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Linear Algebra, Theory And Applications, 2012a

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1.10. SYSTEMS OF EQUATIONS 27<br />

Why is this?<br />

If (x, y) solves (1.4) then it solves the first equation in (1.5). Also, it satisfies aE 1 = af 1<br />

and so, since it also solves E 2 = f 2 it must solve the second equation in (1.5). If (x, y)<br />

solves (1.5) then it solves the first equation of (1.4). Also aE 1 = af 1 and it is given that<br />

the second equation of (1.5) is verified. Therefore, E 2 = f 2 and it follows (x, y) isasolution<br />

of the second equation in (1.4). This shows the solutions to (1.4) and (1.5) are exactly the<br />

same which means they have the same solution set. Of course the same reasoning applies<br />

with no change if there are many more variables than two and many more equations than<br />

two. It is still the case that when one equation is replaced with a multiple of another one<br />

added to itself, the solution set of the whole system does not change.<br />

The other thing which does not change the solution set of a system of equations consists<br />

of listing the equations in a different order. Here is another example.<br />

Example 1.10.1 Find the solutions to the system,<br />

x +3y +6z =25<br />

2x +7y +14z =58<br />

2y +5z =19<br />

(1.6)<br />

To solve this system replace the second equation by (−2) times the first equation added<br />

to the second. This yields. the system<br />

x +3y +6z =25<br />

y +2z =8<br />

2y +5z =19<br />

(1.7)<br />

Now take (−2) times the second and add to the third. More precisely, replace the third<br />

equation with (−2) times the second added to the third. This yields the system<br />

x +3y +6z =25<br />

y +2z =8<br />

z =3<br />

(1.8)<br />

At this point, you can tell what the solution is. This system has the same solution as the<br />

original system and in the above, z =3. Then using this in the second equation, it follows<br />

y + 6 = 8 and so y =2. Now using this in the top equation yields x + 6 + 18 = 25 and so<br />

x =1.<br />

This process is not really much different from what you have always done in solving a<br />

single equation. For example, suppose you wanted to solve 2x +5=3x − 6. You did the<br />

same thing to both sides of the equation thus preserving the solution set until you obtained<br />

an equation which was simple enough to give the answer. In this case, you would add −2x<br />

to both sides and then add 6 to both sides. This yields x =11.<br />

In (1.8) you could have continued as follows. Add (−2) times the bottom equation to<br />

the middle and then add (−6) times the bottom to the top. This yields<br />

x +3y =19<br />

y =6<br />

z =3<br />

Now add (−3) times the second to the top. This yields<br />

x =1<br />

y =6<br />

z =3<br />

,

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