A First Course in Linear Algebra, 2017a

12 Systems of Equations
2. Next we want to prove that the systems 1.1 and 1.3 have the same solution set. That is E 1 = b 1 ,E 2 =
b 2 has the same solution set as the system E 1 = b 1 ,kE 2 = kb 2 provided k ≠ 0. Let (x 1 ,···,x n ) be a
solution of E 1 = b 1 ,E 2 = b 2 ,. We want to show that it is a solution to E 1 = b 1 ,kE 2 = kb 2 . Notice that
the only difference between these two systems is that the second involves multiplying the equation,
E 2 = b 2 by the scalar k. Recall that when you multiply both sides of an equation by the same number,
the sides are still equal to each other. Hence if (x 1 ,···,x n ) is a solution to E 2 = b 2 , then it will also
be a solution to kE 2 = kb 2 . Hence, (x 1 ,···,x n ) is also a solution to 1.3.
Similarly, let (x 1 ,···,x n ) be a solution of E 1 = b 1 ,kE 2 = kb 2 . Then we can multiply the equation
kE 2 = kb 2 by the scalar 1/k, which is possible only because we have required that k ≠ 0. Just as
above, this action preserves equality and we obtain the equation E 2 = b 2 . Hence (x 1 ,···,x n ) is also
a solution to E 1 = b 1 ,E 2 = b 2 .
3. Finally, we will prove that the systems 1.1 and 1.4 have the same solution set. We will show that
any solution of E 1 = b 1 ,E 2 = b 2 is also a solution of 1.4. Then, we will show that any solution of
1.4 is also a solution of E 1 = b 1 ,E 2 = b 2 .Let(x 1 ,···,x n ) be a solution to E 1 = b 1 ,E 2 = b 2 .Then
in particular it solves E 1 = b 1 . Hence, it solves the first equation in 1.4. Similarly, it also solves
E 2 = b 2 . By our proof of 1.3,italsosolveskE 1 = kb 1 . Notice that if we add E 2 and kE 1 , this is equal
to b 2 +kb 1 . Therefore, if (x 1 ,···,x n ) solves E 1 = b 1 ,E 2 = b 2 it must also solve E 2 +kE 1 = b 2 +kb 1 .
Now suppose (x 1 ,···,x n ) solves the system E 1 = b 1 ,E 2 + kE 1 = b 2 + kb 1 . Then in particular it is a
solution of E 1 = b 1 . Again by our proof of 1.3, it is also a solution to kE 1 = kb 1 . Now if we subtract
these equal quantities from both sides of E 2 + kE 1 = b 2 + kb 1 we obtain E 2 = b 2 , which shows that
the solution also satisfies E 1 = b 1 ,E 2 = b 2 .
Stated simply, the above theorem shows that the elementary operations do not change the solution set
of a system of equations.
We will now look at an example of a system of three equations and three variables. Similarly to the
previous examples, the goal is to find values for x,y,z such that each of the given equations are satisfied
when these values are substituted in.
Example 1.9: Solving a System of Equations with Elementary Operations
Find the solutions to the system,
♠
x + 3y + 6z = 25
2x + 7y + 14z = 58
2y + 5z = 19
(1.5)
Solution. We can relate this system to Theorem 1.8 above. In this case, we have
E 1 = x + 3y + 6z, b 1 = 25
E 2 = 2x + 7y + 14z, b 2 = 58
E 3 = 2y + 5z, b 3 = 19
Theorem 1.8 claims that if we do elementary operations on this system, we will not change the solution
set. Therefore, we can solve this system using the elementary operations given in Definition 1.6. First,