1. Section 2.3 Gauss-Jordan Method for General Systems of Equations

2. Recall that we'd like to use row operations on an augmented matrix to get it into the following form:
This is not always possible though. The following are matrices that cannot be put into this form.
Recognize that if we can’t get our matrix to the desired form, then it won’t be as easy to see what the solution to the system of equations will be.

3. For example, this matrix has a solution that is easy to see, (1, 3, 5), because the matrix is in the final form that we want.
This matrix (on the right) has a solution but is not as clear what the solution is. What we can conclude about the solution, (x, y, z), is that the components x, y, and z must obey the equation x + 2y + 3z = 7.
This matrix (on the right) has a solution, but again it is not as clear what it is. What we can conclude about the solution, (x, y, z), is that the components x, y, and z must obey the two equations x + 5z = 2 and y + 6z = 3.
These last two matrices represent systems that do not have a unique solution. Whenever a matrix does not have a unique solution (if it has infinitely many solutions or no solution at all) we will not be able to get our augmented matrix into the form that we really want. When this happens, we want to at least get our matrix as close as possible to this form that we would really like it to be in. When it is as close as it can possibly get, we say it is in reduced row echelon form.

4. Reduced Row Echelon Form
If a matrix cannot be placed into this desired form (seen below), we still try to get it as close to this form as possible. In doing so, we want to use row operations to get as many of our 1’s where we want them – going down diagonally from the upper left of the matrix to the bottom right of the matrix.
Whenever we get a 1 in such a spot, we want to use row operations to “clear out” all the other entries (turning them into 0’s) in that column, thus getting as many of our 0’s where we want them.
Reduced Row Echelon Form
A matrix (any matrix, not just an augmented matrix) is said to be in reduced echelon form if all of the following properties hold true:
1. All rows consisting entirely of zeros are grouped at the bottom.
2. The leftmost nonzero number in each row is 1 (called the leading one).
3. The leading 1 of a row is to the right of the previous row's leading 1.
4. All entries directly above and below a leading 1 are zeros.

5. Reduced Row Echelon Form
Ex. 1 Determine which of the following matrices are in reduced row echelon form.
(a) (b) (c)
Yes, all four properties are true for this matrix.
No, there is a row of all zeros which is not at the bottom of the matrix.
No, the leading one of the third row is not to the right of the leading one of the second row (notice how when this happens the leading ones do not go down diagonally from top left to bottom right).
Reduced Row Echelon Form
1. All rows consisting entirely of zeros are grouped at the bottom.
2. The leftmost nonzero number in each row is 1 (called the leading one).
3. The leading 1 of a row is to the right of the previous row's leading 1.
4. All entries directly above and below a leading 1 are zeros.

6. Reduced Row Echelon Form
Ex. 1 Determine which of the following matrices are in reduced row echelon form.
(d) (e) (f)
Yes.
Yes.
Yes.
Reduced Row Echelon Form
1. All rows consisting entirely of zeros are grouped at the bottom.
2. The leftmost nonzero number in each row is 1 (called the leading one).
3. The leading 1 of a row is to the right of the previous row's leading 1.
4. All entries directly above and below a leading 1 are zeros.

7. Reduced Row Echelon Form
Ex. 1 Determine which of the following matrices are in reduced row echelon form.
(g) (h)
No, the first row does not have a leading one (it has a leading two).
Yes.
Reduced Row Echelon Form
1. All rows consisting entirely of zeros are grouped at the bottom.
2. The leftmost nonzero number in each row is 1 (called the leading one).
3. The leading 1 of a row is to the right of the previous row's leading 1.
4. All entries directly above and below a leading 1 are zeros.

8. Ex. 2 Put the matrix into reduced row echelon form.
Well, we are going to proceed as we did before when we were solving systems of equations. Same process, Gauss–Jordan elimination process using the PIVOT program, the only difference is we might not be able to get the final matrix into the form of the following matrix (but we will get as close to this form as we can – this is precisely what reduced row echelon form is):
R2 – 3R1→R2
R3 – 2R1→R3
–1/13 R2→R2
R1 – 3R2→R1
Reduced row echelon form has been reached!
R3 + 13R2→R3

9. The Calculator And The PIVOT Program.
To enter matrices press the MATRX button then select the EDIT menu.
Once you press the MATRX button
you should see a screen that looks something
like this one. There are three menus at the top.
To create a matrix and store it in the calculator’s
memory, we need to select a matrix from the
EDIT menu. (It doesn’t really matter which
matrix you select, as long as you remember the
name of the one you selected. To make things
easier on myself, I always choose matrix [A]).

10. The Calculator And The PIVOT Program.
After you select [A] from the EDIT menu you
should see a screen somewhat similar to this one.
The top right hand corner of the screen has the
dimensions (number of rows and columns) for
the current matrix.
Let’s enter the matrix from the previous example to the calculator.
First type in the number of rows, followed by the enter button, followed by the number of columns, followed by the enter button again. (So the button sequence for this matrix is
3, ENTER, 3, ENTER.)
Next, start typing in the entries for the matrix, hitting enter after you type in each entry. When you finish typing in the values, your screen should look something like this:

11. The Calculator And The PIVOT Program.
Now we are done creating the matrix, so exit out of the matrix editor (press quit – located at the top left of the calculator). You will now be at the home screen.
We want to begin the PIVOT program and use it to put our matrix [A] into reduced row echelon form. First let me describe what the PIVOT program is going to do.
Pivoting is a process that uses elementary row operations to change a matrix. If we want to pivot on row i, column j, that means we want to use elementary row operations to scale down/up whatever number was in that spot into a one, then use elementary row operations to clear out the other numbers in that column (turning these other numbers into zeros).

12. The Calculator And The PIVOT Program.
The following sequence of matrices shows how we converted column 3 in the first matrix into having a one in the third row and zeros elsewhere. We can do all three of the required elementary row operation involved here with one pivot command.
If we want to place a one in the third row, third column (and clear out zeros elsewhere in the column) then we will say we want to pivot on the third row and third column.
–½ R3→R3
Pivoting on
row 3, column 3.
R2 + 2R3→R2
R1 – 3R3→R1

13. The Calculator And The PIVOT Program.
Ok, now back to the calculator and its PIVOT program.
We want to start this program, and see how it will place our matrix [A] in reduced row echelon form quickly through a series of pivot commands.
While on the home screen, press the PRGM button. You should see a screen somewhat similar to the following.
This is a list of all the programs on the calculator (Your calculator may have more programs on it. Each program on your calculator is listed here). We want to select the PIVOT program (use the up/down arrows to highlight PIVOT then push ENTER).

14. The Calculator And The PIVOT Program.
This pasted the command to run the PIVOT program into the home screen (see the screen shot on the right). Now we must hit ENTER one more time to execute this command.
Once you hit ENTER the program begins. Basically, while you are in the PIVOT program you will want to follow the directions given on the screen.

15. The Calculator And The PIVOT Program.
First we will choose matrix [A] (since that is where we put our matrix earlier).

16. The Calculator And The PIVOT Program.
Next tell the calculator where you want to pivot.
(In putting our matrix into reduced row echelon form we first need to concentrate on the first column, so we begin by pivoting on row one column one.)
Now we see our matrix after the first pivot. Notice that the first column is looking good, so next we will pivot in the second column.

17. The Calculator And The PIVOT Program.
Once we pivot on the 2nd row, 2nd column we see that the process is done (the matrix is now in reduced row echelon form).

18. The Calculator And The PIVOT Program.
To exit the program we follow the direction at the bottom of the screen.

19. Ex. 3 Use the calculator to put the matrix into reduced row echelon form.
This is what we just did in the last several screens.
I do want to mention here what kind of work is expected to be shown (when using the PIVOT program) to justify your answers on an exam.
Whenever you use this program you will need to show the first matrix you typed into the calculator, then the final matrix that the calculator produced before you exit the program.

20. Ex. 4 Solve the following system
Ex. 4 Solve the following system. Then draw a graph of this system and its solution.
x + y = 11
3x – 4y = –6
2x – 7y = –17
There are no solutions to this system because this system consists of equations for three lines which do not all intersect at one common point.
First, we will take the augmented matrix for this system and put it into reduced row echelon form (via the calculator with its PIVOT program).
The result of pivoting on row 1, column 1.
The result of pivoting on row 2, column 2.
This first row states that 1x + 0y = 0 (i.e. x = 0).
The second row states that 0x + 1y = 0 (i.e. y = 0).
So it seems that the solution is x = 0 and y = 0.
But the last row states that 0x + 0y = 1. This is impossible! No values of x and y can satisfy this equation. Thus, there are no values for x and y that solve our system of equations. (This system has no solution.)
The result of pivoting on row 3, column 3.

21. Ex. 5 Solve the following system of equations. 3x1 – 2x2 – 3x3 = –16
The result of pivoting on row 1, column 1.
The result of pivoting on row 2, column 2.
The result of pivoting on row 3, column 3.
Our solution is (1, 2, 5)

22. Ex. 6 Solve the following system of equations. x1 – 4x2 + 7x3 = –3
So, the only information we have for our solution (x1, x2, x3) is
x1 + 47x3 = 1
x2 + 10x3 = 1.
Notice that we can easily solve these equations for x1 and x2 :
x1 = 1 – 47x3
x2 = 1 – 10x3.
Alright. These equations indicate that if we know what the value of x3 is, then we can compute the value of x1 and we can compute the value of x2.
So pick any value for x3, let’s call this k.
Then x3 = k, x1 = 1 – 47k, and x2 = 1 – 10k.
We have infinitely many solutions to this system. The solutions have the parametric form of (1 – 47k, 1 – 10k, k).
Well, our matrix is in reduced row echelon form. All that is left to do is figure out what the solution is.
The 1st row states that 1x1 + 0x2 + 47x3 = 1.
The 2nd row states that 0x1 + 1x2 + 10x3 = 1.
The 3rd row states that 0x1 + 0x2 + 0x3 = 0.
Well, the last row just tells us that 0 = 0. That information is useless when trying to figure out our solution. So, ignore it.

23. Ex. 6 Solve the following system of equations. x1 – 4x2 + 7x3 = –3
(A graph of this system is given below.) –2x1 + 9x2 – 4x3 = 7
x1 – 3x2 + 17x3 = –2
Here's one view of the three planes:
Here's a side view of the three planes:
Systems of equations with two variables have graphs that are lines slicing through two dimensional space. Systems of equations with three variables have graphs that are planes slicing through three dimensional space. That’s what we are seeing here. We have three equations, so there are three planes in the graph. The intersection of these three planes form a line (shown in pink).
Recall that solutions to systems of equations correspond to intersection points on the graph. So, every single point on this line corresponds to a solution to the system of equations. Since this line has infinitely many points on it, there are infinitely many solutions to the system.
Now note that a line is a one dimensional object (embedded in 3D space). This can be seen by observing the parametric form of our solution: (1 – 47k, 1 – 10k, k). This parametric form has one parameter (corresponding to the one dimension for the line, which is the graph of the solution).

24. Ex. 7 Solve the following system of equations. x1 – 2x2 + 3x3 = 4
We will again take the augmented matrix for this system and put it into reduced row echelon form (via the calculator).
The only information we have for our solution (x1, x2, x3) is
x1 – x2 + 3x3 = 4
Notice that we can easily solve this equation for x1:
x1 = 4 + x2 – 3x3.
This equation indicates that if we know what the value of x2 and x3 are, then we can compute the value of x1.
So pick any value for x2 and x3, let’s call the value we pick for x2 to be j and the value we pick for x3 to be k.
Then x2 = j, x3 = k, and x1 = 4 + 2j – 3k.
We have infinitely many solutions to this system. The solutions have the parametric form of (4+2j–3k, j, k).
The 1st row states that 1x1 – 2x2 + 3x3 = 4.
The 2nd row states that 0x1 + 0x2 + 0x3 = 0.
The 3rd row states that 0x1 + 0x2 + 0x3 = 0.
The last two rows just tells us that 0 = 0.
Useless information.
Ignore it.

25. Ex. 7 Solve the following system of equations. x1 – 2x2 + 3x3 = 4
(A graph of this system is given below.) –2x1 + 4x2 – 6x3 = –8
3x1 – 6x2 + 9x3 = 12
Here's one view of the three planes:
Here's a side view of the three planes:
We have three equations, and there really are three planes in the graph. You’re only seeing one because all three planes are exactly the same. The intersection of these three identical planes form a plane (naturally), which is a two dimensional object (embedded in 3D space). This could be seen from our the parametric form of the solution to the system of equations: (4+2j–3k, j, k). This parametric form has two parameters (corresponding to the two dimensions for the plane, which is the graph of the solution).

26. Ex. 8 Solve the following system of equations. 3x1 + 11x2 – 3x3 = 17
Again we take the augmented matrix for this system and put it into reduced row echelon form (via the calculator).
One only needs to look at the last row to determine the solution to this system of equations. The last row states:
0x1 + 0x2 + 0x3 = 1.
Since no values of x1, x2, and x3 can make this equation work we must conclude that there is no solution for this system.

27. Ex. 8 Solve the following system of equations. 3x1 + 11x2 – 3x3 = 17
(A graph of this system is given below.) x1 – 2x2 – 3x3 = 8
11x1 + 7x2 – 9x3 = 10
We have three equations, so there are three planes in the graph. Any two of the planes intersect in a line, but the three planes do not all intersect at the same point (there is no point which happens to be on each plane). Since there is no common intersection point, there is no solution to the system.

28. Ex. 9 Solve the following system of equations. 2x1 + 4x2 – x3 = 0
Again we take the augmented matrix for this system and put it into reduced row echelon form (via the calculator).
Ok. At this point in the PIVOT program we want to pivot on the element in the 2nd row and 2nd column. The calculator won’t let us though since that element is a zero. What we need to do when this happens is think about what we would do without the calculator. What elementary row operation would we use at this point to proceed to put the matrix in reduced row echelon form?
Swapping rows!
Swap rows 2 and 3 so that a nonzero number appears in the 2nd row and 2nd column, then we can put a leading one in this spot with further elementary row operations.
Since this is what we would do without the calculator, this is exactly what we want to do with the calculator. The PIVOT program has an option for switching rows. Use this option here.
This system has a unique solution of
(–2, 1, 14).

29. Ex. 10 A contractor builds houses, duplexes, and apartment units
Ex. 10 A contractor builds houses, duplexes, and apartment units. He has financial backing to build 250 units. He makes a profit of $4,500 on each house, $4,000 on each duplex, and $3,000 on each apartment unit. Each house requires 10 man-months of labor, each duplex requires 12 man-months, and each apartment requires 6 man-months. How many of each should the contractor build if he has 2050 man-months of labor available and wishes to make a total profit of $875,000?
Let x1 = the number of houses to build
x2 = the number of duplexes to build
x3 = the number of apartment units to build
x1 + x2 + x3 = 250
4500x x x3 =
10x1 + 12x2 + 6x3 = 2050
The contractor should build 40houses, 65 duplexes, and 145 apartment units.