“Your proof is wrong. But maybe you are Batman and the statement actually is true.” - from this
Two linear systems are called equivalent if they have the same solution sets. Our approach to solving a system of linear equations is to transform the given system into an equivalent one that is easier to solve.
Solve the system:
Solution:
Starting from the last equation and working backward, we find successively that z = 2, y = 5 - 3(2) = - 1, and x = 2 + ( -1) + 2 = 3. The procedure used to the above system is called back-substitution method. We now turn to the general strategy for transforming a given system into an equivalent one that can be solved easily by back substitution.
Solve the system:
Solution:
We first need to eliminate the variable x from equations 2 and 3. Observe that subtracting appropriate multiples of equation 1 from equations 2 and 3 will do the trick. Next, observe that we are operating on the coefficients, not on the variables, so we can save ourselves some writing if we record the coefficients and constant terms in the matrix where the first three columns contain the coefficients of the variables in order, the final column contains the constant terms, and the vertical bar serves to remind us of the equal signs in the equations. This matrix is called the augmented matrix of the system.
There are various ways to convert the given system into one with the triangular pattern we are after. The steps we will use here are closest in spirit to the more general method described in the next section.We will perform the sequence of operations on the given system and simultaneously on the corresponding augmented matrix. We begin by eliminating x from equations 2 and 3.
This is the same system that we solved using back substitution, where we found that the solution was [3, 1, 2]. This is therefore also the solution to the system given in this example.Why? The calculations above show that any solution of the given system is also a solution of the final one. But since the steps we just performed are reversible, we could recover the original system, starting with the final system. (How?) So any solution of the final system is also a solution of the given one. Thus, the systems are equivalent (as are all of the ones obtained in the intermediate steps above). Moreover, we might just as well work with matrices instead of equations, since it is a simple matter to reinsert the variables before proceeding with the back substitution.
DISCLAIMER
This post is directly copy-pasted from the book "Linear Algebra, A Modern Introduction" by David Poole
