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5
Systems and
Matrices
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5.2
Matrix Solution of Linear Systems
• The Gauss-Jordan Method
• Special Systems
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Matrix Solutions
Since systems of linear equations occur in so
many practical situations, computer methods
have been developed for efficiently solving linear
systems. Computer solutions of linear systems
depend on the idea of a matrix (plural
matrices), a rectangular array of numbers
enclosed in brackets. Each number is an
element of the matrix.
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Gauss-Jordan Method
In this section, we develop a method for
solving linear systems using matrices.
We start with a system and write the
coefficients of the variables and the
constants as an augmented matrix of
the system.
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Gauss-Jordan Method
Linear system of equations
x  3 y  2z  1
2x  y  z  2
xy z2
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Gauss-Jordan Method
The vertical line, which is optional,
separates the coefficients from the
constants. Because this matrix has 3
rows (horizontal) and 4 columns
(vertical), we say its dimension (also
described as order and size) is 3  4
(read “three by four”). The number of
rows is always given first. To refer to a
number in the matrix, use its row and
column numbers.
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Matrix Row Transformations
For any augmented matrix of a system of linear
equations, the following row transformations will
result in the matrix of an equivalent system.
1. Interchange any two rows.
2. Multiply or divide the elements of any row by a
nonzero real number.
3. Replace any row of the matrix by the sum of
the elements of that row and a multiple of the
elements of another row.
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Gauss-Jordan Method
These transformations are restatements in
matrix form of the transformations of
systems discussed in the previous section.
From now on, when referring to the third
transformation, we will abbreviate “a
multiple of the elements of a row” as “a
multiple of a row.”
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Gauss-Jordan Method
The Gauss-Jordan method is a systematic
technique for applying matrix row transformations in
an attempt to reduce a matrix to diagonal form,
with 1s along the diagonal from which the solutions
are easily obtained.
This form is also called reduced-row echelon
form.
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Using the Gauss-Jordan Method to
Put a Matrix into Diagonal Form
Step 1 Obtain 1 as the first element of the first
column.
Step 2 Use the first row to transform the remaining
entries in the first column to 0.
Step 3 Obtain 1 as the second entry in the second
column.
Step 4 Use the second row to transform the
remaining entries in the second column to 0.
Step 5 Continue in this manner as far as possible.
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Note The Gauss-Jordan method
proceeds column by column, from left
to right. In each column, we work to obtain
1 in the appropriate diagonal location, and
then use it to transform the remaining
elements in that column to 0s. When we
are working with a particular column, no
row operation should undo the form of a
preceding column.
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Example 1
USING THE GAUSS-JORDAN
METHOD
Solve the system.
Solution
3 x  4y  1
5 x  2y  19
Both equations are in the same form, with variable
terms in the same order on the left, and constant
terms on the right.
Write the augmented matrix.
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Example 1
USING THE GAUSS-JORDAN
METHOD
The goal is to transform the augmented matrix
into one in which the value of the variables will
be easy to see. That is, since each of the first
two columns in the matrix represents the
coefficients of one variable, the augmented
matrix should be transformed so that it is of the
following form. (Here k and j are real numbers.)
Once the augmented matrix is in this form, the
matrix can be rewritten as a linear system.
This form is our
goal.
xk
y j
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Example 1
USING THE GAUSS-JORDAN
METHOD
It is best to work in columns beginning in each
column with the element that is to become 1.
In the augmented matrix
3 is in the first row, first column position. Use
transformation 2, multiplying each entry in the first
row by 1/3 to get 1 in this position. (This step is
abbreviated as 1 R1 .)
3
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Example 1
USING THE GAUSS-JORDAN
METHOD
Introduce 0 in the second row, first column by
multiplying each element of the first row by –5 and
adding the result to the corresponding element in
the second row, using transformation 3.
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Example 1
USING THE GAUSS-JORDAN
METHOD
Obtain 1 in the second row, second column by
multiplying each element of the second row by
using transformation 2.
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Example 1
USING THE GAUSS-JORDAN
METHOD
Finally, get 0 in the first row, second column by
4
multiplying each element of the second row by
3
and adding the result to the corresponding
element in the first row.
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Example 1
USING THE GAUSS-JORDAN
METHOD
This last matrix corresponds to the system
x3
y  2,
which indicates the solution (3, 2). We can
read this solution directly from the third column
of the final matrix.
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USING THE GAUSS-JORDAN
METHOD
Example 1
CHECK Substitute the solution in both equations of
the original system.
3 x  4y  1
5 x  2y  19
?
?
5(3)  2(2)  19
3(3)  4(2)  1
?
?
15  4  19
9  8 1
1 1
True
19  19
True
Since true statements result, the solution set is {(3,2)}.
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Example 2
USING THE GAUSS-JORDAN
METHOD
Solve the system.
x  y  5z  6
3 x  3 y  z  10
x  3 y  2z  5
Solution
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Example 2
USING THE GAUSS-JORDAN
METHOD
There is already a 1 in the first row, first column.
Introduce 0 in the second row of the first column
by multiplying each element in the first row by – 3
and adding the result to the corresponding
element in the second row.
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Example 2
USING THE GAUSS-JORDAN
METHOD
To change the third element in the first column to
0, multiply each element of the first row by – 1. Add
the result to the corresponding element of the
third row.
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Example 2
USING THE GAUSS-JORDAN
METHOD
Use the same procedure to transform the second and third columns.
For both of these columns, perform the additional step of getting 1 in
the appropriate position of each column. Do this by multiplying the
elements of the row by the reciprocal of the number in that position.
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Example 2
USING THE GAUSS-JORDAN
METHOD
The linear system associated with this final
matrix is
x 1
y 2
z  1.
The solution set is {(1, 2, –1)}. Check the
solution in the original system.
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Example 3
SOLVING AN INCONSISTENT
SYSTEM
Use the Gauss-Jordan method to solve the system.
xy 2
2 x  2y  5
Solution
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Example 3
SOLVING AN INCONSISTENT
SYSTEM
The next step would be to get 1 in the second
row, second column. Because of the 0 there, it is
impossible to go further. Since the second row
corresponds to the equation 0x + 0y = 1, which
has no solution, the system is inconsistent and
the solution set is Ø.
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Example 4
SOLVING A SYSTEM WITH
INFINITELY MANY SOLUTIONS
Use the Gauss-Jordan method to solve the
system.
Solution
2 x  5 y  3z  1
x  2 y  2z  8
Recall from the previous section that a system
with two equations in three variables usually has
an infinite number of solutions. We can use the
Gauss-Jordan method to give the solution with z
arbitrary.
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Example 4
SOLVING A SYSTEM WITH
INFINITELY MANY SOLUTIONS
It is not possible to go further with the Gauss-Jordan method.
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Example 4
SOLVING A SYSTEM WITH
INFINITELY MANY SOLUTIONS
The equations that correspond to the final matrix are
x  16z  38
and
y  7z  15.
Solve these equations for x and y, respectively.
x  16z  38
x  16z  38
y  7z  15
y  7z  15
The solution set, written with z arbitrary, is
16z  38, 7z  15, z .
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Summary of Possible Cases
When matrix methods are used to solve a system of
linear equations and the resulting matrix is written in
diagonal form:
1. If the number of rows with nonzero elements to
the left of the vertical line is equal to the number
of variables in the system, then the system has a
single solution. See Examples 1 and 2.
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Summary of Possible Cases
2. If one of the rows has the form [0 0    0 a] with
a ≠ 0, then the system has no solution.
See Example 3.
3. If there are fewer rows in the matrix containing
nonzero elements than the number of variables,
then the system has either no solution or infinitely
many solutions. If there are infinitely many
solutions, give the solutions in terms of one or
more arbitrary variables. See Example 4.
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