Section 2.2: Using Matrices to Solve Systems of Linear Equations with Two Unknowns

Part C. Row-Reduced Echelon Form

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Part A. Setting Up a System & Doing Row Operations

Part B. Solving a System by Pivoting

Part C: Row-Reduced Echelon Form

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.

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

Question
What is so interesting about row-reduced echelon form?

Answer
Recall the procedure you used in Part B to solve systems of linear equations. (If you don't, press the little pearl to go back. ) 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!

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

Question
Now that we know how to reduce a matrix to row-reduced echelon form, what do we do with it?

Answer
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.

x = 5
y = 2

Consider the system

3x y = 10
0.3x + 0.1y = 1

The reduced form of its associated matrix is:

	3 1 10 0 0 0			1 1/3 10/3 1 1/3 10/3
	1 1/3 10/3 0 0 0

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

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

x

y 3

=

10 3

.

The second row tells us absolutely nothing: that 0 = 0. In other words, we are left with only one equation in two unknowns. If you consult Section 2.1, Example 5 in Finite Mathematics Applied to the Real World or Finite Mathematics and Calculus Applied to the Real World. you will see that this gives infinitely many solutions; one for each choice of y (or x, as was done in that example). To see what these solutions look like, solve the above eqution for x, and write:

x

=

y 3

+

10 3

y arbitrary.

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.

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

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

	0	1	5
	0	0	0

is:

	x = y5 y arbitrary			x arbitrary, y = x5
	no solution			x arbitrary y = 5

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

Now try the rest of the exercises on pp. 137-138 of Finite Mathematics Applied to the Real World, or Finite Mathematics and Calculus Applied to the Real World.

Part A. Setting Up a System & Doing Row Operations

Part B. Solving a System by Pivoting