The following process allows us to use the elementary operations to solve a larger system:

First, set the leading coefficient of the first equation to one using either operation 1 or 2.

Then, eliminate the leading variable of the first equation from subsequent equations using operation 3.

The second equation will have a different leading variable now from equation 1. Repeat the prior steps on the second equation. Do this again

for the third equation, and so on. The last equation will only contain one variable. You can then go backwards through the equations and solve

for each of them.

Example Solve the following system of equations:

First we switch the order so the leading coefficient of the first equation is 1:

Next, we multiply the first equation by 3 and subtract it from the second:

Then we multiply the first equation by 3 and subtract it from the third equation:

Then we multiply the first equation by 4 and subtract it from the last equation:

3x + 2y + z = 10

5x − 2y + 3z = 20

x + y + z = 50

≡

5x − 2y + 3z = 20

3x + 2y + z = 10

x + y + z = 50

3x + 2y + z = 10

5x − 2y + 3z = 20

x + y + z = 50

≡

3x + 2y + z = 10

5x − 2y + 3z = 20

2x + 2y + 2z = 100

3x + 2y + z = 10

5x − 2y + 3z = 20

x + y + z = 50

≡

3x + 2y + z = 10

7x + 5z = 120

x + y + z = 50

3w − 5x − 4y − z = −70

3w + 6x − 2y + 6z = 101

4w − 2x + 3y + 5z = 66

w + x + y + z = 30

9/27/2018 Module 2

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Now we can multiply equation 3 by 3 and add it to equation 2 to set its leading multiplier to 1:

Now we eliminate from equation 3 by subtracting 3 times equation 2:

We go on to eliminate from equation 4 by adding 6 times equation 2

Next we can divide equation 3 by 61:

And add equation 3 × 133 to equation 4:

Solving for in equation 4 and plugging it back into the prior equations we get

We now repeat the process for

Continuing the process, we find and .