a simple $2\times2$ system to split $P_{1a},P_{1b}$.

Answer 2

Write $λP_0 = μP_{1a} + μP_{1b}$ as $$λP_0 - μP_{1a} - μP_{1b} + 0P_2 = 0$$ $μP_{1a} + λP_{1a} = λP_0 + μP_2 $ as $$- λP_0 + (λ-μ)P_{1a} +0P_{1b} - μP_2 = 0$$ etc.

to get the matrix representation of the system as $Mx = a$ where

$M =\begin{pmatrix} λ & -μ & -μ & 0 \\ -λ & λ+μ & 0 & -μ \\ 0 & -λ & -λ & 2μ \\ 1 & 1 & 1 & 1 \end{pmatrix} $, $x = \begin{pmatrix} P_0\\ P_{1a}\\ P_{1b}\\ P_2 \end{pmatrix} $

and $a = \begin{pmatrix} 0\\ 0\\ 0\\ 1 \end{pmatrix} $ We will show how to replace rows of this matrix by combinations with other rows in order to obain a upper triangular matrix: First we see that the second row can be replaced by the sum of itself and the first row. This can be expressed by multiplying with a matrix $D_0 = \begin{pmatrix} 1& 0& 0& 0 \\ 1& 1& 0& 0 \\ 0& 0& 1& 0 \\ 0& 0& 0& 1 \\ \end{pmatrix} $.Since the question generously left us a zero in the third row we only have to get rid of the first element of the fourth row. This is done by the matrix $D_1 = \begin{pmatrix} 1& 0& 0& 0 \\ 0& 1& 0& 0 \\ 0& 0& 1& 0 \\ 1& 0& 0& -λ \\ \end{pmatrix}$. There are still three more matrices, two for the zeroes of the second culumn and one form the zero of the third column (fourth row) they are : $D_2 = \begin{pmatrix} 1& 0& 0& 0 \\ 0& 1& 0& 0 \\ 0& 1& 1& 0 \\ 0& 0& 0& 1 \\ \end{pmatrix}$, $D_3 = \begin{pmatrix} 1& 0& 0& 0 \\ 0& 1& 0& 0 \\ 0& 0& 1& 0 \\ 0& λ+μ& 0& λ \\ \end{pmatrix}$ and $D_3 = \begin{pmatrix} 1& 0& 0& 0 \\ 0& 1& 0& 0 \\ 0& 0& 1& 0 \\ 0& 0& -(λ+μ)& 1 \\ \end{pmatrix}$, this reduces your system to the set of equations $$ λP_0-μP_{1a}-μP_{1b} = 0 $$ $$ λP_{1a}-μP_{1b}-μP_2 = 0$$ $$ -(λ+m)λP_{1b}+μP_2 = 0$$ $$ (-λ^2-2λμ-2μ^2)P_2 = -λ^2 $$ From the last equation you can solve $P_2$, from the third $P_{1b}$, since $P_2$ is already known, and so on.