2.2 Using Matrices to Solve Systems of Equations

Part C: Row-Reduced Echelon Form

(Based on Section 2.2 in Finite Mathematics and Finite Mathematics and Applied Calculus)

Some On-line Resources for This Topic:

Solution of Systems of Equations by Row Operations, Continued
The following discussion is based on Section 2.2 of Finite Mathematics and Finite Mathematics and Applied Calculus.

Definition of Row-Reduced Echelon Form

A matrix is in row-reduced echelon form (or reduced for short) if:

The first non-zero entry ("leading entry") in each row is a 1;
The column of every leading entry is clear; that is, the other entries in the column are all zero;
The rows are arranged so that the leading entries go from left to right as you go down the rows. Further, rows of zeros (if any) must be placed at the bottom.

Q Select which (if any) of the following matrices are in row-reduced echelon form.

	1	0	5
	0	0	1

1	0	5	0
0	1	1	0
0	0	0	1

1	1	5	0	-7
0	1	1	0	-4
0	0	0	1	0
0	0	0	0	0

	1	2	3
	0	0	0

Q What is so interesting about row-reduced echelon form?
A Recall the procedure you used in Part B to solve systems of linear equations. (If you don't, go back to that part of the tutorial.) You used pivoting to clear the columns containing the leading entries, and then, as a final step, you converted the leading entries into 1's. Thus, what you were left with (apart from a possible rearrangement of the rows) was a row-reduced echelon matrix! In other words,

You already know how to reduce any matrix to row-reduced echelon form!

Q Decide whether the given matrix is in row-reduced echelon form. If it is not, finish the reduction.

1	1	5	0	-7
0	1	1	0	-4
0	0	0	1	0
0	0	0	0	0

0	1	0	0
1	0	1/3	1/3
0	1	0	0

	1	0	5
	0	1	2

	2	0	1
	0	1	0

Q Now that we know how to reduce a matrix to row-reduced echelon form, what do we do with it?
A First, look at Choice C directly above. You know by now that it is in row-reduced form. Also, you know from Part B of this tutorial that it happens to represent the following solution to a system of two linear equations in two unknowns.

In other words, once we have reduced the augmented matrix to row-reduced form, we can often read off the solution. However, consider the following little exercise:

Consider the system

Q How do we read off the solution from the reduces form. We never had to deal with a row of zeros before!

A Since we have done all we can with the matrix, we translate the rows of the reduced matrix back into equations, and we get:

The second row tells us absolutely nothing: that 0 = 0. In other words, we are left with only one equation in two unknowns. Since we are left with only one equation in two unknowns, we have infinitely many solutions. To see what these solutions look like, solve the above equation for x, and write:

This is called the general solution. Each choice of y will result in a different particular solution. Thus, for instance, if you choose y = 100, you get the particular solution

x	=	100 3	+	10 3	=	110 3
y	=	100.

Q The general solution of the system whose augmented matrix row-reduces to

	1	2 3	-5
	0	0	0

is:

x = -5
y = 0

x =	2y 3	- 5
y arbitrary

x =	-	2y 3	- 5
y arbitrary

x =	13 3
y arbitrary

Q The general solution of the system whose augmented matrix row-reduces to

	0	1	-5
	0	0	0

is:

	x = -y-5 y arbitrary			x arbitrary, y = -x-5
	no solution			x arbitrary y = -5

Q The general solution of the system whose augmented matrix row-reduces to

	1	1	-5
	0	0	3

is:

	x = -y - 5 y = 3			x = -y - 5 y arbitrary
	no solution			x arbitrary y = 3

Q That is all very well. But what about non-standard forms for larger matrices, such as

1	0	-5	-1
0	1	9	2	?
0	0	0	0

A Since row-reduced echelon form is as far as we can go with the matrix, we translate back into equations and obtain

x		-	5z	=	-1
	y	+	9z	=	2	.

Now we solve each of these equations for the first variable, obtaining

x	=	5z-1
y	=	-9z + 2.

We can now choose z to be any number, and then get corresponding values for x and y according to the formulas, giving infinitely many solutions. Thus, the general solution is

x	=	5z-1
y	=	-9z + 2.
z arbitrary.

We get particular solutions by choosing specific values for z. For example, z = 2 gives the particular solution

The following summary is adapted from Section 2.2 of Finite Mathematics and Finite Mathematics and Applied Calculus (Also, you can press the "Chapter Summary" button on the left to bring up a page with this and other information.)

Solutions of Systems of Linear Equations

One of three things will happen in any system of linear equations. There will be:

No Solutions This occurs when you end up with a row of the form
```
 		[0   0   0  . . .  0   #] ,
```
where # is a non zero number. As soon as you spot such a row, check to see that you haven't made an error and then stop. The given system has no solution, so there is no point in continuing any further.
Exactly One Solution (Unique Solution) This occurs if, at the conclusion of the row reduction, translation back into equations yields a single value for each of the unknowns.
Infinitely Many Solutions This occurs if, at the conclusion of the row reduction, translation back into equations does not yield a single value for each of the unknowns. In this case you can easily solve for the unknowns corresponding to the pivots, which will be the first unknowns in each equation. The other unknowns can be assigned arbitrary values. The arbitrary unknowns are called parameters, and the general solution as we have written it is called a parametrized solution. These are the only things that can happen. It is impossible, for example, to have a system of linear equations with exactly eleven solutions. If it has more than one solution, it must have infinitely many solutions.

Hints

If you end up with one or more rows of zeros, all this means is that one of the equations is redundant, and can thus be ignored.
If a system of linear equations has no solution, it is said to be inconsistent. The reason for this term is that the equations actually contradict one another. Here is an obvious example of such a contradiction.
```
  	x + y + z = 0
	x + y + z = 44
```
A reasonable person could hardly expect x, y, and z to add up to 0 and 44 at the same time!
A system of linear equations where there are fewer equations than unknowns is said to be underdetermined. These are the systems that usually give infinitely many solutions (as in Example 6) but may also result in no solutions if they are contradictory (as in the case immediately above). Such a system can never have a unique solution.
A system of equations in which the number of equations exceeds the number of unknowns is said to be overdetermined. In an overdetermined system, anything can happen, but such a system will usually be inconsistent.

Now try the rest of the exercises in Section 2.2 of Finite Mathematics and Finite Mathematics and Applied Calculus.