Mathematicsss said:
Post moved by moderator from Homework section
Hello, I was curious as to why Gaussian elimination works. I know that if we have two ( or more) systems of two (or more) linear equations, we can write then in terms of a matrix. However, what does it mean when I get the identity on the left hand side and the solution on the right?
Gaussian elimination has been around hundreds of years before the invention of matrices, so matrices are not needed at all (but they make the work cleaner and easier). The method just solves equations successively, each time eliminating a new independent variable.
For example, if we have the system
$$\begin{array}{crcc}
(1):& x + 2y - 3z &=& 10\\
(2):& 2x - y + 4z &=& 20 \\
(3):& 2x - z &=& 8
\end{array}
$$
We can use eq (1) to express ##x## in terms of ##y## and ##z##:
$$(1a): \; x = 10 - 2y + 3z. $$
Substitute (1a) into (2) and (3), to get
$$\begin{array}{crrcc}
(2a):& 20-5y+10z = 20 & \Rightarrow& 5y - 10 z = 0 \\
(3a): & 20-4y+5z = 8 & \Rightarrow& 4y - 5z = 12
\end{array}
$$
So, we have a new system of 2 equations in the two unknowns ##y,z##. We have "eliminated" ##x## and made the problem smaller and easier to deal with. We can continue, for example by using eq. (2a) to express ##y## in terms of ##z##. That will leave us with a single equation (3b) that contains ##y## alone.
The reason it "works" is that each step involves a valid re-writing of the equation or equations, using only standard algebraic rules, and never, ever, performing an illegal step such as dividing by 0. For example, we obtained eq. (2a) by re-writing (2) as ##2(10 - 2y + 3z) -y + 4z = 20## and simplifying.