Set up the matrix Ax = b where the rows of A contain the coefficients of each component (x0, x1, x2, etc) for a specific equation (row 1 = equation 1, row 2 = equation 2, etc).
Once you do that, find the row-reduced echelon form of the matrix to answer your question.
You can do this in MATLAB or Octave by using the rref command.
Hopefully it's clear that there are two linearly independent equations here. We can write this in matrix form as
[tex] \left( \begin{array}{ccc}
1 & 1 & 1 \\
2 & 2 & 2 \\
3 & 3 & 3\\
1 & 0 & 1 \end{array} \right) \left( \begin{array}{c} x\\ y\\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0\\ 0\\ 0 \end{array} \right) [/tex]
It should be clear that the number of independent equations is equal to the number of linearly independent rows of the matrix I wrote down - this is going to be true in general, where you can write your equations in matrix form, and then the number of linearly independent equations is equal to the number of linearly independent rows of the matrix. This number is called the rank of the matrix and there are a number of ways of computing it.
#6
Thank you ! So that means the rank of the matrix you wrote is 2?