This ties into a problem I'm working on in Stochastic Processes (In Statistics) but the basics boils down to Linear Algebra.
Suppose I have something like:
$\frac{1}{2}x_1 + \frac{1}{4}x_3 = x_1$
$\frac{1}{4}x_1 + \frac{1}{4}x_4 = x_4$
$\frac{1}{4}x_1 + \frac{1}{2}x_2 + \frac{1}{4}x_3 + \frac{1}{4}x_4 = x_2$
$\frac{1}{2}x_2 + \frac{1}{2}x_3 + \frac{1}{2}x_4 = x_3$
$x_1 + x_2 + x_3 + x_4 = 1$
This is just a part of the overall system I had but I was wondering how to solve this using linear algebra. Ideally I'd be able to input it into my calculator using RREF and let that solve it.
Note: I know how to reduce this system by hand. Just wondering if my calculator would be able to solve this. Thanks in advance.
In fact even reducing the system is using linear algebra properties (by reducing the system, you multiply the matrix of the system by other matrices on his left side).
But if you want to use matrice to represent the problem, it is very simple. You have four variables: $x_1, x_2, x_3 $ and $x_4$, that you will group into a vector $X = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4\end{bmatrix}$. You are researching the solution of the system $AX = X$ with the constraint on $X$ that $x_1 + x_2 + x_3 + x_4 = 1$. If you now a bit about matrix multiplication you can verify that
$A = \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{4} & 0 \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} & \frac{1}{4} \\ 0 & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & 0 & 0& \frac{1}{4}\end{bmatrix} $
corresponds to what we are researching, and is unique!! I Just put the coefficient of $x_i$ in the equation with result $x_j$ into $A_{ij}$.
To solve this system, you now must find the eigenvectors with eigenvalues one of the system. This can be done easily with your calculator, normally you will have a function for this.
I will now explain some math to find the solution by hand (on the last paragraph I explain how to respect the constraint, which is very easy).
In order to find them you have to find the vectors X such that $A * X = I * X$ or $(A-I)*X = 0$ (I is the indentity matrix). I prefer the last version.
So let's find the vectors $X$ such that $\begin{bmatrix} -\frac{1}{2} & 0 & \frac{1}{4} & 0\\ \frac{1}{4} & -\frac{1}{2} & \frac{1}{4} & \frac{1}{4} \\ 0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{4} & 0 & 0& -\frac{3}{4}\end{bmatrix} * X= 0$.
In linear algebra the set of this vector is named the kernel space (or zero space) of A. It is a vectorial space. So you know that zero is a solution of this equation. But it doesn't respect the constraint $x_1 + x_2 + x_3 + x_4 = 1$.
We need a non-zero solution of the equation. In fact, to solve this by hand, you need to find a linear combination of the columns of $A-I$ such that it equals to zero (if you can't there is no solution). You can notice that $3*C_1 + 5*C_2 + 6*C_3 + C_4 = 0$. I just make it works so that the coefficient in for $C_4$ is 1, so on the last row in order to have $0$, you need to have a $3$ for $C_1$. The on the first row, in order to have $0$, you need to have a $6$ for $C_3$. Then you complete in order to have $0$ everywhere with a $5$ on $C_2$. You have to keep in mind that it won't always be so easy.
In this case I supposed that the eigenvalue 1 has only an eigenspace of dimension 1. It could be more, and you can find more than one eigenvector. In this case, any linear combination of all of this vector is a solution.
If you don't like this way, you will need to reduce the system, sorry.
So $\begin{bmatrix} 3 \\ 5 \\ 6 \\ 1 \end{bmatrix}$ is a solution of $A*X = X$ (you will normally get a multiple of this vector). You now just have to know that any solution is a linear multiple of this vector. You want to find now a number $\alpha$ such that $\alpha*X$ is a solution of the equation representing the constraint. Here $\alpha$ is the invert of the sum of $X$ components. So you need to divide the vector $X$ by the sum of its components in order to respect the constraint, the final solution is: $\begin{bmatrix} \frac{3}{15} \\ \frac{5}{15} \\ \frac{6}{15} \\ \frac{1}{15} \end{bmatrix} = \begin{bmatrix} \frac{1}{5} \\ \frac{1}{3} \\ \frac{2}{5} \\ \frac{1}{15} \end{bmatrix}$.