# Determining independent linear systems of equations

I am not if I am using the right terms. But I am looking for a method to determine if equations of a linear system are independent (decoupled) or not. For example consider the following system:

$\begin{bmatrix} a_{11} & a_{12} & 0 & 0 \\ a_{21} & a_{22} & 0 & 0 \\ 0 & 0 & a_{33} & a_{34} \\ 0 & 0 & a_{43} & a_{44} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{bmatrix}$.

The variables $x_1$ and $x_2$ in this system are completely independent of $x_3$ and $x_4$; i.e., if for example the value of $b_1$ changes, it only changes the values of $x_1$ and $x_2$ and has not any impact on the values of $x_3$ and $x_4$.

The problem I am looking for, is not only about coefficient matrix itself, but it is about the whole system. For example consider this system:

$\begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} +1 \\ \alpha \\ -1 \end{bmatrix}$.

Looking at the coefficient matrix, one may say it is a coupled system. But the RHS says that the value of $x_2$ depends only on $\alpha$.

Is there any method to determine if equations of a linear system are independent (decoupled)?

Yes: we use Gaussian elimination to place the matrix of coefficients in row echelon form. See this: http://en.m.wikipedia.org/wiki/Gaussian_elimination Now this is not an efficient way because we have to do it for all the $n!$ possible orderings of the variables. For the right ordering, when the system admits decoupling: you will see the blocks of zeros as in your first example.
• You are right. My answer is wrong or incomplete at the best. The brute force way I see is to do Guaussian elimination for every possible ordering of the variables, $n!$ and then for the right ordering you would see the blocks/decoupled components as in your example. – Sergio Parreiras Nov 15 '13 at 14:41