I am not sure whether my approach is correct or wonder if there are other ways. Also it would be very nice if there is a textbook or some chapters that I can study for the question.
I am not even sure how to call the following problem.
Find coefficients of $\lambda_1,\lambda_2,\lambda_3$ in $x$, where
$ x(\lambda_1,y,z) = \lambda_1 + \alpha_by+\alpha_cz \\ y(\lambda_2,x,z) = \lambda_2 +\beta_ax+\beta_cz \\ z(\lambda_3,x,y) = \lambda_3 +\gamma_ax+\gamma_by $
and
$0<\alpha_b,\alpha_c,\beta_a,\beta_c,\gamma_a,\gamma_c<1$
My approach is to build a matrix that distribute each $x,y,z$ with given values.
For the first step, we need a matrix to distribute $x$, which can be represented as
$A= \begin{bmatrix} 0 & \alpha_b & \alpha_c\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
Then we need a matrix for distributing $y$ and $z$
$B= \begin{bmatrix} 0 & 0 & 0 \\ \beta_a & 0 & \beta_c \\ \gamma_a & \gamma_b & 0 \end{bmatrix}$.
After the second step, we have to distribute $x,y,z$ at the same time so we need
$C=A+B= \begin{bmatrix} 0 & \alpha_b & \alpha_c \\ \beta_a & 0 & \beta_c \\ \gamma_a & \gamma_b & 0 \end{bmatrix}$.
The repeated substitution can be represented as the following
$AB+ABC+ABC^2+\cdots+ABC^n \\ = AB\left(I+C+C^2\cdots+C^n \right) \\ = AB(I-C)^{-1} \text{ as $n \rightarrow \infty$}$
Adding initial values,
$\begin{bmatrix} 1 & \alpha_b & \alpha_c \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + AB(I-C)^{-1}.$
The first row of the final matrix is the coefficients of $\lambda_1,\lambda_2,\lambda_3$ in $x$
Edit:
The question can be rephrased as
- Product X contains some sugar $\lambda_1$ and some Product Y and Z
- Product Y contains some salt $\lambda_2$ and some Product X and Z
- Product Z contains some pepper $\lambda_3$ and some Product X and Y
Then how many sugar, salt, pepper in Product X?
Goal is to keep substituting $x,y,z$ in $x$ until only $\lambda_i$ remain in $x$
\begin{aligned} x &= \lambda_1 + \alpha_by + \alpha_cz \\ &= \lambda_1 + \alpha_b(\lambda_2 +\beta_ax+\beta_cz) + \alpha_c(\lambda_3 +\gamma_ax+\gamma_by) \\ &= \lambda_1 + \alpha_b\lambda_2+\alpha_c\lambda_3+(\alpha_b\beta_a+\alpha_c\gamma_a)x+\gamma_by+\alpha_b\beta_cz \\ &\vdots \\ &= \phi_1\lambda_1 + \phi_2\lambda_2 + \phi_3\lambda_3 \end{aligned}
Still not sure to understand, so I will go with that and perhaps update it when I get more information.
We can rewrite the system of linear equations as
$$\begin{bmatrix}1 & -\alpha_b & -\alpha_c\\ -\beta_a & 1 & -\beta_c\\ -\gamma_a & -\gamma_b & 1\end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}\lambda_1\\\lambda_2\\\lambda_3\end{bmatrix}.$$
Then, one can solve for $(x,y,z)$ as a function of $(\lambda_1,\lambda_2,\lambda_3)$ when it is possible using standard linear algebra.
For some values of the parameters, the above matrix is invertible, then there exists a unique solution $(x,y,z)$ to your problem which is given by
$$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1 & -\alpha_b & -\alpha_c\\ -\beta_a & 1 & -\beta_c\\ -\gamma_a & -\gamma_b & 1\end{bmatrix}^{-1}\begin{bmatrix}\lambda_1\\\lambda_2\\\lambda_3\end{bmatrix}.$$
When the matrix is non-invertible, then this will have to be done for all those cases separately and you may have either no solution or an infinite number of them.