Systems of Linear Equations:
An Introduction
11 22 33 44 55 66
66
55
44
33
22
11
1 1
yy
xx
(2, 3)(2, 3)
2 1x y 2 1x y
3 2 12x y 3 2 12x y
Systems of Equations
Recall that a system of two linear equations in two
variables may be written in the general form
where a, b, c, d, h, and k are real numbers and neither
a and b nor c and d are both zero.
Recall that the graph of each equation in the system is a
straight line in the plane, so that geometrically, the
solution to the system is the point(s) of intersection of the
two straight lines L1 and L2, represented by the first and
second equations of the system.
ax by h
cx dy k
Systems of Equations
Given the two straight lines L1 and L2, one and only one of
the following may occur:
1. L1 and L2 intersect at exactly one point.
y
x
L1
L2
Unique
solution
(x1, y1)(x1, y1)
x1
y1
Systems of Equations
Given the two straight lines L1 and L2, one and only one of
the following may occur:
2. L1 and L2 are coincident.
y
x
L1, L2
Infinitely
many
solutions
Systems of Equations
Given the two straight lines L1 and L2, one and only one of
the following may occur:
3. L1 and L2 are parallel.
y
x
L1L2
No
solution
Example:
A System of Equations With Exactly One Solution
Consider the system
Solving the first equation for y in terms of x, we obtain
Substituting this expression for y into the second equation
yields
2 1
3 2 12
x y
x y
2 1y x
3 2(2 1) 12
3 4 2 12
7 14
2
x x
x x
x
x
Example:
A System of Equations With Exactly One Solution
Finally, substituting this value of x into the expression for y
obtained earlier gives
Therefore, the unique solution of the system is given by
x = 2 and y = 3.
2 1
2(2) 1
3
y x
1 2 3 4 5 6
6
5
4
3
2
1
1
Example:
A System of Equations With Exactly One Solution
Geometrically, the two lines represented by the two
equations that make up the system intersect at the
point (2, 3):
y
x
(2, 3)
2 1x y
3 2 12x y
Example:
A System of Equations With Infinitely Many Solutions
Consider the system
Solving the first equation for y in terms of x, we obtain
Substituting this expression for y into the second equationyields
which is a true statement.
This result follows from the fact that the second equationis equivalent to the first.
2 1
6 3 3
x y
x y
2 1y x
6 3(2 1) 3
6 6 3 3
0 0
x x
x x
Example:
A System of Equations With Infinitely Many Solutions
Thus, any order pair of numbers (x, y) satisfying the
equation y = 2x 1 constitutes a solution to the system.
By assigning the value t to x, where t is any real number,
we find that y = 2t 1 and so the ordered pair (t, 2t 1)
is a solution to the system.
The variable t is called a parameter.
For example:
Setting t = 0, gives the point (0, 1) as a solution of the system.
Setting t = 1, gives the point (1, 1) as another solution of the system.
65
4
3
2
1
1 1 2 3 4 5 6
Example:
A System of Equations With Infinitely Many Solutions
Since t represents any real number, there are infinitely many solutions of the system.
Geometrically, the two equations in the system represent the same line, and all solutions of the system are points lying on the line:
y
x
2 1
6 3 3
x y
x y
Example:
A System of Equations That Has No Solution
Consider the system
Solving the first equation for y in terms of x, we obtain
Substituting this expression for y into the second equationyields
which is clearly impossible.
Thus, there is no solution to the system of equations.
2 1
6 3 12
x y
x y
2 1y x
6 3(2 1) 12
6 6 3 12
0 9
x x
x x
1 2 3 4 5 6
Example:
A System of Equations That Has No Solution
To interpret the situation geometrically, cast both equations in the slope-intercept form, obtaining
y = 2x 1 and y = 2x 4
which shows that the lines are parallel.
Graphically:6
5
4
3
2
1
1
y
x
2 1x y
6 3 12x y
Systems of Linear Equations:
Unique Solutions
3 2 8 9
2 2 1 3
1 2 3 8
3 2 8 9
2 2 1 3
1 2 3 8
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
1 0 0 3
0 1 0 4
0 0 1 1
1 0 0 3
0 1 0 4
0 0 1 1
The Gauss-Jordan Method
The Gauss-Jordan elimination method is a technique for solving systems of linear equations of any size.
The operations of the Gauss-Jordan method are
1. Interchange any two equations.
2. Replace an equation by a nonzero constant multiple of itself.
3. Replace an equation by the sum of that equation and a constant multiple of any other equation.
Example
Solve the following system of equations:
Solution
First, we transform this system into an equivalent system in which the coefficient of x in the first equation is 1:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
Multiply the
equation by 1/2
Example
Solve the following system of equations:
Solution
First, we transform this system into an equivalent system in which the coefficient of x in the first equation is 1:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
2 3 11
3 8 5 27
2 2
x y z
x y z
x y z
Multiply the first
equation by 1/2
Example
Solve the following system of equations:
Solution
Next, we eliminate the variable x from all equations except the first:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
2 3 11
3 8 5 27
2 2
x y z
x y z
x y z
Replace by the sum of
3 X the first equation
+ the second equation
Example
Solve the following system of equations:
Solution
Next, we eliminate the variable x from all equations except the first:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
2 3 11
2 4 6
2 2
x y z
y z
x y z
Replace by the sum of
3 the first equation + the second equation
Example
Solve the following system of equations:
Solution
Next, we eliminate the variable x from all equations except the first:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
2 3 11
2 4 6
2 2
x y z
y z
x y z
Replace by the sum of the first equation
+ the third equation
Example
Solve the following system of equations:
Solution
Next, we eliminate the variable x from all equations except the first:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
2 3 11
2 4 6
3 5 13
x y z
y z
y z
Replace by the sum of the first equation
+ the third equation
Example
Solve the following system of equations:
Solution
Then we transform so that the coefficient of y in the second equation is 1:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
2 3 11
2 4 6
3 5 13
x y z
y z
y z
Multiply the second
equation by 1/2
Example
Solve the following system of equations:
Solution
Then we transform so that the coefficient of y in the second equation is 1:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
2 3 11
2 3
3 5 13
x y z
y z
y z
Multiply the second
equation by 1/2
Example
Solve the following system of equations:
Solution
We now eliminate y from all equations except the second:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
2 3 11
2 3
3 5 13
x y z
y z
y z
Replace by the sum of
the first equation +
(2) the second equation
Example
Solve the following system of equations:
Solution
We now eliminate y from all equations except the second:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
7 17
2 3
3 5 13
x z
y z
y z
Replace by the sum of
the first equation +
(2) the second equation
Example
Solve the following system of equations:
Solution
We now eliminate y from all equations except the second:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
7 17
2 3
3 5 13
x z
y z
y z
Replace by the sum of the third equation +
(3) the second equation
Example
Solve the following system of equations:
Solution
We now eliminate y from all equations except the second:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
7 17
2 3
11 22
x z
y z
z
Replace by the sum of the third equation +
(3) the second equation
Example
Solve the following system of equations:
Solution
Now we transform so that the coefficient of z in the third equation is 1:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
7 17
2 3
11 22
x z
y z
z
Multiply the third equation by 1/11
Example
Solve the following system of equations:
Solution
Now we transform so that the coefficient of z in the third equation is 1:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
7 17
2 3
2
x z
y z
z
Multiply the third equation by 1/11
Example
Solve the following system of equations:
Solution
We now eliminate z from all equations except the third:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
Replace by the sum of
the first equation +
(7) the third equation
7 17
2 3
2
x z
y z
z
Example
Solve the following system of equations:
Solution
We now eliminate z from all equations except the third:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
3
2 3
2
x
y z
z
Replace by the sum of
the first equation +
(7) the third equation
Example
Solve the following system of equations:
Solution
We now eliminate z from all equations except the third:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
3
2 3
2
x
y z
z
Replace by the sum of
the second equation +
2 the third equation
Example
Solve the following system of equations:
Solution
We now eliminate z from all equations except the third:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
3
1
2
x
y
z
Replace by the sum of
the second equation +
2 the third equation
Example
Solve the following system of equations:
Solution
Thus, the solution to the system is x = 3, y = 1, and z = 2.
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
3
1
2
x
y
z
Augmented Matrices
Matrices are rectangular arrays of numbers that can aid us by eliminating the need to write the variables at each step of the reduction.
For example, the system
may be represented by the augmented matrixCoefficient
Matrix
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
2 4 6 22
3 8 5 27
1 1 2 2
Matrices and Gauss-Jordan
Every step in the Gauss-Jordan elimination method can be
expressed with matrices, rather than systems of equations,
thus simplifying the whole process:
Steps expressed as systems of equations:
Steps expressed as augmented matrices:
2 4 6 22
3 8 5 27
2 2
x y z
x y z
x y z
2 4 6 22
3 8 5 27
1 1 2 2
Matrices and Gauss-Jordan
Every step in the Gauss-Jordan elimination method can be
expressed with matrices, rather than systems of equations,
thus simplifying the whole process:
Steps expressed as systems of equations:
Steps expressed as augmented matrices:
1 2 3 11
3 8 5 27
1 1 2 2
2 3 11
3 8 5 27
2 2
x y z
x y z
x y z
Matrices and Gauss-Jordan
Every step in the Gauss-Jordan elimination method can be
expressed with matrices, rather than systems of equations,
thus simplifying the whole process:
Steps expressed as systems of equations:
Steps expressed as augmented matrices:
1 2 3 11
0 2 4 6
1 1 2 2
2 3 11
2 4 6
2 2
x y z
y z
x y z
Matrices and Gauss-Jordan
Every step in the Gauss-Jordan elimination method can be
expressed with matrices, rather than systems of equations,
thus simplifying the whole process:
Steps expressed as systems of equations:
Steps expressed as augmented matrices:
1 2 3 11
0 2 4 6
0 3 5 13
2 3 11
2 4 6
3 5 13
x y z
y z
y z
Matrices and Gauss-Jordan
Every step in the Gauss-Jordan elimination method can be
expressed with matrices, rather than systems of equations,
thus simplifying the whole process:
Steps expressed as systems of equations:
Steps expressed as augmented matrices:
1 2 3 11
0 1 2 3
0 3 5 13
2 3 11
2 3
3 5 13
x y z
y z
y z
Matrices and Gauss-Jordan
Every step in the Gauss-Jordan elimination method can be
expressed with matrices, rather than systems of equations,
thus simplifying the whole process:
Steps expressed as systems of equations:
Steps expressed as augmented matrices:
1 0 7 17
0 1 2 3
0 3 5 13
7 17
2 3
3 5 13
x z
y z
y z
Matrices and Gauss-Jordan
Every step in the Gauss-Jordan elimination method can be
expressed with matrices, rather than systems of equations,
thus simplifying the whole process:
Steps expressed as systems of equations:
Steps expressed as augmented matrices:
1 0 7 17
0 1 2 3
0 0 11 22
7 17
2 3
11 22
x z
y z
z
Matrices and Gauss-Jordan
Every step in the Gauss-Jordan elimination method can be
expressed with matrices, rather than systems of equations,
thus simplifying the whole process:
Steps expressed as systems of equations:
Steps expressed as augmented matrices:
1 0 7 17
0 1 2 3
0 0 1 2
7 17
2 3
2
x z
y z
z
Matrices and Gauss-Jordan
Every step in the Gauss-Jordan elimination method can be
expressed with matrices, rather than systems of equations,
thus simplifying the whole process:
Steps expressed as systems of equations:
Steps expressed as augmented matrices:
1 0 0 3
0 1 2 3
0 0 1 2
3
2 3
2
x
y z
z
Matrices and Gauss-Jordan
Every step in the Gauss-Jordan elimination method can be
expressed with matrices, rather than systems of equations,
thus simplifying the whole process:
Steps expressed as systems of equations:
Steps expressed as augmented matrices:
1 0 0 3
0 1 0 1
0 0 1 2
3
1
2
x
y
z
Row Reduced Form
of the Matrix
Row-Reduced Form of a Matrix
Each row consisting entirely of zeros lies below all
rows having nonzero entries.
The first nonzero entry in each nonzero row is 1
(called a leading 1).
In any two successive (nonzero) rows, the leading 1
in the lower row lies to the right of the leading 1 in
the upper row.
If a column contains a leading 1, then the other
entries in that column are zeros.
Row Operations
1. Interchange any two rows.
2. Replace any row by a nonzero constant
multiple of itself.
3. Replace any row by the sum of that row
and a constant multiple of any other row.
Terminology for the
Gauss-Jordan Elimination Method
Unit Column
A column in a coefficient matrix is in unit form
if one of the entries in the column is a 1 and the
other entries are zeros.
Pivoting
The sequence of row operations that transforms
a given column in an augmented matrix into a
unit column.
Notation for Row Operations
Letting Ri denote the ith row of a matrix, we write
Operation 1: Ri Rj to mean:
Interchange row i with row j.
Operation 2: cRi to mean:
replace row i with c times row i.
Operation 3: Ri + aRj to mean:
Replace row i with the sum of row i
and a times row j.
Example
Pivot the matrix about the circled element
Solution
3 5 9
2 3 5
3 5 9
2 3 5
113
R 53
31
52 3
2 12R R53
13
1 3
0 1
The Gauss-Jordan Elimination Method
1. Write the augmented matrix corresponding to
the linear system.
2. Interchange rows, if necessary, to obtain an
augmented matrix in which the first entry in
the first row is nonzero. Then pivot the matrix
about this entry.
3. Interchange the second row with any row below
it, if necessary, to obtain an augmented matrix
in which the second entry in the second row is
nonzero. Pivot the matrix about this entry.
4. Continue until the final matrix is in row-
reduced form.
Example
Use the Gauss-Jordan elimination method to solve the system of equations
Solution
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
3 2 8 9
2 2 1 3
1 2 3 8
1 2R R
Example
Use the Gauss-Jordan elimination method to solve the system of equations
Solution
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
1 0 9 12
2 2 1 3
1 2 3 8
2 12R R
3 1R R1 2R R
Example
Use the Gauss-Jordan elimination method to solve the system of equations
Solution
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
1 0 9 12
0 2 19 27
0 2 12 4
2 12R R
3 1R R2 3R R
Example
Use the Gauss-Jordan elimination method to solve the system of equations
Solution
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
2 3R R1
22R
1 0 9 12
0 2 12 4
0 2 19 27
Example
Use the Gauss-Jordan elimination method to solve the system of equations
Solution
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
122
R
1 0 9 12
0 1 6 2
0 2 19 27
3 2
R R
Example
Use the Gauss-Jordan elimination method to solve the system of equations
Solution
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
1 0 9 12
0 1 6 2
0 0 31 31
3 2R R1
331R
Example
Use the Gauss-Jordan elimination method to solve the system of equations
Solution
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
1 0 9 12
0 1 6 2
0 0 1 1
1331
R1 39R R
2 36R R
Example
Use the Gauss-Jordan elimination method to solve the system of equations
Solution
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
1 0 0 3
0 1 0 4
0 0 1 1
2 36R R
1 39R R
Example
Use the Gauss-Jordan elimination method to solve the system of equations
Solution
The solution to the system is thus x = 3, y = 4, and z = 1.
3 2 8 9
2 2 3
2 3 8
x y z
x y z
x y z
1 0 0 3
0 1 0 4
0 0 1 1