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This Lesson (GAUSS-JORDAN ELIMINATION METHOD FOR SOLVING LINEAR EQUATIONS) was created by by Theo(3060)
  About Theo:
This lesson reviews the Gauss-Jordan Elimination Method for solving linear equations.
REFERENCES http://www.karlscalculus.org/cgi-bin/linear.pl http://www.gregthatcher.com/Mathematics/GaussJordan.aspx http://www.mathwords.com/g/gauss-jordan_elimination.htm http://www.krellinst.org/AiS/textbook/unit2/example_projects/starter/math/matrix/gauss.html LINEAR EQUATIONS A linear equation is an equation with the highest order of exponent equal to 1. The variables in a linear equation can be raised to the 0 power or to the 1 power. Since any variable to the 0 power = 1, then the coefficient of that variable becomes a constant and the variable name is not used because the value of that variable has already been defined. Example: 5x^0 = 5*1 = 5 SOLVING A SYSTEM OF LINEAR EQUATIONS SIMULTANEOUSLY When you solve a system of linear equations simultaneously, you have a unique solution for each variable that applies to every equation in the system where the variable is present. An example:
x + y = 5
2x + 3y = 13
The unique solution to this system of equations is x = 2 and y = 3. The same value for x is applied to every equation in the system where it is present, and the same value for y is applied to every equation in the system where it is present. When you substitute 2 for x and 3 for y, then ... the first equation becomes 2 + 3 = 5 which is true. the second equation becomes 4 + 9 = 13 which is also true. Your unique solution for each variable makes all equations in the system true. Another example:
x + y = 7
y = 6
The unique solution to this system of equations is x = 1 and y = 6. The same value for x is applied to every equation in the system where it is present, and the same value for y is applied to every equation in the system where it is present. When you substitute 1 for x and 6 for y, then ... the first equation becomes 1 + 6 = 7 which is true. the second equation becomes 6 = 6 which is also true. Your unique solution for each variable makes all equations in the system true. STANDARD FORM OF A SYSTEM OF LINEAR EQUATIONS The standard form of a system of linear equations has the coefficients and the variables on the left side of the equal sign and the constant terms on the right side of the equal side. An example of the standard form of a system of linear equations in four dimensions could look like this.
4w + 1x + 2y + 3z = 6
3w - 6x + 5y + 4z = 1
9w + 7x - 7y + 8z = 4
1w + 1x + 1y - 1z = 1
3w + 4x + 3y + 2z = 5
w,x,y,z are the variables. coefficients and variables are on the left side of the equal sign. constant terms are on the right side of the equal sign. AUGMENTED MATRIX From a system of linear equations, we make what is called an augmented matrix. The augmented matrix has a column for the coefficients of each variable plus a column for the result (constant terms). The augmented matrix created from the four dimension system of equations above is shown here:
4 1 2 3 | 6
3 -6 5 4 | 1
9 7 -7 8 | 4
1 1 1 -1 | 1
3 4 3 2 | 5
The coefficients of each variable have their own column on the left side of the vertical line in the matrix. The constant terms have their own column on the right side of the vertical line in the matrix. Each equation has its own row. In this matrix: Column 1 is for the coefficients of w. Column 2 is for the coefficients of x. Column 3 is for the coefficients of y. Column 4 is for the coefficients of z. Column 5 is for the constant terms. Row 1 contains first equation. Row 2 contains second equation. Row 3 contains third equation. Row 4 contains fourth equation. The vertical line between the columns containing the coefficient and the column containing the constant term are there to show you that this is an augmented matrix. If there were only the matrix containing the coefficients without a column showing the constant term, then this would be called a coefficient matrix. This vertical line should be displayed for clarity, but some mechanized tools may not display it. As long as you know that you are dealing with an augmented matrix, this is not a problem. The mechanized tool we will be using will not display it. Just be aware that if you are showing an augmented matrix, you should have the vertical line separating the coefficient part of the matrix from the results part of the matrix, but you should also be aware that it is not always shown. Usually you will be told if it is an augmented matrix or not. MATRIX AND POSITION The matrix requires that the coefficients of each variable have their own column, and that the constant terms have their own column. This is easy if all the terms of the system of linear equations are explicit as shown above. If not, then you have to make sure you place the coefficients of each variable in the proper column and row. Example: Here's a system of linear equations where the translation to matrix form requires some adjustment. z = -4 x - y = 3 v + z = 2 7w - 8x = -14 v = 2 This system has 5 linear equations with 5 unknowns, so the matrix you will be creating will have 5 rows and 6 columns. The matrix will look like this:
0 0 0 0 1 | -4
0 0 1 -1 0 | 3
1 0 0 0 1 | 2
0 7 -8 0 0 | -14
2 0 0 0 0 | 2
The variables in this system of equations are v,w,x,y,z. The columns for each variable have been placed in an alphabetic order in the matrix. Column 1 contains all the coefficients for v. Column 2 contains all the coefficients for w. Column 3 contains all the coefficients for x. Column 4 contains all the coefficients for y. Column 5 contains all the coefficients for z. Column 6 contains all the coefficients for the results (constant terms). Row 1 contains the first equation. Row 2 contains the second equation. Row 3 contains the third equation. Row 4 contains the fourth equation. Row 5 contains the fifth equation. GAUSS-JORDAN ELIMINATION METHOD AND THE AUGMENTED MATRIX The Gauss-Jordan Elimination method works with the augmented matrix in order to solve the system of equations. The goal of the Gauss-Jordan Elimination method is to convert the matrix into this form (four dimensional matrix is used for demonstration purposes).
1 0 0 0 | r1
0 1 0 0 | r2
0 0 1 0 | r3
0 0 0 1 | r4
r1,r2,r3,r4 represent the results of each equation (constant terms). Once you have the matrix in this form, your problem is solved. The only thing you have to figure out is how to convert the matrix into this form. REQUIREMENTS FOR A UNIQUE SOLUTION TO A SYSTEM OF LINEAR EQUATIONS USING THE GAUSS-JORDAN ELIMINATION METHOD The requirements for a unique solution to a system of linear equations using the Gauss-Jordan Elimination Method are that the number of unknowns must equal the number of equations. When the number of equations and the number of unknowns are the same, you will obtain an augmented matrix where the number of columns is equal to the number of rows plus 1. For example, if you have a system of 4 linear equations in 4 unknowns, then the number of rows must be equal to 4 and the number of columns must be equal to 5 (4 columns for the coefficients and 1 column for the results). The vertical line between the coefficients part of the matrix and the results part of the matrix (the constant terms are in the results part of the matrix) does not count as a column. It's there for display purposes only if you have the capability to display it. When you create your matrix, just make sure that the number of columns is equal to the number of rows plus 1. Note that it is possible to get a unique solution to a system of linear equations where the number of equations is greater than the number of unknowns. An example of this would be 5 lines in a plane all intersecting in the same point. There is a unique solution for the x and y variables that makes all the equations in the system true. This type of situation, however, is not conducive to solving using the Gauss-Jordan Elimination Method since that method requires the number of equations and the number of unknowns to be the same. Note also that it is not possible to get a unique solution to a system of linear equations where the number of equations is less than the number of unknowns. If you use the Gauss-Jordan Elimination Method, just make sure that the number of equations is equal to the number of unknowns and the method will work just fine. ROW OPERATIONS ON A MATRIX You are allowed to do the following without changing the equality of any equation in the matrix and without impacting the solution to the system of linear equations:
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> YOU CAN SWAP ANY ROW IN THE MATRIX WITH ANY OTHER ROW IN THE MATRIX.
Example:
Old row 1 equals: 0 5 3 2 | 1
Old row 3 equals: 0 -2 2 3 | -3
SWAP ROW 3 WITH ROW 1 --------------------------------
New row 1 equals: 0 -2 2 3 | -3 (equals old row 3)
New row 3 equals: 0 5 3 2 | 1 (equals old row 1)
-----------------------------------------------------------------------------------------------
> YOU CAN MULTIPLY OR DIVIDE ANY ROW IN THE MATRIX BY ANY CONSTANT.
Example:
Old row 3 equals: 0 5 3 2 | 1
MULTIPLY ROW 3 BY 5 ----------------------------------
New row 3 equals: 0 25 15 10 | 5 (equals old row 3 * 5)
-----------------------------------------------------------------------------------------------
> YOU CAN ADD OR SUBTRACT ANY ROW FROM ANY OTHER ROW IN THE MATRIX.
Example:
Old row 1 equals: 0 5 3 2 | 1
Old row 3 equals: 0 -2 2 3 | -3
ADD ROW 1 TO ROW 3 -----------------------------------
New row 1 equals: 0 5 3 2 | 1 (equals old row 1 unchanged)
New row 3 equals: 0 3 5 5 | -2 (equals old row 1 + old row 3)
-----------------------------------------------------------------------------------------------
> YOU CAN MULTIPLY OR DIVIDE ANY ROW BY ANY CONSTANT AND THEN ADD OR SUBTRACT THAT ROW FROM
ANY OTHER ROW IN THE MATRIX. (This is the next to last rule referenced below).
Example:
Old row 1 equals: 0 5 3 2 | 1
Old row 3 equals: 0 -2 2 3 | -3
MULTIPLY ROW 1 BY 2 AND ADD IT TO ROW 3 --------------
New row 1 equals: 0 10 6 4 | 2 (equals old row 1 * 2)
New row 3 equals: 0 8 8 7 | -1 (equals old row 3 + new row 1)
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> YOU CAN ADD OR SUBTRACT ANY ROW IN THE MATRIX MULTIPLIED OR DIVIDED BY ANY CONSTANT FROM ANY
OTHER ROW IN THE MATRIX. (This is the last rule referenced below).
Example:
Old row 1 equals: 0 5 3 2 | 1
Old row 3 equals: 0 -2 2 3 | -3
ADD 2 * ROW 1 TO ROW 3 -------------------------------
New row 1 equals: 0 5 3 2 | 1 (equals old row 1 unchanged)
New row 3 equals: 0 8 8 7 | -1 (equals old row 3 + old row 1 * 2)
-----------------------------------------------------------------------------------------------
The last two rules give you the same result for the new row 3 but provide different results for the new row 1. MULTIPLY ROW 1 BY 2 AND ADD IT TO ROW 3 (next to last rule shown above) makes new row 1 different from old row 1 (changes row 1). ADD 2 * ROW 1 TO ROW 3 (last rule shown above) makes new row 1 the same as old row 1 (leaves row 1 unchanged). When we go through the displays from the mechanized tool, you will see that the tool uses the last rule when making the coefficients of the remaining rows in the same column equal to 0. EXAMPLE OF SOLVING A SYSTEM OF LINEAR EQUATIONS IN FOUR DIMENSIONS USING THE GAUSS-JORDAN ELIMINATION METHOD The system of linear equations that we want to solve is shown below:
4z = 5
3y = 4
5w + 6x + 7y + 8z = 9
4x + 3y + 2z = 1
From this system of linear equations, we create our augmented matrix as shown below:
0 0 0 4 | 5
0 0 3 0 | 4
5 6 7 8 | 9
0 4 3 2 | 1
Column 1 contains all the coefficients of w. Column 2 contains all the coefficients of x. Column 3 contains all the coefficients of y. Column 4 contains all the coefficients of z. Column 5 contains all the results (constant terms). Row 1 contains the first equation. Row 2 contains the second equation. Row 3 contains the third equation. Row 4 contains the fourth equation. We execute the tool by clicking on the following link: http://www.gregthatcher.com/Mathematics/GaussJordan.aspx We enter the number of rows (4) and the number of columns (5). We fill in the matrix. We hit the "solve" button. We are presented with a series of displays of the original matrix we entered and the step by step results of the procedures the mechanized tool used to solve the problem. the step by step displays are shown below. Don't try to follow it through just yet. We will go into the details further down in this lesson. For now, just look at display number 1 and display number 13. 
Display Number 1 shows the original matrix as it was entered into the mechanized tool. Display Number 13 shows the matrix in its final form. As you can see, the goal of the Gauss-Jordan Elimination Method has been achieved and the problem is solved. That goal was to get the matrix into this form:
1 0 0 0 | c1
0 1 0 0 | c2
0 0 1 0 | c3
0 0 0 1 | c4
From Display Number 13 you can see that the matrix is in this final form. Here's display number 13.
1 0 0 0 -5/12
0 1 0 0 -11/8
0 0 1 0 4/3
0 0 0 1 5/4
Column 1 contains all the coefficients of the w variable. Column 2 contains all the coefficients of the x variable. Column 3 contains all the coefficients of the y variable. Column 4 contains all the coefficients of the z variable. Column 5 contains all the constant terms. Row 1 contains all the coefficients and constant terms of our first equation. That row translates to 1w + 0x + 0y + 0z = -5/12 which becomes w = -5/12. Row 2 contains all the coefficients and constant terms of our second equation. That row translates to 0w + 1x + 0y + 0z = -11/8 which becomes x = -11/8. Row 3 contains all the coefficients and constant terms of our third equation. That row translates to 0w + 0x + 1y + 0z = 4/3 which becomes y = 4/3. Row 4 contains all the coefficients and constant terms of our fourth equation. That row translates to 0w + 0x + 0y + 1z = 5/4 which becomes z = 5/4. The final form of the matrix is the solution to this system of linear equations. The solution is: w = -5/12 x = -11/8 y = 4/3 z = 5/4 How the tool went from the original form of the matrix to the final form of the matrix is shown in displays number 2 through 13. Before we show you that, however, we'll provide you with the general strategy used by the mechanized tool. This is the same general strategy that you would use if you were solving the system of linear equations manually using the Gauss-Jordan Elimination method. Here's the general strategy: Column 1 forward processing: <1.1> Start from row 1 column 1. <1.2> Check to see if the coefficient of column 1 of row 1 is not zero. <1.3> If it is not zero then go to step 1.6. If it is zero, then go to step 1.4. <1.4> Look for any row below row 1 that contains a nonzero value in column 1. <1.5> When found, swap that row with row 1. Row 1 now contains a nonzero value in column 1. <1.6> Divide row 1 by the coefficient in row 1 column 1 so that the coefficient in row 1 column 1 is equal to 1. <1.7> Use row 1 to make all the coefficients of column 1 equal to zero in all rows that are below row 1. <1.8> When row 1 contains a coefficient of 1 in column 1 and all the other rows below it contain a coefficient of 0 in column 1, then you are done with column 1 forward processing and can move on to column 2 forward processing. Column 2 forward processing: <2.1> Start from row 2 column 2. <2.2> Check to see if the coefficient of column 2 of row 2 is not zero. <2.3> If it is not zero then go to step 2.6. If it is zero, then go to step 2.4. <2.4> Look for any row below row 2 that contains a nonzero value in column 2. <2.5> When found, swap that row with row 2. Row 2 now contains a nonzero value in column 1. <2.6> Divide row 2 by the coefficient in row 2 column 2 so that the coefficient in row 2 column 2 is equal to 1. <2.7> Use row 2 to make all the coefficients of column 2 equal to zero in all rows that are below row 2. <2.8> When row 2 contains a coefficient of 1 in column 2 and all the other rows below it contain a coefficient of 0 in column 2, then you are done with column 2 forward processing and can move on to column 3 forward processing. Column 3 forward processing: <3.1> Start from row 3 column 3. <3.2> Check to see if the coefficient of column 3 of row 3 is not zero. <3.3> If it is not zero then go to step 3.6. If it is zero, then go to step 3.4. <3.4> Look for any row below row 3 that contains a nonzero value in column 3. <3.5> When found, swap that row with row 3. Row 3 now contains a nonzero value in column 3. <3.6> Divide row 3 by the coefficient in row 3 column 3 so that the coefficient in row 3 column 3 is equal to 1. <3.7> Use row 3 to make all the coefficients of column 3 equal to zero in all rows that are below row 3. <3.8> When row 3 contains a coefficient of 1 in column 3 and all the other rows below it contain a coefficient of 0 in column 3, then you are done with column 3 forward processing and can move on to column 4 forward processing. Column 4 forward processing: <4.1> Start from row 4 column 4. <4.2> Divide row 4 by the coefficient in row 4 column 4 so that the coefficient in row 4 column 4 is equal to 1. <4.3> You are done with column 4 forward processing and can move on to column 4 backward processing. Column 4 backward processing: <5.1> Start from row 4 column 4. <5.2> Use row 4 to make all the coefficients of column 4 equal to zero in all rows that are above row 4. <5.3> When all the other rows above row 4 contain a coefficient of 0 in column 4, then you are done with column 4 backward processing and can move on to column 3 backward processing. Column 3 backward processing: <6.1> Start from row 3 column 3. <6.2> Use row 3 to make all the coefficients of column 3 equal to zero in all rows that are above row 3. <6.3> When all the other rows above row 3 contain a coefficient of 0 in column 3, then you are done with column 3 backward processing and can move on to column 2 backward processing. Column 2 backward processing: <7.1> Start from row 2 column 2. <7.2> Use row 2 to make all the coefficients of column 2 equal to zero in all rows that are above row 2. <7.3> When all the other rows above row 2 contain a coefficient of 0 in column 2, then you are done. The Gauss-Jordan Elimination Method for solving this system of four linear equations in four unknowns is complete. We will now go through the step by step procedures that the Gauss-Jordan Elimination Mechanized Tool used to solve our system of 4 linear equations in 4 unknowns. We'll start the whole process again beginning with the system of equations that we want to solve so that referencing back will be easier. The system of linear equations that we want to solve is shown below:
4z = 5
3y = 4
5w + 6x + 7y + 8z = 9
4x + 3y + 2z = 1
From this system of linear equations, we create our augmented matrix as shown below:
0 0 0 4 | 5
0 0 3 0 | 4
5 6 7 8 | 9
0 4 3 2 | 1
Column 1 contains all the coefficients of w. Column 2 contains all the coefficients of x. Column 3 contains all the coefficients of y. Column 4 contains all the coefficients of z. Column 5 contains all the results (constant terms). Row 1 contains the first equation. Row 2 contains the second equation. Row 3 contains the third equation. Row 4 contains the fourth equation. We execute the tool by clicking on the following link: http://www.gregthatcher.com/Mathematics/GaussJordan.aspx We enter the number of rows (4) and the number of columns (5). We fill in the matrix. We hit the "solve" button. We are presented with a series of displays of the original matrix we entered and the step by step results of the procedures the mechanized tool used to solve the problem. The step by step displays are shown below.
Display Number 1 shows you the original matrix as we entered it into the mechanized tool. Display Number 2 shows you the results of the tool starting with row 1 column 1 and scanning down until it finds a row with a nonzero entry in column 1 and then making that row the new row number 1. In this particular case, row 3 was the first row that contained a nonzero element in column 1. That's why row 3 was swapped with row 1. Display Number 3 shows you the results of the tool dividing row 1 by 5 so that the coefficient in row 1 column 1 becomes 1. Note that every element in row 1 is divided by 5. Note also that column 1 forward processing is now finished because row 1 contains a coefficient of 1 and every other row below row 1 contains a coefficient of 0 in column 1. Display Number 4 shows you the results of the tool starting with row 2 column 2 and scanning down until it finds a row with a nonzero entry in column 2 and then making that row the new row number 2. In this particular case, row 4 was the first row that contained a nonzero element in column 2. That's why row 4 was swapped with row 2. Display Number 5 shows you the results of the tool dividing row 2 by 4 so that the coefficient in row 2 column 2 becomes 1. Note also that column 2 forward processing is now finished because row 2 contains a coefficient of 1 in column 2 and every other row below row 2 contains a coefficient of 0 in column 2. Display Number 6 shows you the results of the tool starting with row 3 column 3 and scanning down until it finds a row with a nonzero entry in column 3 and then making that row the new row number 3. In this particular case, row 4 was the first row that contained a nonzero element in column 3. That's why row 4 was swapped with row 3. Display Number 7 shows you the results of the tool dividing row 3 by 3 so that the coefficient in row 3 column 3 becomes 1. Note also that column 3 forward processing is now finished because row 3 contains a coefficient of 1 in column 3 and every other row below row 3 contains a coefficient of 0 in column 3. Display Number 8 shows you the results of the tool dividing row 4 by 4 so that the coefficient in row 4 column 4 becomes 1. Since this was the last row in the matrix, forward processing is done and the tool moves on to backward processing. Display Number 9 shows you the results of the tool adding row 4 multiplied by (-1/2) to row 2 in order to make the coefficient in row 2 column 4 equal to 0. The tool skipped row 3 because row 3 column 4 already had a coefficient equal to 0. Display Number 10 shows you the results of the tool adding row 4 multiplied by (-8/5) to row 1 in order to make the coefficient in row 1 column 4 equal to 0. Since all the rows above row 4 column 4 have a coefficient of 0 in column 4, then column 4 backward processing is done and the tool moves on to column 3 backward processing. Display Number 11 shows you the results of the tool adding row 3 multiplied by (-3/4) to row 2 in order to make the coefficient in row 2 column 3 equal to 0. Display Number 12 shows you the results of the tool adding row 3 multiplied by (-7/5) to row 1 in order to make the coefficient in row 1 column 3 equal to 0. Since all the rows above row 3 column 3 have a coefficient of 0 in column 3, then column 3 backward processing is done and the tool moves on to column 2 backwards processing. Display Number 13 shows you the results of the tool adding row 2 multiplied by (-6/5) to row 1 in order to make the coefficient in row 1 column 2 equal to 0. Since all the rows above row 2 column 2 have a coefficient of 0 in column 2, then column 2 backward processing is done and the tool is finished. Display Number 13 shows you the final form of the matrix. Column 1 contains a 0 coefficient in all rows except row 1. Column 2 contains a 0 coefficient in all rows except row 2. Column 3 contains a 0 coefficient in all rows except row 3. Column 4 contains a 0 coefficient in all rows except row 4. All the nonzero coefficients are equal to 1. From display number 13, you can immediately go to the solution of the system of linear equations which is: w = -5/12 x = -11/8 y = 4/3 z = 5/4 UNIQUE SOLUTIONS TO A SYSTEM OF LINEAR EQUATIONS If your matrix is in the final form shown below, then you have a unique solution to the system of linear equations.
1 0 0 0 | r1
0 1 0 0 | r2
0 0 1 0 | r3
0 0 0 1 | r4
r1,r2,r3,r4 are the results of each equation (constant terms). If column 1 represents w variable and column 2 represents x variable and column 3 represents y variable and column 4 represents z variable and column 5 represents the constant terms (results), then: w = r1 x = r2 y = r3 z = r4 represent the unique solution to this system of equations. If your matrix is not in the final form shown above, then you do not have a unique solution to the system of linear equations. As you progress through the steps of your matrix, if you encounter a row that has all zeroes in the coefficient part of the matrix, that's a clue that tells you that you will not have a unique solution to your matrix. Here's an example of a system of linear equations that does not have a unique solution. Your equations are:
x + y + z = 5
3x + 3y + 3z = 20
2x - 3y - 4z = 6
Your matrix looks like this:
1 1 1 | 5
3 3 3 | 20
2 -3 -4 | 6
The mechanized Gauss-Jordan Elimination Method Tool looks like this: 
Your inputs are in the top array above the numbered displays. Display Number 1 shows that you already have a problem. All the columns in the coefficient part of the matrix show 0 which means that the mechanized tool will not be able to find a unique solution to this system of equations. Display Number 4 shows that the problem can't be resolved any further and is a logical point for stopping the processing since going any further will clearly not produce a unique solution to the problem. CAUSES OF NOT BEING ABLE TO FIND A UNIQUE SOLUTION TO A SYSTEM OF LINEAR EQUATIONS With Linear Equations in 2 dimensions, the causes are clear. The lines represented by the equations are either parallel (no solution), or intersecting in a point (unique solution), or identical (infinite number of solutions). There is no in between. With Linear Equations in 3 dimensions, the causes get a little muddier. The possible causes are:
> All 3 planes are parallel to each other.
There are no points of intersection.
This system would be classified as inconsistent.
> All 3 planes are identical to each other.
There are an infinite number of points of intersection.
This system would be classified as dependent.
> All 3 planes intersect with each other in a point.
There is one unique point of intersection.
This system would be classified as independent.
> Two planes are parallel and the third plane intersects with each of them in a line.
This system does not a unique point of intersection.
Two of the equations are inconsistent with each other.
The third equation is independent of the other two because it does intersect
with each of them only the intersection is in a line and not a point.
> Two planes are identical to each other and the third plane intersects with each of them
in the same line.
Two of the equations are dependent on each other (they are identical).
The third plane is independent of the other two because it does intersect
with each of them only the intersection is in a line and not a point.
You can see that the causes of not having a unique solution are getting more complicated to determine. With Linear Equations in more than 3 dimensions, the all parallel to each other analogy is still there, and the all identical to each other analogy is still there, even though you can't see it. The mixed solutions get more complicated and difficult, if not impossible to visualize. Once you get systems of linear equations in 3 dimensions or more, the causes of not finding a unique solution become complicated and you are not usually asked to determine them. It is sufficient to know that you either have a unique solution or you don't. Once you get a row of zeroes in all columns of the coefficient part of the matrix, that's enough to tell you that you will not have a unique solution. Display Number 2 in the example was that point. Regardless, if the final form of the matrix is not the way it's supposed to look, you do not have a unique solution to the system of equations. This assumes you entered the appropriate number of rows and columns. There is another tool using the Gauss-Jordan Elimination Method worth mentioning, that does not allow you to enter the incorrect number of rows and columns. This is because it simply asks you for the number of equations. This tool also will stop the processing when it becomes apparent that you will not have a unique solution to the problem. The following link will take you to this tool. http://www.karlscalculus.org/cgi-bin/linear.pl We will process this same system of linear equations that did not have a unique solution to show you how this tool would handle it. Your equations that do not have a unique solution are (once again):
x + y + z = 5
3x + 3y + 3z = 20
2x - 3y - 4z = 6
Your matrix looks like this:
1 1 1 | 5
3 3 3 | 20
2 -3 -4 | 6
The displays from this second tool are shown below: 
You can see that this tool stopped when it discovered that there would not be a unique solution to this problem. Both tools are good training tools and good solution checking tools. Each has its own unique advantages. You are encouraged to try both of them. They can be found at the following links. http://www.karlscalculus.org/cgi-bin/linear.pl http://www.gregthatcher.com/Mathematics/GaussJordan.aspx Comments or questions regarding this lesson can be directed to dtheophilis@yahoo.com This lesson has been accessed 21020 times. |
