5
Systems of Linear Equations and Matrices
 Systems of Linear Equations:
✦ An Introduction
✦ Unique Solutions
✦ Underdetermined and
Overdetermined Systems
 Matrices
 Multiplication of Matrices
 The Inverse of a Square Matrix
5.1
Systems of Linear Equations:
An Introduction
y
6
2x  y  1
5
4
(2, 3)
3
2
3x  2 y  12
1
–1
1
2
3
4
5
6
x
Systems of Equations
 Recall that a system of two linear equations in two
variables may be written in the general form
ax  by  h
cx  dy  k
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.
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
L1
y1
Unique
solution
(x1, y1)
(x1, y1)
x1
x
L2
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
L1, L2
Infinitely
many
solutions
x
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
L1
L2
No
solution
x
Example:
A System of Equations With Exactly One Solution
 Consider the system
2x  y  1
3 x  2 y  12
 Solving the first equation for y in terms of x, we obtain
y  2x  1
 Substituting this expression for y into the second equation
yields
3x  2(2 x  1)  12
3x  4 x  2  12
7 x  14
x2
Example:
A System of Equations With Exactly One Solution
 Finally, substituting this value of x into the expression for y
obtained earlier gives
y  2x 1
 2(2)  1
3
 Therefore, the unique solution of the system is given by
x = 2 and y = 3.
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
6
2x  y  1
5
4
(2, 3)
3
2
3x  2 y  12
1
–1
1
2
3
4
5
6
x
Example:
A System of Equations With Infinitely Many Solutions
 Consider the system
2x  y  1
6x  3y  3
 Solving the first equation for y in terms of x, we obtain
y  2x  1
 Substituting this expression for y into the second equation
yields
6 x  3(2 x  1)  3
6x  6x  3  3
00
which is a true statement.
 This result follows from the fact that the second equation
is equivalent to the first.
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.
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
6
5
2x  y  1
6x  3y  3
4
3
2
1
–1
1
2
3
4
5
6
x
Example:
A System of Equations That Has No Solution
 Consider the system
2x  y  1
6 x  3 y  12
 Solving the first equation for y in terms of x, we obtain
y  2x  1
 Substituting this expression for y into the second equation
yields
6 x  3(2 x  1)  12
6 x  6 x  3  12
09
which is clearly impossible.
 Thus, there is no solution to the system of equations.
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:
y
6
5
2x  y  1
6 x  3 y  12
4
3
2
1
–1
1
2
3
4
5
6
x
5.2
Systems of Linear Equations:
Unique Solutions
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
 3 2 8

1
 2 2
 1 2 3
9

3
8 
 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:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 First, we transform this system into an equivalent system
in which the coefficient of x in the first equation is 1:
Toggle slides
back and forth to
compare before
and changes
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Multiply the
equation by 1/2
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 First, we transform this system into an equivalent system
in which the coefficient of x in the first equation is 1:
Toggle slides
back and forth to
compare before
and changes
x  2 y  3z  11
3x  8 y  5z  27
 x  y  2z  2
Multiply the first
equation by 1/2
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 Next, we eliminate the variable x from all equations except
the first:
Toggle slides
back and forth to
compare before
and changes
x  2 y  3z  11
3x  8 y  5z  27
 x  y  2z  2
Replace by the sum of
– 3 X the first equation
+ the second equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 Next, we eliminate the variable x from all equations except
the first:
Toggle slides
back and forth to
compare before
and changes
x  2 y  3z  11
2 y  4 z  6
 x  y  2z  2
Replace by the sum of
– 3 ☓ the first equation
+ the second equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 Next, we eliminate the variable x from all equations except
the first:
Toggle slides
back and forth to
compare before
and changes
x  2 y  3z  11
2 y  4 z  6
 x  y  2z  2
Replace by the sum
of the first equation
+ the third equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 Next, we eliminate the variable x from all equations except
the first:
Toggle slides
back and forth to
compare before
and changes
x  2 y  3z  11
2 y  4 z  6
3 y  5z  13
Replace by the sum
of the first equation
+ the third equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 Then we transform so that the coefficient of y in the
second equation is 1:
Toggle slides
back and forth to
compare before
and changes
x  2 y  3z  11
2 y  4 z  6
3 y  5z  13
Multiply the second
equation by 1/2
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 Then we transform so that the coefficient of y in the
second equation is 1:
Toggle slides
back and forth to
compare before
and changes
x  2 y  3z  11
y  2 z  3
3 y  5z  13
Multiply the second
equation by 1/2
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 We now eliminate y from all equations except the second:
Toggle slides
back and forth to
compare before
and changes
x  2 y  3z  11
y  2 z  3
3 y  5z  13
Replace by the sum of
the first equation +
(–2) ☓ the second equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 We now eliminate y from all equations except the second:
Toggle slides
back and forth to
compare before
and changes
x
 7 z  17
y  2 z  3
3 y  5z  13
Replace by the sum of
the first equation +
(–2) ☓ the second equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 We now eliminate y from all equations except the second:
Toggle slides
back and forth to
compare before
and changes
x
 7 z  17
y  2 z  3
3 y  5z  13
Replace by the sum of
the third equation +
(–3) ☓ the second equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 We now eliminate y from all equations except the second:
Toggle slides
back and forth to
compare before
and changes
x
 7 z  17
y  2 z  3
11z  22
Replace by the sum of
the third equation +
(–3) ☓ the second equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 Now we transform so that the coefficient of z in the third
equation is 1:
Toggle slides
back and forth to
compare before
and changes
x
 7 z  17
y  2 z  3
11z  22
Multiply the third
equation by 1/11
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 Now we transform so that the coefficient of z in the third
equation is 1:
Toggle slides
back and forth to
compare before
and changes
x
 7 z  17
y  2 z  3
z2
Multiply the third
equation by 1/11
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 We now eliminate z from all equations except the third:
Toggle slides
back and forth to
compare before
and changes
x
 7 z  17
y  2 z  3
z2
Replace by the sum of
the first equation +
(–7) ☓ the third equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 We now eliminate z from all equations except the third:
Toggle slides
back and forth to
compare before
and changes
x
3
y  2 z  3
z2
Replace by the sum of
the first equation +
(–7) ☓ the third equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 We now eliminate z from all equations except the third:
Toggle slides
back and forth to
compare before
and changes
x
3
y  2 z  3
z2
Replace by the sum of
the second equation +
2 ☓ the third equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 We now eliminate z from all equations except the third:
Toggle slides
back and forth to
compare before
and changes
x
y
3
1
z2
Replace by the sum of
the second equation +
2 ☓ the third equation
Example
 Solve the following system of equations:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
Solution
 Thus, the solution to the system is x = 3, y = 1, and z = 2.
x
y
3
1
z2
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
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
may be represented by the augmented matrix
Coefficient
Matrix
 2 4 6

 3 8 5
 1 1 2
22 

27 
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:
2 x  4 y  6 z  22
3x  8 y  5z  27
 x  y  2z  2
 Steps expressed as augmented matrices:
Toggle slides
back and forth to
compare before
and changes
 2 4 6

 3 8 5
 1 1 2
22 

27 
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:
x  2 y  3z  11
3x  8 y  5z  27
 x  y  2z  2
 Steps expressed as augmented matrices:
Toggle slides
back and forth to
compare before
and changes
 1 2 3

 3 8 5
 1 1 2
11 

27 
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:
x  2 y  3z  11
2 y  4 z  6
 x  y  2z  2
 Steps expressed as augmented matrices:
Toggle slides
back and forth to
compare before
and changes
3
 1 2

 0 2 4
 1 1 2
11 

6 
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:
x  2 y  3z  11
2 y  4 z  6
3 y  5z  13
 Steps expressed as augmented matrices:
Toggle slides
back and forth to
compare before
and changes
3
1 2

 0 2 4
 0 3 5
11 

6 
13 
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:
x  2 y  3z  11
y  2 z  3
3 y  5z  13
 Steps expressed as augmented matrices:
Toggle slides
back and forth to
compare before
and changes
3
1 2

 0 1 2
0 3 5
11 

3
13 
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:
x
 7 z  17
y  2 z  3
3 y  5z  13
 Steps expressed as augmented matrices:
Toggle slides
back and forth to
compare before
and changes
1 0 7

 0 1 2
0 3 5
17 

3
13 
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:
x
 7 z  17
y  2 z  3
11z  22
 Steps expressed as augmented matrices:
Toggle slides
back and forth to
compare before
and changes
1 0 7

 0 1 2
0 0 11
17 

3
22 
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:
x
 7 z  17
y  2 z  3
z2
 Steps expressed as augmented matrices:
Toggle slides
back and forth to
compare before
and changes
1 0 7

 0 1 2
0 0
1
17 

3
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:
x
3
y  2 z  3
z2
 Steps expressed as augmented matrices:
Toggle slides
back and forth to
compare before
and changes
1 0 0

 0 1 2
0 0
1
3

3
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:
x
y
3
1
z2
 Steps expressed as augmented matrices:
Toggle slides
back and forth to
compare before
and changes
1 0 0

0 1 0
0 0 1
3

1
2 
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
 3 5 9


2
3
5


Solution
 3 5 9


2
3
5


1
3
R1
 1 53 3 R2  2R1


5
2
3


5
1
3
3


1
0


1
3


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 rowreduced form.
Example
 Use the Gauss-Jordan elimination method to solve the
system of equations
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
Solution
 3 2 8

1
 2 2
 1 2 3
9

3 R1  R2
8 
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forth to compare before
and after matrix changes
Example
 Use the Gauss-Jordan elimination method to solve the
system of equations
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
Solution
 1 0 9 12 

 R2  2R1
1 3  R1  R2
 2 2
R  R1
 1 2 3 8  3
Toggle slides back and
forth to compare before
and after matrix changes
Example
 Use the Gauss-Jordan elimination method to solve the
system of equations
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
Solution
9
1 0

0 2 19
0 2 12
12 
 R2  2R1
27  R2  R3
R3  R1

4 
Toggle slides back and
forth to compare before
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Example
 Use the Gauss-Jordan elimination method to solve the
system of equations
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
Solution
9
1 0

0 2 12
0 2 19
12 

R3
4  R12 2R
2
27 
Toggle slides back and
forth to compare before
and after matrix changes
Example
 Use the Gauss-Jordan elimination method to solve the
system of equations
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
Solution
1 0 9

 0 1 6
 0 2 19
12 

2  R12 3R2 R2
27 
Toggle slides back and
forth to compare before
and after matrix changes
Example
 Use the Gauss-Jordan elimination method to solve the
system of equations
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
Solution
1 0 9

 0 1 6
0 0 31
12 
 1
2  R
3 R2
313R
31 
Toggle slides back and
forth to compare before
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Example
 Use the Gauss-Jordan elimination method to solve the
system of equations
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
Solution
1 0 9

 0 1 6
0 0
1
12 
 R11  9R3
2  31 R3
R  6R3
1 2
Toggle slides back and
forth to compare before
and after matrix changes
Example
 Use the Gauss-Jordan elimination method to solve the
system of equations
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
Solution
1 0 0

0 1 0
0 0 1
3

4
1 
Toggle slides back and
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R1  9R3
R2  6R3
Example
 Use the Gauss-Jordan elimination method to solve the
system of equations
3x  2 y  8 z  9
2 x  2 y  z  3
x  2 y  3z  8
Solution
1 0 0

0 1 0
0 0 1
3

4
1 
 The solution to the system is thus x = 3, y = 4, and z = 1.
5.3
Systems of Linear Equations:
Underdetermined and Overdetermined systems
x  2 y  3z  2
3x  y  2 z  1
2 x  3 y  5z  3
xz 0
y  z  1
 1 2 3

 3 1 2
 2 3 5
xz
y  z 1
2 

1
3
A System of Equations
with an Infinite Number of Solutions
 Solve the system of equations given by
x  2 y  3z  2
3x  y  2 z  1
2 x  3 y  5z  3
Solution
 1 2 3

 3 1 2
 2 3 5
2 
 R2  3R1
1
R3  2R1

3
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A System of Equations
with an Infinite Number of Solutions
 Solve the system of equations given by
x  2 y  3z  2
3x  y  2 z  1
2 x  3 y  5z  3
Solution
 1 2 3

 0 7 7
0 1 1
2 
R2  3R1
 
7  17 R2
R3  2R1

1
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A System of Equations
with an Infinite Number of Solutions
 Solve the system of equations given by
x  2 y  3z  2
3x  y  2 z  1
2 x  3 y  5z  3
Solution
 1 2 3

0 1 1
0 1 1
2  R  2 R
  11 R 2
1 7 2
R3  R2

1
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A System of Equations
with an Infinite Number of Solutions
 Solve the system of equations given by
x  2 y  3z  2
3x  y  2 z  1
2 x  3 y  5z  3
Solution
 1 0 1

0 1 1
0 0 0
0 R  2R
2
 1
1
R3  R2

0
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A System of Equations
with an Infinite Number of Solutions
 Solve the system of equations given by
x  2 y  3z  2
3x  y  2 z  1
2 x  3 y  5z  3
Solution
 1 0 1

0 1 1
0 0 0
0

1
0 
 Observe that row three reads 0 = 0, which is true but
of no use to us.
A System of Equations
with an Infinite Number of Solutions
 Solve the system of equations given by
x  2 y  3z  2
3x  y  2 z  1
2 x  3 y  5z  3
Solution
 This last augmented matrix is in row-reduced form.
 Interpreting it as a system of equations gives a system of
two equations in three variables x, y, and z:
xz 0
y  z  1
 1 0 1

0 1 1
0 0 0
0

1
0 
A System of Equations
with an Infinite Number of Solutions
 Solve the system of equations given by
x  2 y  3z  2
3x  y  2 z  1
2 x  3 y  5z  3
Solution
 Let’s single out a single variable –say, z– and solve for x
and y in terms of it.
 If we assign a particular value of z –say, z = 0– we obtain
x = 0 and y = –1, giving the solution (0, –1, 0).
xz 0
y  z  1
x  z  (0)  0
y  z  1  (0)  1  1
A System of Equations
with an Infinite Number of Solutions
 Solve the system of equations given by
x  2 y  3z  2
3x  y  2 z  1
2 x  3 y  5z  3
Solution
 Let’s single out a single variable –say, z– and solve for x
and y in terms of it.
 If we instead assign z = 1, we obtain the solution (1, 0, 1).
xz 0
y  z  1
x  z  (1)  1
y  z  1  (1)  1  0
A System of Equations
with an Infinite Number of Solutions
 Solve the system of equations given by
x  2 y  3z  2
3x  y  2 z  1
2 x  3 y  5z  3
Solution
 Let’s single out a single variable –say, z– and solve for x
and y in terms of it.
 In general, we set z = t, where t represents any real number
(called the parameter) to obtain the solution (t, t – 1, t).
xz 0
y  z  1
 (t )  t
xz
y  z  1  (t )  1  t  1
A System of Equations That Has No Solution
 Solve the system of equations given by
x y z 1
3x  y  z  4
x  5 y  5z  1
Solution
1 1 1

3 1 1
 1 5 5
1
 R2  3R1
4
R3  R1

1
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A System of Equations That Has No Solution
 Solve the system of equations given by
x y z 1
3x  y  z  4
x  5 y  5z  1
Solution
1 1 1

 0  4 4
0 4 4
1
R2  3R1
 R
1 3  R2
R3  R1

2 
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forth to compare before
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A System of Equations That Has No Solution
 Solve the system of equations given by
x y z 1
3x  y  z  4
x  5 y  5z  1
Solution
1 1 1

 0  4 4
0 0 0
1

1 R3  R2
1
Toggle slides back and
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A System of Equations That Has No Solution
 Solve the system of equations given by
x y z 1
3x  y  z  4
x  5 y  5z  1
Solution
1 1 1

 0  4 4
0 0 0
1

1
1
 Observe that row three reads 0x + 0y + 0z = –1 or 0 = –1!
 We therefore conclude the system is inconsistent and has
no solution.
Systems with no Solution
 If there is a row in the augmented matrix
containing all zeros to the left of the vertical line
and a nonzero entry to the right of the line, then
the system of equations has no solution.
Theorem 1
a. If the number of equations is greater than or
equal to the number of variables in a linear
system, then one of the following is true:
i. The system has no solution.
ii. The system has exactly one solution.
iii. The system has infinitely many solutions.
b. If there are fewer equations than variables in
a linear system, then the system either has no
solution or it has infinitely many solutions.
5.4
Matrices
2 X  B  3A
 3 4  3 2
2 X  3A  B  3 



 1 2  1 2
 9 12  3 2  6 10






 3 6  1 2  2 4
1  6 10  3 5

X 

2  2 4  1 2
Matrix
 A matrix is an ordered rectangular array of numbers.
 A matrix with m rows and n columns has size m ☓ n.
 The entry in the ith row and jth column is denoted by aij.
Applied Example: Organizing Production Data
 The Acrosonic Company manufactures four different
loudspeaker systems at three separate locations.
 The company’s May output is as follows:
Model A
Model B
Model C
Model D
Location I
320
280
460
280
Location II
480
360
580
0
Location III
540
420
200
880
 If we agree to preserve the relative location of each entry
in the table, we can summarize the set of data as follows:
 320 280 460 280 
 480 360 580
0


 540 420 200 880 
Applied Example: Organizing Production Data
 We have Acrosonic’s May output expressed as a matrix:
 320 280 460 280 
P   480 360 580
0 
 540 420 200 880 
a. What is the size of the matrix P?
Solution
Matrix P has three rows and four columns and hence
has size 3 ☓ 4.
Applied Example: Organizing Production Data
 We have Acrosonic’s May output expressed as a matrix:
 320 280 460 280 
P   480 360 580
0 
 540 420 200 880 
b. Find a24 (the entry in row 2 and column 4 of the
matrix P) and give an interpretation of this number.
Solution
The required entry lies in row 2 and column 4, and is
the number 0. This means that no model D
loudspeaker system was manufactured at location II
in May.
Applied Example: Organizing Production Data
 We have Acrosonic’s May output expressed as a matrix:
 320 280 460 280 
P   480 360 580
0 
 540 420 200 880 
c. Find the sum of the entries that make up row 1 of P
and interpret the result.
Solution
The required sum is given by
320 + 280 + 460 + 280 = 1340
which gives the total number of loudspeaker systems
manufactured at location I in May as 1340 units.
Applied Example: Organizing Production Data
 We have Acrosonic’s May output expressed as a matrix:
 320 280 460 280 
P   480 360 580
0 
 540 420 200 880 
d. Find the sum of the entries that make up column 4 of
P and interpret the result.
Solution
The required sum is given by
280 + 0 + 880 = 1160
giving the output of Model D loudspeaker systems at
all locations in May as 1160 units.
Equality of Matrices
 Two matrices are equal if they have the same size
and their corresponding entries are equal.
Example
 Solve the following matrix equation for x, y, and z:
1
2

3 1 4 z 


y  1 2 2 1 2
x
Solution
 Since the corresponding elements of the two matrices must
be equal, we find that x = 4, z = 3, and y – 1 = 1, or y = 2.
Addition and Subtraction of Matrices
 If A and B are two matrices of the same size, then:
1. The sum A + B is the matrix obtained by adding
the corresponding entries in the two matrices.
2. The difference A – B is the matrix obtained by
subtracting the corresponding entries in B from
those in A.
Applied Example: Organizing Production Data
 The total output of Acrosonic for May is
Model A
Model B
Model C
Model D
Location I
320
280
460
280
Location II
480
360
580
0
Location III
540
420
200
880
 The total output of Acrosonic for June is
Model A
Model B
Model C
Model D
Location I
210
180
330
180
Location II
400
300
450
40
Location III
420
280
180
740
 Find the total output of the company for May and June.
Applied Example: Organizing Production Data
Solution
 Expressing the output for May and June as matrices:
✦ The total output of Acrosonic for May is
 320 280 460 280
A   480 360 580
0


 540 420 200 880
✦ The total output of Acrosonic for June is
 210 180 330 180
B   400 300 450 40


 420 280 180 740
Applied Example: Organizing Production Data
Solution
 The total output of the company for May and June is given
by the matrix
 320
A  B   480

 540
530
 880

960
280 460 280  210 180 330 180
360 580
0   400 300 450 40
 

420 200 880  420 280 180 740
460 790 460
660 1030
40

700 380 1620
Laws for Matrix Addition
 If A, B, and C are matrices of the same size, then
1. A + B = B + A
2. (A + B) + C = A + (B + C)
Commutative law
Associative law
Transpose of a Matrix
 If A is an m ☓ n matrix with elements aij,
then the transpose of A is the n ☓ m matrix
AT with elements aji.
Example
 Find the transpose of the matrix
 1 2 3
A  4 5 6


 7 8 9 
Solution
 The transpose of the matrix A is
1 4 7
AT   2 5 8 


 3 6 9 
Scalar Product
 If A is a matrix and c is a real number, then
the scalar product cA is the matrix obtained
by multiplying each entry of A by c.
Example
 Given
 3 4
A 


1
2


and
 3 2
B


1
2


find the matrix X that satisfies 2X + B = 3A
Solution
2 X  B  3A
2 X  3A  B
 3 4  3 2
 3




1
2

1
2

 

 9 12  3 2  6 10







2
4

3
6

1
2


 
 
1  6 10  3 5

X 

2  2 4  1 2
Applied Example: Production Planning
 The management of Acrosonic has decided to increase its
July production of loudspeaker systems by 10%
(over June output).
 Find a matrix giving the targeted production for July.
Solution
 We have seen that Acrosonic’s total output for June may
be represented by the matrix
 210 180 330 180
B   400 300 450 40


 420 280 180 740
Applied Example: Production Planning
 The management of Acrosonic has decided to increase its
July production of loudspeaker systems by 10%
(over June output).
 Find a matrix giving the targeted production for July.
Solution
 The required matrix is given by
 210 180 330 180 
(1.1) B  1.1  400 300 450 40 


 420 280 180 740 
 231 198 363 198
  440 330 495 44 


 462 308 198 814 
5.5
Multiplication of Matrices
a
a13 
a
A   11 12

a
a
a
23 
 21 22
 b11 b12 b13 b14 
B  b21 b22 b23 b24 


 b31 b32 b33 b34 
Same
Size of A (2 ☓ 3)
(3 ☓ 4) Size of B
(2 ☓ 4)
Size of AB
Multiplying a Row Matrix by a Column Matrix
 If we have a row matrix of size 1☓ n,
A  [a1 a2 a3  an ]
 And a column matrix of size n ☓ 1,
 b1 
b 
 2
B   b3 
 
 
bn 
 Then we may define the matrix product of A and B, written
AB, by
AB  [a1 a2
a3
 b1 
b 
 2
 an ] b3   a1b1  a2b2  a3b3    anbn
 
 
bn 
Example
 Let
A  [1 2 3 5]
and
 2
 3
B 
 0
 
 1
 Find the matrix product AB.
Solution
 2
 3
AB  [1 2 3 5]    (1)(2)  ( 2)(3)  (3)(0)  (5)( 1)  9
 0
 
 1
Dimensions Requirement
for Matrices Being Multiplied
 Note from the last example that for the multiplication to
be feasible, the number of columns of the row matrix A
must be equal to the number of rows of the column
matrix B.
Dimensions of the Product Matrix
 From last example, note that the product matrix AB has
size 1 ☓ 1.
 This has to do with the fact that we are multiplying a row
matrix with a column matrix.
 We can establish the dimensions of a product matrix
schematically:
Size of AB
(1 ☓ 1)
Size of A
(1 ☓ n)
(n ☓ 1)
Same
Size of B
Dimensions of the Product Matrix
 More generally, if A is a matrix of size m ☓ n and B is a
matrix of size n ☓ p, then the matrix product of A and B,
AB, is defined and is a matrix of size m ☓ p.
 Schematically:
Size of AB
(m ☓ p)
Size of A
(m ☓ n)
(n ☓ p)
Size of B
Same
 The number of columns of A must be the same as the
number of rows of B for the multiplication to be feasible.
Mechanics of Matrix Multiplication
 To see how to compute the product of a 2 ☓ 3 matrix A and
a 3 ☓ 4 matrix B, suppose
 a11 a12
A 
 a21 a22
a13 
a23 
 b11 b12
B  b21 b22

 b31 b32
b13
b23
b33
b14 
b24 

b34 
 From the schematic
Size of A (2 ☓ 3)
Size of AB
(2 ☓ 4)
(3 ☓ 4)
Size of B
Same
we see that the matrix product C = AB is feasible (since the
number of columns of A equals the number of rows of B) and
has size 2 ☓ 4.
Mechanics of Matrix Multiplication
 To see how to compute the product of a 2 ☓ 3 matrix A and
a 3 ☓ 4 matrix B, suppose
 a11 a12
A 
 a21 a22
 Thus,
 b11 b12
B  b21 b22

 b31 b32
a13 
a23 
 c11 c12
C
 c21 c22
c13
c23
b13
b23
b33
b14 
b24 

b34 
c14 
c24 
 To see how to calculate the entries of C consider entry c11:
c11  [a11 a12
 b11 
a13 ] b21   a11b11  a12b21  a13b31
 
 b31 
Mechanics of Matrix Multiplication
 To see how to compute the product of a 2 ☓ 3 matrix A and
a 3 ☓ 4 matrix B, suppose
 a11 a12
A 
 a21 a22
 Thus,
a13 
a23 
 c11 c12
C
 c21 c22
 b11 b12
B  b21 b22

 b31 b32
c13
c23
b13
b23
b33
b14 
b24 

b34 
c14 
c24 
 Now consider calculating the entry c12:
c12  [a11 a12
 b12 
a13 ] b22   a11b12  a12b22  a13b32
 
 b32 
Mechanics of Matrix Multiplication
 To see how to compute the product of a 2 ☓ 3 matrix A and
a 3 ☓ 4 matrix B, suppose
 a11 a12
A 
 a21 a22
 Thus,
a13 
a23 
 c11 c12
C
 c21 c22
 b11 b12
B  b21 b22

 b31 b32
c13
c23
b13
b23
b33
b14 
b24 

b34 
c14 
c24 
 Now consider calculating the entry c21:
c21  [a21 a22
 b11 
a23 ] b21   a21b11  a22b21  a23b31
 
 b31 
Mechanics of Matrix Multiplication
 To see how to compute the product of a 2 ☓ 3 matrix A and
a 3 ☓ 4 matrix B, suppose
 a11 a12
A 
 a21 a22
 Thus,
a13 
a23 
 c11 c12
C
 c21 c22
 b11 b12
B  b21 b22

 b31 b32
c13
c23
b13
b23
b33
c14 
c24 
 Other entries are computed in a similar manner.
b14 
b24 

b34 
Example
 Let
 3 1 4
A 


1
2
3


 1 3 3
B   4 1 2 


 2 4 1 
 Compute AB.
Solution
 Since the number of columns of A is equal to the number
of rows of B, the matrix product C = AB is defined.
 The size of C is 2 ☓ 3.
Example
 Let
 3 1 4
A 


1
2
3


 1 3 3
B   4 1 2 


 2 4 1 
 Compute AB.
Solution
 Thus,
 1 3 3
 c11 c12
 3 1 4 

C  AB  
4 1 2  


 1 2 3 2 4 1   c21 c22


c13 
c23 
 Calculate all entries for C:
1
c11  [3 1 4]  4   (3)(1)  (1)(4)  (4)(2)  15
 
 2 
Example
 Let
 3 1 4
A 


1
2
3


 1 3 3
B   4 1 2 


 2 4 1 
 Compute AB.
Solution
 Thus,
 1 3 3
 15 c12
 3 1 4 

C  AB  
4 1 2  


 1 2 3 2 4 1   c21 c22


c13 
c23 
 Calculate all entries for C:
 3
c12  [3 1 4]  1  (3)(3)  (1)( 1)  (4)(4)  24
 
 4 
Example
 Let
 3 1 4
A 


1
2
3


 1 3 3
B   4 1 2 


 2 4 1 
 Compute AB.
Solution
 Thus,
 1 3 3
 15 24 c13 
 3 1 4 

C  AB  
4 1 2  




c
c
c

1
2
3

 2 4 1
21
22
23 

 
 Calculate all entries for C:
 3
c13  [3 1 4]  2   (3)( 3)  (1)(2)  (4)(1)  3
 
 1
Example
 Let
 3 1 4
A 


1
2
3


 1 3 3
B   4 1 2 


 2 4 1 
 Compute AB.
Solution
 Thus,
 1 3 3
 15 24 3
 3 1 4 

C  AB  
4 1 2  




c
c
c

1
2
3

 2 4 1
21
22
23 

 
 Calculate all entries for C:
 1
c21  [ 1 2 3]  4   ( 1)(1)  (2)(4)  (3)(2)  13
 
 2 
Example
 3 1 4
A 


1
2
3


 Let
 1 3 3
B   4 1 2 


 2 4 1 
 Compute AB.
Solution
 Thus,
 1 3 3
15 24 3
 3 1 4 

C  AB  
4 1 2  




13
c
c

1
2
3

 2 4 1
22
23 

 
 Calculate all entries for C:
 3
c22  [ 1 2 3]  1  ( 1)(3)  (2)( 1)  (3)(4)  7
 
 4 
Example
 3 1 4
A 


1
2
3


 Let
 1 3 3
B   4 1 2 


 2 4 1 
 Compute AB.
Solution
 Thus,
 1 3 3
15 24 3
 3 1 4 

C  AB  
4 1 2  




13
7
c

1
2
3

 2 4 1
23 

 
 Calculate all entries for C:
 3
c23  [1 2 3]  2   ( 1)( 3)  (2)(2)  (3)(1)  10
 
 1
Example
 Let
 3 1 4
A 


1
2
3


 1 3 3
B   4 1 2 


 2 4 1 
 Compute AB.
Solution
 Thus,
 1 3 3
 3 1 4 
15 24 3

C  AB  
4 1 2  





1
2
3
13
7
10

 2 4 1




Laws for Matrix Multiplication
 If the products and sums are defined for the
matrices A, B, and C, then
1. (AB)C = A(BC)
Associative law
2. A(B + C) = AB + AC
Distributive law
Identity Matrix
 The identity matrix of size n is given by
1 0  0
0 1  0
 n rows
In  




0
0

1


n columns
Properties of the Identity Matrix
 The identity matrix has the properties that
✦ In A = A for any n ☓ r matrix A.
✦ BIn = B for any s ☓ n matrix B.
✦ In particular, if A is a square matrix of
size n, then
I n A  AI n  A
Example
 Let
 1 3 1
A   4 3 2 


 1 0 1 
 Then
1 0 0  1 3 1   1 3 1 
I 3 A  0 1 0  4 3 2    4 3 2   A


 

0 0 1   1 0 1   1 0 1 
 1 3 1  1 0 0   1 3 1 
AI 3   4 3 2  0 1 0   4 3 2   A


 

 1 0 1  0 0 1   1 0 1 
 So, I3A = AI3 = A.
Matrix Representation
 A system of linear equations can be expressed in the form
of an equation of matrices. Consider the system
2x  4 y  z  6
3x  6 y  5z  1
x  3y  7z  0
 The coefficients on the left-hand side of the equation can
be expressed as matrix A below, the variables as matrix X,
and the constants on right-hand side of the equation as
matrix B:
1
 2 4
A   3 6 5


 1 3 7 
 x
X   y
 
 z 
 6
B   1
 
 0
Matrix Representation
 A system of linear equations can be expressed in the form
of an equation of matrices. Consider the system
2x  4 y  z  6
3x  6 y  5z  1
x  3y  7z  0
 The matrix representation of the system of linear
equations is given by AX = B, or
1  x   6
 2 4
 3 6 5  y    1

   
 1 3 7   z   0
Matrix Representation
 A system of linear equations can be expressed in the form
of an equation of matrices. Consider the system
2x  4 y  z  6
3x  6 y  5z  1
x  3y  7z  0
 To confirm this, we can multiply the two matrices on the
left-hand side of the equation, obtaining
 2 x  4 y  z   6
 3 x  6 y  5 z    1

  
 x  3 y  7 z   0
which, by matrix equality, is easily seen to be equivalent to
the given system of linear equations.
5.6
The Inverse of a Square Matrix
x
 y 
 
 z 
 (3)(1)  ( 1)(2)  ( 1)( 1)   2 
( 4)(1)  (2)(2)  (1)( 1)    1

  
 ( 1)(1)  (0)(2)  (1)( 1)   2 
Inverse of a Matrix
 Let A be a square matrix of size n.
 A square matrix A–1 of size n such that
A1 A  AA1  I n
is called the inverse of A.
 Not every matrix has an inverse.
✦ A square matrix that has an inverse is
said to be nonsingular.
✦ A square matrix that does not have an
inverse is said to be singular.
Example: A Nonsingular Matrix
1
 2
1 2
1
 The matrix A  
has a matrix A   3
1


3
4
2
 2


as its inverse.
 This can be demonstrated by multiplying them:
1 1 0
1 2  2
AA  

I



3
1
3 4   2  2  0 1 
1
1 1 2 1 0
 2
A A  3

I


1
 2  2  3 4  0 1
1
Example: A Singular Matrix
0 1
 The matrix B  

0
0


does not have an inverse.
a b 
 If B had an inverse given by B  
where

c d 
1
a, b, c, and d are some appropriate numbers, then by
definition of an inverse we would have BB–1 = I.
 That is
 0 1   a b  1 0 
0 0  c d   0 1 


 

 c d  1 0 
0 0   0 1 

 

implying that 0 = 1, which is impossible!
Finding the Inverse of a Square Matrix
 Given the n ☓ n matrix A:
1. Adjoin the n ☓ n identity matrix I to obtain
the augmented matrix [A | I ].
2. Use a sequence of row operations to reduce
[A | I ] to the form [I | B] if possible.
 Then the matrix B is the inverse of A.
Example
 Find the inverse of the matrix
Solution
 We form the augmented matrix
2 1 1
A   3 2 1


 2 1 2 
 2 1 1 1 0 0


3
2
1
0
1
0


 2 1 2 0 0 1
Example
 Find the inverse of the matrix
2 1 1
A   3 2 1


 2 1 2 
Solution
 And use the Gauss-Jordan elimination method to reduce it
to the form [I | B]:
Toggle slides
back and forth to
compare before
and changes
 2 1 1 1 0 0


3
2
1
0
1
0


 2 1 2 0 0 1
R1  R2
Example
 Find the inverse of the matrix
2 1 1
A   3 2 1


 2 1 2 
Solution
 And use the Gauss-Jordan elimination method to reduce it
to the form [I | B]:
Toggle slides
back and forth to
compare before
and changes
 1 1 0 1 1 0  R1

 R  R
3R
3
2
1
0
1
0
23

 12
 2 1 2 0 0 1 R3  2 R1
Example
 Find the inverse of the matrix
2 1 1
A   3 2 1


 2 1 2 
Solution
 And use the Gauss-Jordan elimination method to reduce it
to the form [I | B]:
Toggle slides
back and forth to
compare before
and changes
 1 1 0 1 1 0 R1R1R2

 R R3R
0

1
1
3

2
0

 2 2 3
0 1 2 2 2 1 RR332RR21
Example
 Find the inverse of the matrix
2 1 1
A   3 2 1


 2 1 2 
Solution
 And use the Gauss-Jordan elimination method to reduce it
to the form [I | B]:
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 1 0 1 2 1 0 RR1  RR2

 1 R 3
2
0 1 1 3 2 0 R2  R
3
0 0 1 1 0 1 R3  R2
Example
 Find the inverse of the matrix
2 1 1
A   3 2 1


 2 1 2 
Solution
 And use the Gauss-Jordan elimination method to reduce it
to the form [I | B]:
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3 1 1
1 0 0

 R1  R3
0 1 0 4 2 1 R2  R3
0 0 1 1 0 1
In
B
Example
 Find the inverse of the matrix
2 1 1
A   3 2 1


 2 1 2 
Solution
 Thus, the inverse of A is the matrix
 3 1 1
A1   4 2 1


 1 0 1
A Formula for the Inverse of a 2 ☓ 2 Matrix
 Let
a b 
A

c
d


 Suppose D = ad – bc is not equal to zero.
 Then A–1 exists and is given by
1  d b
A  
D  c a 
1
Example
 Find the inverse of
1 2
A 

3
4


Solution
 We first identify a, b, c, and d as being 1, 2, 3, and 4
respectively.
 We then compute
D = ad – bc = (1)(4) – (2)(3) = 4 – 6 = – 2
Example
 Find the inverse of
1 2
A 

3
4


Solution
 Next, we substitute the values 1, 2, 3, and 4 instead of
a, b, c, and d, respectively, in the formula matrix
 d b 
 c a 


to obtain the matrix
 4 2
 3 1


Example
 Find the inverse of
1 2
A 

3
4


Solution
 Finally, multiplying this matrix by 1/D, we obtain
1
1  d b 1  4 2  2
A  

 3



D  c a  2  3 1  2  12 
1
Using Inverses to Solve Systems of Equations
 If AX = B is a linear system of n equations
in n unknowns and if A–1 exists, then
X = A–1B
is the unique solution of the system.
Example
 Solve the system of linear equations
2x  y  z  1
3x  2 y  z  2
2 x  y  2 z  1
Solution
 Write the system of equations in the form AX = B where
2 1 1
A   3 2 1


 2 1 2 
 x
X   y
 
 z 
 1
B   2
 
 1
Example
 Solve the system of linear equations
2x  y  z  1
3x  2 y  z  2
2 x  y  2 z  1
Solution
 Find the inverse matrix of A:
2 1 1
A   3 2 1


 2 1 2 
 x
X   y
 
 z 
 1
B   2
 
 1
 3 1 1
A1   4 2 1 
 1 0 1 
Example
 Solve the system of linear equations
2x  y  z  1
3x  2 y  z  2
2 x  y  2 z  1
Solution
 Finally, we write the matrix equation X = A–1B and multiply:
x
 y 
 
 z 
 3 1 1  1
 4 2 1   2 

 
 1 0 1   1
Example
 Solve the system of linear equations
2x  y  z  1
3x  2 y  z  2
2 x  y  2 z  1
Solution
 Finally, we write the matrix equation X = A–1B and multiply:
x
 y 
 
 z 
 (3)(1)  ( 1)(2)  ( 1)( 1)   2 
( 4)(1)  (2)(2)  (1)( 1)    1

  
 ( 1)(1)  (0)(2)  (1)( 1)   2 
 Thus, the solution is x = 2, y = –1, and z = –2.
End of
Chapter