Sometimes one may be asked to solve a system with three variables. This means that there will have to be at least 3 equations. Although it is possible to use the methods used for those with two variables, it is much easier to solve those with 3 variables by using the row elimination method. It is a very lengthy method but it is also the easiest way. Keep in mind that at any time while working one of these problems, a simple mistake with addition or subtraction will result in the wrong answer and backtracking the steps can be very tedious and may require just starting the problem over again. Make sure to go slow and get it right the first time to avoid wasting a lot of time.

 

Row elimination uses a matrix to solve a system. The most important thing to be able to know about matrices for this application is the recognition of the identity matrix. As a reminder of what the 3x3 identity matrix looks like, this is it.

 

 

There are 3 main steps in solving systems of degree 3.

1)      Line up.

2)      Get Identity Matrix.

3)      Write Answer.

 

In Step 2 solve a system by creating a matrix and change it around to get the identity matrix. The answer will be in the result of that matrix. Now that sounds very confusing so here is a walkthrough of a problem.

 

 

 

Example

 

Solve 2a – 4b +3c = 17

a – b + c = 6

a + 2b – c = -3

 

The first step says to line everything up. This is done the same way as in the elimination procedure earlier. The only difference here is that there are three equations and three variables. This system is already lined up.

 

Create a matrix by using the numbers in the system. Keep in mind if a variable is by itself, like the a, b, or c in the second equation. It is understood that there is a 1 in front of it. When writing this new matrix, copy down the numbers in order as they appear in the system.

 

 

 

 

 

2a – 4b +3c = 17

a – b + c = 6

a + 2b – c = -3

 

 

To create this matrix, copy down the numbers in order. Take the matrix and draw a line in between the 3^rd and 4^th column.

 

 

 

What this actually does is divides the matrix so when looking at the left side and ignoring the fourth column there is a 3x3 matrix. It is this 3x3 matrix that will eventually turn into the 3x3 identity matrix.

 

→

 

The process used to do this will be explained. The fourth column will change as the problem moves on and will eventually be the answers.

 

To change the matrix into the identity matrix, it is easiest to follow a certain pattern in order. The order to go by is

 

→.

 

What this illustration is saying is to start with the first element, or the 2, turn it into a 1 just like the identity matrix. From there, move down and turn the 1 underneath it into a 0. Then the 1 underneath that to a zero, and so on. How to turn one number into another is the heart of the problem and will be explained. There are certain things that one can do to change numbers they are;

1)      Switch rows.

2)      Multiply a row by any number and add it to another row.

3)      Divide a row by any number.

4)      Multiply a row by any number

The first thing to do is to turn the 2 into a 1. This is easily done with this matrix because one can switch the position of any of the rows at any time, as stated in #1 above. That is done here by switching Rows 1 and 2.

R₁↔ R₂

 

 

By stitching the rows, notice that it made the first element a 1.

 

Note- In the above illustration, there was some sub-notation, R₁ and R₂. They stand for Row 1 and Row 2. The arrow in between them says that one is switching Row 1 and Row 2. This type of notation will be used for all of these types of problems. It is pretty standard to use this notation and a teacher may require it used to show work.

 

Since the first element is a 1, follow the pattern by going down and turning the 2 into a 0. Do not change row 1 or row 3 here only row 2, so go ahead and copy down what will not change and leave row 2 blank.

 

 

 

It may get confusing from here and needs to be emphasized that row 1 and row 3 will not change with the next step.

 

To change the 2 to a zero multiply row 1 by –2 and add it to row 2. Remember, of row 1 will not change.

 

Multiplying row 1 by –2 will result in the values

Add it to row 2

 

-2 2 -2 -12

+ 2 -4 3 17

0 -2 1 5

 

 

 

This result is a new row 2, so plug it into the matrix. It is shown here with the proper notation.

 

R₁(-2)+ R₂

 

 

The notation is read as “row 1 times –2 added to row 2”. The reason 2 was used was to get opposites, just like in elimination.

 

Continuing on the pattern, look at row 3 and specifically the 1 at the beginning of it. Continue to use row 1 to help turn that 1 into a 0. Multiply row 1 by –1 and add it to row 3, it will turn that 1 into a 0.

 

Multiply row 1 by –1

Add to row 3

 

-1 1 -1 -6

+ 1 2 -1 -3

0 3 -2 -9

This is the new row 3. Here it what the step will look like

 

R₁(-1)+ R₃

 

 

The first part of the pattern says to go down the first column. Specifically, to turn the first element into a 1, the next to a 0, and the third to 0. This is the first column of the identity matrix. Next, the pattern says to go right and try to turn the 3 into a zero. This step is always best done by using row 2. First get opposites with row 2 and row 3 and then add them. Look directly at the -2 in the second row and the 3 in the third row since they are the ones that will be added. One cannot multiply the -2 to get –3 without using fractions, multiply both equations by something (it can be different) to get opposites. This is applying the rule where one can multiply any row by any number. Multiply row 2 by 3 and multiply row 3 by 2. Here is the result.

R₂(3), R₃(2)

 

 

To turn the 6 in row 3 into a zero add row 2 and row 3. Keeping in mind this will only change row 3.

R₂ + R₃

 

 

That term is now 0, so go to the next element in the pattern which is just to the right at the –1. Turn it into a 1. Apply the rule, divide any row by any number. Divide row 3 by –1.

 

R₃/(-1)

 

 

 

The pattern says to go up. So turn the 3 into a zero. Always use row 3 for this step. Multiply row 3 by –3 and add it to row 2.

 

R₃(-3)+ R₂

 

 

At this point break from the pattern for 1 step. Make the –6 into a 1. This is done by dividing row 2 by –6.

R₂ /(-6)

 

 

Getting back on with the pattern. Go up to the 1 in the first row. Always use row 3 for this step. Multiply row 3 by –1 and add it to row 1.

 

R₃(-1)+ R₁

 

 

Note- In many of the steps, including the previous one, there were some elements multiplied that were zero’s and then added to another row. In those particular elements it did not change anything. This is a good thing and is why certain rows are used for certain steps. Do not change anything that looks like the identity matrix, and using these particular rows insures that.

 

Look at the pattern or look at the matrix as a whole, notice all that is left to do is to change the –1 to a 0. Always use the second row for this step. Since there are opposites, to create a zero just add row 2 and row 1.

 

R₂ + R₁

 

 

Looking at the result, notice the identity matrix. The hard work is done and the answer is here. The last column contains the answers in order from top to bottom. It says that a = 2, b = -1, and c = 3. Write the answer in proper form.

(2, -1, 3)

 

Answer!

 

 

 

 

 

To summarize what has been done.

The pattern followed was

.

 

 

 

The first element became1 by swapping row 1 and row 2. Then to get the ones below it zero, row 1 was used.

 

 

 

R₁ R₂ R₁(-2)+ R₂

 

R₁(-1)+ R₃

 

 

 

The next step was to multiply row 2 by 3 and row 3 by 2 to get opposites. From there, row 2 was added to row 3.

 

 

R₂(3), R₃(2) R₂ + R₃

 

 

To turn the –1 into a 1 in row 3, row 3 was divided by –1. From there go up, row 3 was multiplied by –3 and the result was added to row 2.

 

R₃/(-1) R₃(-3)+ R₂

 

 

 

Next turn the –6 into a 1 by dividing row 2 by –6. Continuing to move up and using row 3, multiply row 3 by –1 and add the result to row 1.

 

 

R₂ /(-6) R₃(-1)+ R₁

 

 

 

The last step to get to the identity matrix was to add row 2 and row 1. Row 2 did not have to be multiplied by anything since there were already opposites.

 

R₂ + R₁

 

 

 

Now that there is the identity matrix, all that is left is to write the answer.

 

(2,-1,3)

 

Answer!

 

Not only is the pattern very important, but also the row that is used for each step is equally important. Assuming there is already a 1 as the first element, here is diagram that shows which row to use to get each element to the value (either 0 or 1) needed.

 

 

 

 

Perhaps this will help to visualize what to do along with the pattern.

 

 

Example

 

Solve x + 2y + z = 9

2x – y + 2z = -2

-x + 3y – 3z = 7

 

Everything is lined up so create the matrix.

 

 

 

The first element is already a 1 so move on to making the 2 a 0, and the –1 a zero.

 

R₁(-2)+R₂ R₁+R₃

 

 

 

Now there is a 1,0, and 0 in the first column. Make the 5 in the third row a zero. After that, make the –2 a 1.

 

R₂+R₃ R₃/-2

 

 

 

 

 

Following the pattern make the element above where the –2 was a 0. It is already a zero, so do not do anything there. After that, ignore the pattern for 1 step to get the –5 to a 1. Continuing on, get the 1 that is in the first row and the third column to a zero.

 

R₂/-5 R₃(-1)+R₁

 

 

 

Now this is the last step, which is to get the 2 in the first row to a zero.

 

 

R₂(-2)+R₁

 

 

 

Write the answer.

 

(-1,4,2) says that x = -1, y = 4, and z = 2.

 

 

 

 

 

Example

 

Solve 2x₁- x₂ + x₃ = 1

x₁= 3x₂ + 2x₃ +13

x₁+ x₂ + x₃ = -2

 

These equations are not lined up. The first and third equations are fine, but the second needs to be changed a little. Subtract 3x₂ and 2x₃ from both sides of the equation.

 

2x₁- x₂ + x₃ = 1

x₁- 3x₂ - 2x₃ =13

x₁+ x₂ + x₃ = -2

 

Now the equations are lined up. Create the matrix and work the problem. The following way that this problem is worked should be similar to the work on other problems. In other words, explanations are left out and just showing here how the work should appear on paper.

 

 

 

 

 

R₁ R₂ R₁(-2)+ R₂

 

R₁(-1)+ R₃ R₂(-4), R₃(5)

 

R₂+ R₃ R₃/-5

 

R₃(20)+ R₂R₂/-20

 

R₃(2)+ R₁ R₂(3)+ R₁

 

 

 

Done!

 

(3, 0, -5) Is the answer.

 

So x₁ = 3, x₂ = 0, and x₃ = -5.