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Systems of Linear Equations:
An Introduction
11 22 33 44 55 66
66
55
44
33
22
11
– – 11
y y
x x
(2, 3)(2, 3)
2 1 x y 2 1 x y
3 2 12 x y 3 2 12 x y
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Systems of Equations
Recall that a
system of two linear equations in twovariables
may be written in the general form
where
a
,
b
,
c
,
d
,
h
, and
k
are
real numbers
and neithera
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 thetwo straight lines
L
1
and
L
2
, represented by the first and
second equations of the system.
ax by h
cx dy k
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Systems of Equations
Given the two straight lines
L
1
and
L
2
,
one and only one
ofthe following may occur:
1.
L
1
and
L
2
intersect at
exactly one point
.
y
x
L
1
L
2
Unique
solution(
x
1
,
y
1
)(
x
1
,
y
1
)
x
1
y
1
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Systems of Equations
Given the two straight lines
L
1
and
L
2
,
one and only one
ofthe following may occur:
2.
L
1
and
L
2
are
coincident
.
y
x
L
1
,
L
2
Infinitely
many
solutions
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Systems of Equations
Given the two straight lines
L
1
and
L
2
,
one and only one
ofthe following may occur:
3.
L
1
and
L
2
are
parallel
.
y
x
L
1
L
2
No
solution
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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 1 y x
3 2(2 1) 12
3 4 2 12
7 14
2
x x
x x
x
x
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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
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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 1 x y
3 2 12 x y
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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 equation
is equivalent to the first.
2 1
6 3 3
x y
x y
2 1 y x
6 3(2 1) 3
6 6 3 3
0 0
x x
x x
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Example:
A System of Equations With Infinitely Many Solutions
Thus, any order pair of numbers ( x, y) satisfying theequation y = 2 x – 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.
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6
5
4
3
2
1
– 11 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 representthe same line, and all solutions of the system are pointslying on the line:
y
x
2 1
6 3 3
x y
x y
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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 1 y x
6 3(2 1) 12
6 6 3 12
0 9
x x
x x
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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 = 2 x – 1 and y = 2 x – 4
which shows that the lines are parallel.
Graphically:
6
5
4
3
2
1
– 1
y
x
2 1 x y
6 3 12 x y
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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
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The Gauss-Jordan Method
The Gauss-Jordan elimination method is a technique forsolving 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 aconstant multiple of any other equation.
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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
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Example
Solve the following system of equations:
Solution
First, we transform this system into an equivalent systemin 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
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Example
Solve the following system of equations:
Solution
Next, we eliminate the variable x from all equations exceptthe 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
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Example
Solve the following system of equations:
Solution
Next, we eliminate the variable x from all equations exceptthe 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
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Example
Solve the following system of equations:
Solution
Next, we eliminate the variable x from all equations exceptthe 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 sumof the first equation
+ the third equation
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Example
Solve the following system of equations:
Solution
Next, we eliminate the variable x from all equations exceptthe 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
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Example
Solve the following system of equations:
Solution
Then we transform so that the coefficient of y in thesecond 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
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Example
Solve the following system of equations:
Solution
Then we transform so that the coefficient of y in thesecond 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
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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
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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
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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
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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
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Example
Solve the following system of equations:
Solution
Now we transform so that the coefficient of z in the thirdequation 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
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Example
Solve the following system of equations:
Solution
Now we transform so that the coefficient of z in the thirdequation 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
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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
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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
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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
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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
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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
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Augmented Matrices
Matrices are rectangular arrays of numbers that can aid
us by eliminating the need to write the variables at eachstep 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
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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
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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
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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
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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
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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
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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
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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
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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
i G
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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
M i d G
J d
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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
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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 otherentries in that column are zeros.
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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 rowand a constant multiple of any other row.
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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.
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Notation for Row Operations
Letting Ri denote the ith row of a matrix, we write
Operation 1: Ri ↔ R j to mean:
Interchange row i with row j.
Operation 2: cRi to mean:
replace row i with c times row i.
Operation 3: Ri + aR j to mean:
Replace row i with the sum of row i
and a times row j.
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Example
Pivot the matrix about the circled element
Solution
3 5 9
2 3 5
3 5 9
2 3 5
1
13 R 5
3 31
52 3
2 12 R R
5
3
1
3
1 3
0 1
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The Gauss-Jordan Elimination Method
1. Write the augmented matrix corresponding tothe 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 matrixabout 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.
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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 2 R R
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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 12 R R
3 1 R R
1 2 R R
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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 12 R R
3 1 R R
2 3 R R
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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 3 R R
1
22 R
1 0 9 12
0 2 12 4
0 2 19 27
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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
22 R
1 0 9 12
0 1 6 2
0 2 19 27
3 2 R R
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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 2 R R
1
331 R
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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
1
331
R
1 39 R R
2 36 R R
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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 36 R R
1 39 R R
E l
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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