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

Part B. Solving a System by Pivoting

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Part A. Setting Up a System & Doing Row Operations

Part C. Row-Reduced Echelon Form

Part B: Solving by Pivoting

Solution of Systems of Equations by Row Operations
(The following discussion is based on pp. 126-130 of Finite Mathematics Applied to the Real World and Finite Mathematics and Calculus Applied to the Real World. )

Here, we put row operations to work for us in solving systems of equations. (If you are not familiar with row operations, go to Part A of this tutorial.) First of all, let's start with a complicated looking system of equations, such as

x 2	+	2y 3	=	1 6
2x	+	y 4	=	1 2

This system has the following augmented matrix.

	1 2	2 3	1 6
	2	1 4	1 2

We shall now reduce the matrix with our bare hands in a few simple steps by performing row operations to obtain a new matrix that will instantly tell us the solution of the system.

First Step: Clear all fractions and/or decimals (if any) by multiplying rows that contain them by a suitable number.

To clear the fractions, we multiply the first row by 6 and the second row by 4 as follows.

1 2 2 3 1 6 6 R1 2 1 4 1 2 4 R2

3 4 1 8 1 2

Second Step: Designate the first non-zero entry in the first row as the pivot.

This gives the 3 as the pivot:

first row		3	4	1
		8	1	2

Third Step: Use the pivot to clear its column using operations of type 3. By clearing a column, or pivoting,we mean winding up with a matrix in which the pivot is the only non-zero number in its column. This the goal is to end up with the 8 "cleared away:"

Goal:
cleared pivot column


	3	4	1
	0	*	*

The asterisks stand for whatever numbers come out of the calculation. We must accomplish that by replacing Row 2 using an operation of the third type listed above. Here, we do the following:

3 4 1 8 1 2 3 R2 + 8 R1

3 4 1 0 35 14

Question
Wait a minute! Just how did you come up with that row operation?

Answer
Press the little pearl to open a new window showing a method that always works, and requires little or no thought!

The next step is one that can be performed at any time.

Simplification Step: If at any stage of the process, all the numbers in a row are multiples of an integer, divide by that integer -- a type 2 operation.

Here, we notice that the entries in Row 2 are divisible by 7, so we divide that row by 7.

3 4 1 0 35 14 1 7 R2

3 4 1 0 5 2

Fourth Step: Select the first non-zero number in the next row (Row 2) as pivot, and go to the Third Step.

This is a two-step process: First select a pivot on the second row. Next, perform the pivoting operation; that is, clear the pivot column (the pivot column is now the second column).

Goal:
clear the pivot column


3 4 1 5 R14 R2 0 5 2

15 0 3 0 5 2

Question
So how did you come up with that row operation?

Answer
Press the little pearl to find out! Note that all the row operations we use to pivot have the following form:

a R_c ± b R_p

Here, a and b are always positive numbers.
R_c is the row you are changing. This is also the row next to which the instruction is written.
R_p is the pivot row -- the row containing the pivot.
Note that the only place a minus sign ever appears is between R_c and R_p.

We have one last step:

Final Step: Turn all the pivots into 1's by dividing each of the rows by the value of its pivot.

15 0 3 1 15 R1 0 5 2 1 5 R2

1 0 1 5 0 1 2 5

And we're done! The first row now corresponds to the equation

1(x) + 0(y) =		1 5
or
x =		1 5

The second row corresponds to

0(x) + 1(y) =	2 5
or
y =	2 5

Press the little pearl to bring up a new window showing the complete sequence of row operations we used.

Press the "summary" button on the sidebar for a convenient summary of this procedure and also other material...

Consider the following system.

x 2		5y 2	=	1
3x 5	+	y	=	1

Which of the following corresponds to the augmented matrix after fractions are cleared?

	1 5 2 3 5 5			1 5 1 3 1 1
	5 5 1 6 2 1

The instruction required to perform the first pivot operation is:

	R2 + 3 R1, written next to Row 2
	3 R2R1, written next to Row 2
	R23 R1, written next to Row 1
	R23 R1, written next to Row 2

The result of the above pivot operation is

	1 5 2 0 10 1			1 5 2 0 20 1
	1 5 2 0 20 1			0 20 1 3 5 5

The instruction required to perform the next pivot operation is:

	4 R1 R2, written next to Row 1
	4 R1 + R2, written next to Row 1
	20 R1 + 5 R2, written next to Row 1

The result of the above pivot operation is

	1 0 7 0 20 1			1 15 1 0 20 1
	20 0 35 0 20 1			4 0 7 0 20 1

The solution of the original system of equations is

x

=

7

,y

=

1

x

=

4 7

,y

=

20

x

=

7 4

,y

=

1 20

x

=

1 4

,y

=

7 20

Note Sometimes, things don't work out so smoothly. For instance, you might wind up with a row of zeros, or a row like this: [0 0 9]. Just how to deal with this is discussed in Part C of this tutorial.

Tired of tutorials for now? Try some 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 and then go on to Part C to help answer unresolved questions.

Part A. Setting Up a System & Doing Row Operations

Part C. Row-Reduced Echelon Form