4.1. SYSTEMS OF EQUATIONS, ALGEBRAIC PROCEDURES 57

If (x1, · · · ,xn) solves B then (x1, · · · ,xn) solves the first equation of A. Also (x1, · · · ,xn)solves the second equation of B so, since E1 = f1, it follows that aE1 = a f1 and these canbe subtracted from each side of the second equation in B to find E2 = f2 which is the secondequation of A. ■

Stated simply, the above theorem shows that the elementary operations do not changethe solution set of a system of equations.

Exercise 4.1.6 Identify what was done to get the following equivalent systems of equations.

x− y = 2x+ z = 1

−x+2y+2z =−1,

x− y = 2x+ z = 1y+2z = 1

,

x− y = 2y+ z =−1y+2z = 1

,

x− y = 2y+ z =−1

z = 2,

x+ z = 1y+ z =−1

z = 2,

x =−1y =−3z = 2

4.1.2 Gauss EliminationA less cumbersome way to represent a linear system is to write it as an augmented matrix.For example the linear system, 4.2 can be written as 1 3 6 | 25

2 7 14 | 580 2 5 | 19

 .

It has exactly the same information as the original system but here it is understood there

is an x column,

 120

 , a y column,

 372

 and a z column,

 6145

 . The rows corre-

spond to the equations in the system. Thus the top row in the augmented matrix correspondsto the equation, x+ 3y+ 6z = 25. Now when you replace an equation with a multiple ofanother equation added to itself, you are just taking a row of this augmented matrix andreplacing it with a multiple of another row added to it. Thus the first step in solving 4.2would be to take (−2) times the first row of the augmented matrix above and add it to thesecond row,  1 3 6 | 25

0 1 2 | 80 2 5 | 19

 .

Note how this corresponds to 4.3. Next take (−2) times the second row and add to thethird,  1 3 6 | 25

0 1 2 | 80 0 1 | 3

This augmented matrix corresponds to the system

x+3y+6z = 25y+2z = 8

z = 3