Download Page 1 Solutions to Section 1.2 Homework Problems S. F.

Solutions to Section 1.2 Homework Problems
S. F. Ellermeyer
Practice Problem 1. The linear system whose augmented matrix is
1 3 5 0
0
1
1
3
is equivalent to the linear system whose augmented matrix is
1 0 2 9
0 1
1
.
3
Since the latter matrix has echelon form, it can be seen that these systems
are consistent and have one free variable. The system corresponding to the
latter matrix is
x 1  2x 3  9
x2  x3  3
and the general solution of this system (and hence of the original system) is
x 1  9  2x 3
x2  3  x3
x 3 is free.
1.
a.
b.
c.
d.
Reduced Echelon Form
Reduced Echelon Form
Not Echelon Form
Echelon Form but not Reduced Echelon Form
3.
1
2 3 4
4
5 6 7
6
7 8 9
1


2
3
4
1
0 3 6 9

2
3
4
0 3
6
9
0 5 10 15
6
7
8
9
1
2
3
4
0
1
2
3
0 5 10 15
1 2 3 4

0 1 2 3
0 0 0 0

1
0 1 2
0
1
2
3
0
0
0
0
The pivot columns are columns 1 and 2.
5.
The possible echelon forms of a non–zero 2  2 matrix are
 
0 
1
,
 
0 0
0 
and
.
0 0
 
6.
The possible echelon forms of a non–zero 3  2 matrix are
0 0
,
0 0
 
0 
0
 
,
0
0
0 
, and

.
0 0
0 
0 0
7.
1 3 4 7
3 9 7 6
~
1 3
4
7
1 3 4 7

0 0 5 15
1 3 0 5

0 0 1 3
0 0 1
3
.
The general solution to the systems represented by these augmented
matrices is
x 1  5  3x 2
x 2  free
x 3  3.
9.
1
6
5
1 2
7
6
0
~
1 0 5 4
0 1 6 5
.
The general solution to the systems represented by these augmented
matrices is
x 1  4  5x 3
x 2  5  6x 3
x 3  free.
11.
3
4
2
0
9 12 6 0
6
8
~
4 0
1  43
2
3
0
0
0 0
0
0
0 0
0
.
The general solution to the systems represented by these augmented
matrices is
x1  4 x2  2 x3
3
3
x 2  free
x 3  free
2
13.
1 3 0 1
0
2
0
1
0
0
4
1
0
0
0
1
9
4
0
0
0
0
0
0
1 0 0 0 3 5
0 1 0 0 4 1
~
0 0 0 1
9
4
0 0 0 0
0
0
.
The general solution to the systems represented by these augmented
matrices is
x 1  5  3x 5
x 2  1  4x 5
x 3  free
x 4  4  9x 5
x 5  free.
15.
a.
b.
System is consistent and solution is unique.
System is inconsistent.
a.
b.
System is consistent and solution is unique.
System is consistent and there are infinitely many solutions (in
other words, not a unique solution).
16.
17.
2 3 h
2 3
~
4 6 7
h
0 0 2h  7
.
The systems represented by these augmented matrices are consistent if and
only is 2h  7  0 (that is h  72 ).
19. The augmented matrix for the system
x 1  hx 2  2
4x 1  8x 2  k
is
1 h 2
4 8 k
~
1
h
2
0 4h  8 8  k
.
If h  2 and k is any value except 8, then the systems represented
by these augmented matrices are inconsistent.
b. If h is any value but 2 and k is any value, then the systems
represented by these augmented matrices have a unique solution.
c. If h  2 and k  8, then the systems represented by these
augmented matrices have infinitely many solutions.
a.
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21.
In some cases, a matrix may be reduced to more than one matrix in
reduced echelon form, using different sequences of row operations.
False.
For any given matrix, A, there is exactly one matrix with reduced
echelon form that is equivalent to A. (See page 15 of the textbook.)
b. The row reduction algorithm applies only to augmented matrices for
a linear system. False.
The row reduction algorithm applies to any matrix. (See page 14 of
the textbook.)
c. A basic variable in a linear system is a variable that corresponds to
a pivot column in the coefficient matrix. True.
See page 20 of the textbook.
d. Finding a parametric description of the solution set of a linear
system is the same thing as solving the system. True.
See page 22 of the textbook.
e. If one row in an echelon form of an augmented matrix is
0 0 0 5 0 , then the associated linear system is inconsistent.
a.
False. Or better to say not necessarily true.
The system might be inconsistent because there might be another
row in the echelon form that looks like
0 0 0 0 b
(where b  0), but we cannot conclude that the system is
inconsistent just based on the knowledge that there is a row
0 0 0 5 0
.
22.
The echelon form of a matrix is unique. False.
The reduced echelon form of a matrix is unique.
b. The pivot positions in a matrix depend on whether row interchanges
are used in the row reduction process. False.
A pivot position in a matrix, A , is a position in the matrix that
corresponds to a row–leading 1 in the reduced echelon form of A.
Since the reduced echelon form of A is unique, the pivot positions
are uniquely determined and do not depend on whether or not row
interchanges are performed in the reduction process.
c. Reducing a matrix to echelon form is called the forward phase of
the row reduction process. True.
See page 20 of the textbook.
d. Whenever a system has free variables, the solution set contains
many solutions. True.
See Theorem 2 on page 24 of the textbook.
a.
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A general solution of a system is an explicit description of all
solutions of the system. True.
See page 21 of the textbook.
23. Suppose a 3  5 coefficient matrix of a linear system has three pivot columns.
Is the system consistent? Why or why not?
The system is consistent because each row of the coefficient matrix has a
pivot. The augmented matrix for this system is a 3  6 matrix and it is not
possible that the 6th column in this matrix is a pivot column because the
maximum possible number (3) of pivots are already “used up” in the
coefficient matrix.
25. If the coefficient matrix of a linear system has a pivot position in every row,
then the reduced echelon form of this coefficient matrix has no row consisting
entirely of zeros. This means that the reduced echelon form of the
augmented matrix for this system cannot have a row that looks like
e.
0 0  0 b
.
Thus, the system must be consistent.
27. If a linear system is consistent, then the solution is unique if and only if every
column of the coefficient matrix of the system is a pivot column.
29. A linear system with fewer equations than unknowns has a coefficient matrix
with fewer rows than columns. This means that not every column of this
coefficient matrix can be a pivot column. Thus, if the system is consistent,
then based on what was said in problem 27, there must be an infinite number
of solutions.
31. Yes. A linear system with more equations than unknowns can be consistent.
Here is an example:
x1  x2  3
2x 1  2x 2  6
4x 1  8x 2  0.
Note that
1
1
3
2
2
6
4 8 0
1 0 2
~
0 1 1
0 0 0
indicating consistency since the last column of the augmented matrix is not a
pivot column.
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