Solution Manual For Elementary Linear Algebra Applications Version 11th Edition by Anton

1.
1 Introduction to Systems of Linear Equations 1
CHAPTER 1: SYSTEMS OF LINEAR EQUATIONS AND MATRICES
1.1 Introduction to Systems of Linear Equations
1. (a) This is a linear equation in , , and .

(b) This is not a linear equation in , , and because of the term .
(c) We can rewrite this equation in the form therefore it is a linear equation in
, , and .
(d) This is not a linear equation in , , and because of the term .
(e) This is not a linear equation in , , and because of the term .
(f) This is a linear equation in , , and .

2. (a) This is a linear equation in and .
(b) This is not a linear equation in and because of the terms and .
(c) This is a linear equation in and .
(d) This is not a linear equation in and because of the term .
(e) This is not a linear equation in and because of the term .

(f) We can rewrite this equation in the form thus it is a linear equation in and .
3. (a)
(b)
(c)
4. (a) (b) (c)

2 Chapter 1: Systems of Linear Equations and Matrices
5. (a) (b)
6. (a) (b)
7. (a) (b) (c)
8. (a) (b) (c)
9. The values in (a), (d), and (e) satisfy all three equations – these 3-tuples are solutions of the system.
The 3-tuples in (b) and (c) are not solutions of the system.
10. The values in (b), (d), and (e) satisfy all three equations – these 3-tuples are solutions of the system.
The 3-tuples in (a) and (c) are not solutions of the system.
11. (a) We can eliminate from the second equation by adding times the first equation to the
second. This yields the system
The second equation is contradictory, so the original system has no solutions. The lines
represented by the equations in that system have no points of intersection (the lines are
parallel and distinct).
(b) We can eliminate from the second equation by adding times the first equation to the
The second equation does not impose any restriction on and therefore we can omit it. The
lines represented by the original system have infinitely many points of intersection. Solving the
1.1 Introduction to Systems of Linear Equations 3
first equation for we obtain . This allows us to represent the solution using
parametric equations
where the parameter is an arbitrary real number.

(c) We can eliminate from the second equation by adding times the first equation to the
From the second equation we obtain . Substituting for into the first equation
results in . Therefore, the original system has the unique solution
The represented by the equations in that system have one point of intersection: .
12. We can eliminate from the second equation by adding times the first equation to the second.
This yields the system
If (i.e., ) then the second equation imposes no restriction on and ;

consequently, the system has infinitely many solutions.
If (i.e., ) then the second equation becomes contradictory thus the system has no
solutions.
There are no values of and for which the system has one solution.
13. (a) Solving the equation for we obtain therefore the solution set of the original
equation can be described by the parametric equations
(b) Solving the equation for we obtain therefore the solution set of the
original equation can be described by the parametric equations
where the parameters and are arbitrary real numbers.
(c) Solving the equation for we obtain therefore the solution set of
the original equation can be described by the parametric equations
where the parameters , , and are arbitrary real numbers.
(d) Solving the equation for we obtain therefore the solution set of the
where the parameters , , , and are arbitrary real numbers.

14. (a) Solving the equation for we obtain therefore the solution set of the original

(b) Solving the equation for we obtain therefore the solution set of the
(c) Solving the equation for we obtain therefore the solution set of
the original equation can be described by the parametric equations
where the parameters , , and are arbitrary real numbers.

(d) Solving the equation for we obtain therefore the solution set of the
where the parameters , , , and are arbitrary real numbers.

15. (a) We can eliminate from the second equation by adding times the first equation to the
The second equation does not impose any restriction on and therefore we can omit it.
Solving the first equation for we obtain . This allows us to represent the solution
using parametric equations

(b) We can see that the second and the third equation are multiples of the first: adding times the
first equation to the second, then adding the first equation to the third yields the system
The last two equations do not impose any restriction on the unknowns therefore we can omit
them. Solving the first equation for we obtain . This allows us to
represent the solution using parametric equations

16. (a) We can eliminate from the first equation by adding times the second equation to the
first. This yields the system
The first equation does not impose any restriction on and therefore we can omit it. Solving
the second equation for we obtain . This allows us to represent the solution
using parametric equations

(b) We can see that the second and the third equation are multiples of the first: adding times the
first equation to the second, then adding times the first equation to the third yields the system
The last two equations do not impose any restriction on the unknowns therefore we can omit
them. Solving the first equation for we obtain . This allows us to represent
the solution using parametric equations
17. (a) Add times the second row to the first to obtain .
(b) Add the third row to the first to obtain

(another solution: interchange the first row and the third row to obtain ).
18. (a) Multiply the first row by to obtain .
(b) Add the third row to the first to obtain
(another solution: add times the second row to the first to obtain ).
19. (a) Add times the first row to the second to obtain which corresponds to the
system
If then the second equation becomes , which is contradictory thus the system
becomes inconsistent.
If then we can solve the second equation for and proceed to substitute this value into
the first equation and solve for .
Consequently, for all values of the given augmented matrix corresponds to a consistent
linear system.
(b) Add times the first row to the second to obtain which corresponds to the
system
If then the second equation becomes , which does not impose any restriction on
and therefore we can omit it and proceed to determine the solution set using the first
equation. There are infinitely many solutions in this set.
If then the second equation yields and the first equation becomes .
Consequently, for all values of the given augmented matrix corresponds to a consistent linear
system.
20. (a) Add times the first row to the second to obtain which corresponds to the
system
If then the second equation is contradictory thus the system becomes inconsistent.
Consequently, the given augmented matrix corresponds to a consistent linear system only when
.
(b) Add the first row to the second to obtain which corresponds to the system
If then the second equation yields and the first equation becomes .
Consequently, for all values of the given augmented matrix corresponds to a consistent linear
system.
21. Substituting the coordinates of the first point into the equation of the curve we obtain
Repeating this for the other two points and rearranging the three equations yields
This is a linear system in the unknowns , , and . Its augmented matrix is .
23. Solving the first equation for we obtain therefore the solution set of the original

Substituting these into the second equation yields
which can be rewritten as

This equation must hold true for all real values , which requires that the coefficients associated with
the same power of on both sides must be equal. Consequently, and .
24. (a) The system has no solutions if either
 at least two of the three lines are parallel and distinct or
 each pair of lines intersects at a different point (without any lines being parallel)
(b) The system has exactly one solution if either
 two lines coincide and the third one intersects them or
 all three lines intersect at a single point (without any lines being parallel)
(c) The system has infinitely many solutions if all three lines coincide.
25.
26. We set up the linear system as discussed in Exercise 21:
i.e.
One solution is expected, since exactly one parabola passes through any three given points ,
, if , , and are distinct.
27.
True-False Exercises
(a) True. is a solution.
(b) False. Only multiplication by a nonzero constant is a valid elementary row operation.
(c) True. If then the system has infinitely many solutions; otherwise the system is inconsistent.
(d) True. According to the definition, is a linear equation if the 's are not
all zero. Let us assume . The values of all 's except for can be set to be arbitrary parameters,
and the equation can be used to express in terms of those parameters.
(e) False. E.g. if the equations are all homogeneous then the system must be consistent. (See True-False
Exercise (a) above.)
(f) False. If then the new system has the same solution set as the original one.
(g) True. Adding times one row to another amounts to the same thing as subtracting one row from
another.
(h) False. The second row corresponds to the equation , which is contradictory.
1.2 Gaussian Elimination 9
1.2 Gaussian Elimination
1. (a) This matrix has properties 1-4. It is in reduced row echelon form, therefore it is also in row
echelon form.
(b) This matrix has properties 1-4. It is in reduced row echelon form, therefore it is also in row
echelon form.
(c) This matrix has properties 1-4. It is in reduced row echelon form, therefore it is also in row
echelon form.
(d) This matrix has properties 1-4. It is in reduced row echelon form, therefore it is also in row
echelon form.
(e) This matrix has properties 1-4. It is in reduced row echelon form, therefore it is also in row
echelon form.
(f) This matrix has properties 1-4. It is in reduced row echelon form, therefore it is also in row
echelon form.
(g) This matrix has properties 1-3 but does not have property 4: the second column contains a
leading 1 and a nonzero number ( ) above it. The matrix is in row echelon form but not
reduced row echelon form.
2. (a) This matrix has properties 1-3 but does not have property 4: the second column contains a
leading 1 and a nonzero number (2) above it. The matrix is in row echelon form but not reduced
row echelon form.
(b) This matrix does not have property 1 since its first nonzero number in the third row (2) is not a
1. The matrix is not in row echelon form, therefore it is not in reduced row echelon form either.
(c) This matrix has properties 1-3 but does not have property 4: the third column contains a
row echelon form.
(d) This matrix has properties 1-3 but does not have property 4: the second column contains a
row echelon form.
(e) This matrix does not have property 2 since the row that consists entirely of zeros is not at the
bottom of the matrix. The matrix is not in row echelon form, therefore it is not in reduced row
echelon form either.
(f) This matrix does not have property 3 since the leading 1 in the second row is directly below the
leading 1 in the first (instead of being farther to the right). The matrix is not in row echelon
form, therefore it is not in reduced row echelon form either.
(g) This matrix has properties 1-4. It is in reduced row echelon form, therefore it is also in row
echelon form.
3. (a) The linear system
can be rewritten as
and solved by back-substitution:
therefore the original linear system has a unique solution: , , .

(b) The linear system
can be rewritten as
Let . Then
therefore the original linear system has infinitely many solutions:

, , ,
where is an arbitrary value.
(c) The linear system
can be rewritten: , , .
Let and . Then
therefore the original linear system has infinitely many solutions:
where and are arbitrary values.

(d) The system is inconsistent since the third row of the augmented matrix corresponds to the
equation
4. (a) A unique solution: , , .

(b) Infinitely many solutions: , , , where is an arbitrary

value.
(c) Infinitely many solutions: , , , , where
and are arbitrary values.
(d) The system is inconsistent since the third row of the augmented matrix corresponds to the
equation
5. The augmented matrix for the system.
The first row was added to the second row.
times the first row was added to the third row.
The second row was multiplied by .
10 times the second row was added to the third row.
The third row was multiplied by .
The system of equations corresponding to this augmented matrix in row echelon form is
and can be rewritten as
Back-substitution yields
The linear system has a unique solution: , , .
The first row was multiplied by .

times the first row was added to the second row.
Solve the equations for the leading variables
then substitute the second equation into the first
If we assign an arbitrary value , the general solution is given by the formulas
The first row was added to the third row.

times the first row was added to the fourth row.
times the second row was added to the third row.
times the second row was added to the fourth row.
If we assign and the arbitrary values and , respectively, the general solution is given by the
formulas
The first and second rows were interchanged.

The system of equations corresponding to this augmented matrix in row echelon form
is clearly inconsistent.
5 times the third row was added to the second row.

times the third row was added to the first row.
times the second row was added to the first row.
The linear system has a unique solution: , , .
Infinitely many solutions: , , where is an arbitrary value.

the first row was added to the third row.
the second row was added to the first row.
formulas

times the third row was added to the second row.
The last row corresponds to the equation
therefore the system is inconsistent.

(Note: this was already evident after the fifth elementary row operation.)
13. Since the number of unknowns (4) exceeds the number of equations (3), it follows from Theorem
1.2.2 that this system has infinitely many solutions. Those include the trivial solution and infinitely
many nontrivial solutions.
14. The system does not have nontrivial solutions.
(The third equation requires , which substituted into the second equation yields Both
of these substituted into the first equation result in .)
15. We present two different solutions.

Solution I uses Gauss-Jordan elimination
The augmented matrix for the system.
The third row was added to the second row

and times the third row was added to the first row
Unique solution: , , .
Solution II. This time, we shall choose the order of the elementary row operations differently in order
to avoid introducing fractions into the computation. (Since every matrix has a unique reduced row
echelon form, the exact sequence of elementary row operations being used does not matter – see part
1 of the discussion “Some Facts About Echelon Forms” on p. 21)
The first and second rows were interchanged

(to avoid introducing fractions into the first row).
The third row was added to the second row.
16. We present two different solutions.
Solution I uses Gauss-Jordan elimination

times the third row was added to the second row

and times the third row was added to the first row
Solution II. This time, we shall choose the order of the elementary row operations differently in order
to avoid introducing fractions into the computation. (Since every matrix has a unique reduced row
echelon form, the exact sequence of elementary row operations being used does not matter – see part
1 of the discussion “Some Facts About Echelon Forms” on p. 21)
The first and third rows were interchanged

The second row was added to the third row.

If we assign and the arbitrary values and , respectively, the general solution is given by
the formulas
.
(Note that fractions in the solution could be avoided if we assigned instead, which along with
would yield , , , .)

times the first row was added to the third row

and times the first row was added to the fourth row.
times the second row was added to the third row

and the second row was added to the fourth row.
formulas

and times the first row was added to the fourth row.

times the second row was added to the third and

times the third row was added to the fourth row.
times the third row was added to the second and

If we assign an arbitrary value the general solution is given by the formulas

.
times the first row was added to the second row,

times the first row was added to the fourth row,
and times the first row was added to the fifth row.
times the second row was added to the third row,

times the second row was added to the fourth row,
and times the second row was added to the fifth row.

times the third row was added to the fourth row

and times the third row was added to the fifth row.
The fourth row was multiplied by .
times the fourth row was added to the fifth row.
The augmented matrix in row echelon form corresponds to the system
Using back-substitution, we obtain the unique solution of this system

.
The first and second rows were interchanged


times the first row was added to the third row,
and times the first row was added to the fourth.
times the second row was added to the third row and
The fourth row was added to the third row,

times the fourth row was added to the second,
and times the fourth row was added to the first.
times the third row was added to the second row,

and times the third row was added to the first row.
Unique solution: , , , .
The first and third rows were interchanged.

The first row was added to the second row

and times the first row was added to the last row.
The second and third rows were interchanged.
If we assign and the arbitrary values and , respectively, the general solution is given by
the formulas
.
23. (a) The system is consistent; it has a unique solution (back-substitution can be used to solve for all
three unknowns).
(b) The system is consistent; it has infinitely many solutions (the third unknown can be assigned an
arbitrary value , then back-substitution can be used to solve for the first two unknowns).
(c) The system is inconsistent since the third equation is contradictory.
(d) There is insufficient information to decide whether the system is consistent as illustrated by
these examples:
 For the system is consistent with infinitely many solutions.

 For the system is inconsistent (the matrix can be reduced to ).
24. (a) The system is consistent; it has a unique solution (back-substitution can be used to solve for all
three unknowns).
(b) The system is consistent; it has a unique solution (solve the first equation for the first unknown,
then proceed to solve the second equation for the second unknown and solve the third equation
last.)
(c) The system is inconsistent (adding times the first row to the second yields ;
the second equation is contradictory).

(d) There is insufficient information to decide whether the system is consistent as illustrated by
these examples:
 For the system is consistent with infinitely many solutions.
 For the system is inconsistent (the matrix can be reduced to ).
times the first row was added to the second row

and times the first row was added to the third row.
The system has no solutions when (since the third row of our last matrix would then
correspond to a contradictory equation ).
The system has infinitely many solutions when (since the third row of our last matrix would
then correspond to the equation ).
For all remaining values of (i.e., and ) the system has exactly one solution.


The system has no solutions when or (since the third row of our last matrix would
then correspond to a contradictory equation).
For all remaining values of (i.e., and ) the system has exactly one solution.
There is no value of for which this system has infinitely many solutions.
If then the linear system is consistent. Otherwise (if ) it is

inconsistent.
The first row was added to the second row and

If then the linear system is consistent. Otherwise (if ) it is

inconsistent.

The system has exactly one solution: and .

The system has exactly one solution: , , and .
31. Adding times the first row to the second yields a matrix in row echelon form .
Adding times its second row to the first results in , which is also in row echelon form.
32.

33. We begin by substituting , , and so that the system becomes


and the first row was added to the third row.
times the third row was added to the second row and
This system has exactly one solution

On the interval , the equation has three solutions: , , and .
On the interval , the equation has two solutions: and .
On the interval , the equation has three solutions: , , and .

Overall, solutions can be obtained by combining the values of , , and listed
above: , etc.



The second row was added to the first row.
This system has exactly one solution

The only angles and that satisfy the inequalities , , and the
equations
are and .

The second and third rows were interchanged

(to avoid introducing fractions into the second row).

We obtain

and the first row was added to the third row.
Using back-substitution, we obtain
37. Each point on the curve yields an equation, therefore we have a system of four equations
e uation correspondin to


and times the first row was added to the third.
times the fourth row was added to the third row,

times the fourth row was added to the second row,
and times the fourth row was added to the first.
The linear system has a unique solution: , , , . These are the coefficient
values required for the curve to pass through the four given points.
38. Each point on the curve yields an equation, therefore we have a system of three equations
The augmented matrix of this system has the reduced row echelon form
(For instance, letting the free variable have the value yields , , and .)
39. Since the homogeneous system has only the trivial solution, its augmented matrix must be possible to
reduce via a sequence of elementary row operations to the reduced row echelon form .
Applying the same sequence of elementary row operations to the augmented matrix of the
nonhomogeneous system yields the reduced row echelon form where , , and are
some real numbers. Therefore, the nonhomogeneous system has one solution.
40. (a) 3 (this will be the number of leading 1's if the matrix has no rows of zeros)
(b) 5 (if all entries in are 0)
(c) 2 (this will be the number of rows of zeros if each column contains a leading 1)
41. (a) There are eight possible reduced row echelon forms:
, , , , , , , and
where and can be any real numbers.

(b) There are sixteen possible reduced row echelon forms:
, , , , , ,
, , , , , ,
, , , and .
where , , , and can be any real numbers.

42. (a) Either the three lines properly intersect at the origin, or two of them completely overlap and the
other one intersects them at the origin.
(b) All three lines completely overlap one another.
43. (a) We consider two possible cases: (i) , and (ii) .
(i) If then the assumption implies that and . Gauss-Jordan
elimination yields
We assumed
The rows were interchanged.
The first row was multiplied by and

the second row was multiplied by (Note that
(ii) If then we perform Gauss-Jordan elimination as follows:

(Note that both and are nonzero.)
In both cases ( as well as ) we established that the reduced row echelon form of
is provided that .
(b) Applying the same elementary row operation steps as in part (a) the augmented matrix
will be transformed to a matrix in reduced row echelon form where
and are some real numbers. We conclude that the given linear system has exactly one
solution: , .
(a) True. A matrix in reduced row echelon form has all properties required for the row echelon form.
(b) False. For instance, interchanging the rows of yields a matrix that is not in row echelon form.
(c) False. See Exercise 31.

(d) True. In a reduced row echelon form, the number of nonzero rows equals to the number of leading
1's. The result follows from Theorem 1.2.1.
(e) True. This is implied by the third property of a row echelon form (see p. 11).
(f) False. Nonzero entries are permitted above the leading 1's in a row echelon form.
(g) True. In a reduced row echelon form, the number of nonzero rows equals to the number of leading
1's. From Theorem 1.2.1 we conclude that the system has free variables, i.e. it has only the
trivial solution.
(h) False. The row of zeros imposes no restriction on the unknowns and can be omitted. Whether the
system has infinitely many, one, or no solution(s) depends solely on the nonzero rows of the reduced
row echelon form.
(i) False. For example, the following system is clearly inconsistent:
1.3 Matrices and Matrix Operations
1. (a) Undefined (the number of columns in does not match the number of rows in )
(b) Defined; matrix
(c) Defined; matrix
(d) Defined; matrix
(e) Defined; matrix
(f) Defined; matrix
2. (a) Defined; matrix
(b) Undefined (the number of columns in does not match the number of rows in )
(c) Defined; matrix
(d) Defined; matrix
(e) Defined; matrix
(f) Undefined ( is a matrix, which cannot be added to a matrix )
3. (a)
(b)
(c)
(d)
(e) Undefined (a matrix cannot be subtracted from a matrix )
(f)
(g)
(h)
(i)
(j)
(k)
(l) Undefined (trace is only defined for square matrices)
4. (a)
(b)
1.3 Matrices and Matrix Operations 39
(c)
(d) Undefined (a matrix cannot be added to a matrix )
(e)
(f)
(g)
(h)
(i)
(j) Undefined (a matrix cannot be multiplied by a matrix )
(k)
46
(l) Undefined ( is a matrix; trace is only defined for square matrices)

5. (a)
(b) Undefined (the number of columns of does not match the number of rows in )
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
6. (a)
(b) Undefined (a matrix cannot be added to a matrix )
(c)
(d)
(e)
(f)
7. (a) first row of [first row of ]
(b) third row of [third row of ]
(c) second column of [second column of ]
(d) first column of [first column of ]
(e) third row of [third row of ]
(f) third column of [third column of ]
8. (a) first column of [first column of ]
(b) third column of [third column of ]
(c) second row of [second row of ]

(d) first column of [first column of ]
(e) third column of [third column of ]
(f) first row of [first row of ]
9. (a) first column of
second column of
third column of
(b) first column of
second column of
third column of
10. (a) first column of
second column of
third column of
(b) first column of

second column of
third column of
11. (a) , , ; the matrix equation:
(b) , , ; the matrix equation:
12. (a) , , ; the matrix equation:
(b) , , ; the matrix equation:
13. (a) (b)
14. (a) (b)
15.
The only value of that satisfies the equation is .
16.
The values of that satisfy the equation are and .
17.
18.
19.
20.
21.
22.
23. The given matrix equation is equivalent to the linear system
After subtracting first equation from the fourth, adding the second to the third, and back-substituting,
we obtain the solution: , , , and .
24. The given matrix equation is equivalent to the linear system
After subtracting first equation from the second, adding the third to the fourth, and back-substituting,
we obtain the solution: , , , and .
25. (a) If the th row vector of is then it follows from Formula (9) in Section 1.3 that
th row vector of
(b) If the th column vector of is then it follows from Formula (8) in Section 1.3 that
the th column vector of
26. (a) (b)

(c) (d)
27. Setting the left hand side equal to
yields
Assuming the entries of are real numbers that do not depend on , , and , this requires that the
coefficients corresponding to the same variable on both sides of each equation must match.
Therefore, the only matrix satisfying the given condition is .
28. Setting the left hand side equal to yields
Assuming the entries of are real numbers that do not depend on , , and , it follows that no real
numbers , , and exist for which the first equation is satisfied for all , , and . Therefore no
matrix with real number entries can satisfy the given condition.
(Note that if were permitted to depend on , , and , then solutions do exist e.g.,
.)
29. (a) and
(b) Four square roots can be found: , , , and .
32. (a) (b) (c)
the total cost of items purchased in anuar

the total cost of items purchased in Februar
33. The given matrix product represents .
the total cost of items purchased in arch
the total cost of items purchased in April
34. (a) The matrix represents sales over the two month period.
(b) The matrix represents the decrease in sales of each item from May to June.
(c)
(d)
(e) The entry in the matrix represents the total number of items sold in May.
(a) True. The main diagonal is only defined for square matrices.
(b) False. An matrix has row vectors and column vectors.
(c) False. E.g., if and then does not equal .
(d) False. The th row vector of can be computed by multiplying the th row vector of by .
(e) True. Using Formula (14), .
(f) False. E.g., if and then the trace of is , which does not equal
.
(g) False. E.g., if and then does not equal .
(h) True. The main diagonal entries in a square matrix are the same as those in .
(i) True. Since is a matrix, it follows from being a matrix that must be a
matrix. Consequently, is a matrix.
(j) True.
(k) True. The equality of the matrices and implies that for all and .
Adding to both sides yields for all and . Consequently, the matrices and are
equal.
(l) False. E.g., if and then even though .
(m) True. If is a matrix and is an matrix then being defined requires and
being defined requires . For the matrix to be possible to add to the matrix ,
we must have .
(n) True. If the th column vector of is then it follows from Formula (8) in Section 1.3 that
the th column vector of .
(o) False. E.g., if and then does not have a column of zeros even though
does.
1.4 Inverses; Algebraic Properties of Matrices
1. (a) (b)
(c) (d)
2. (a)
(b)
(c)
(d)
3. (a) (b)
4. (a) (b)
5. The determinant of , , is nonzero. Therefore is invertible and its
inverse is .

inverse is .
inverse is .
inverse is .
9. The determinant of ,
is
nonzero. Therefore is invertible and its inverse is .
10. The determinant of the matrix is . Therefore the matrix is

invertible and its inverse is .
11. ;
12. ;
13. ;
14. ; ;
15. From part (a) of Theorem 1.4.7 it follows that the inverse of is .
Thus . Consequently, .
Thus Consequently, .
1.4 Inverses; Algebraic Properties of Matrices 51
Thus .
Consequently,
18. From part (a) of Theorem 1.4.7 we have . Therefore .
19. (a)
(b)
(c)
20. (a)
(b)
(c)
21. (a) (b) (c)
22. (a) (b) (c)
23. ; .
The matrices and commute if , i.e.
Therefore, and commute if and .

formulas
24. ; .
The matrices and commute if , i.e.
Therefore, and commute if and .

formulas
25. ,
26. ,
27. ,
28. ,
29. ,
30.
Theorem 1.4.1(e)
Theorem 1.4.1(i)
Theorem 1.4.1(m)
Property on p. 43
Theorem 1.4.1(b)
31. (a) If and then does not
equal .
(b) Using the properties in Theorem 1.4.1 we can write
(c) If the matrices and commute (i.e., ) then .

32. We can let be one of the following eight matrices: , , ,
, , , , .
Note that these eight are not the only solutions - e.g., can be , etc.
33. (a) We can rewrite the equation
which shows that is invertible and .

(b) Let with . The equation can be rewritten
as
which shows that is invertible and .
34. If then it follows that therefore must be invertible (

35. If the th row vector of is then it follows from Formula (9) in Section 1.3 that
th row vector of .
Consequently no matrix can be found to make the product thus does not have an
inverse.
If the th column vector of is then it follows from Formula (8) in Section 1.3 that
inverse.
36. If the th and th row vectors of are equal then it follows from Formula (9) in Section 1.3 that
th row vector of th row vector of .
inverse.
If the th and th column vectors of are equal then it follows from Formula (8) in Section 1.3 that
the th column vector of the th column vector of
inverse.
37. Letting , the matrix equation becomes
Setting the first columns on both sides equal yields the system
Subtracting the second and third equations from the first leads to . Therefore
and (after substituting this into the remaining equations) .
The second and the third columns can be treated in a similar manner to result in
. We conclude that invertible and its inverse is .
38. Letting , the matrix equation becomes
Although this corresponds to a system of nine equations, it is sufficient to examine just the three
equations corresponding to the first column
to see that subtracting the second and third equations from the first leads to a contradiction .
We conclude that is not invertible.
39.
Theorem 1.4.6
Theorem 1.4.7(a)
Theorem 1.4.1(c)
Formula (1) in Section 1.4
Property on p. 43
40.
Theorem 1.4.6
Theorem 1.4.7(a)
Theorem 1.4.1(c)
Property on p. 43
41. If and then and
.
42. Yes, it is true. From part (e) of Theorem 1.4.8, it follows that . This
statement can be extended to factors (see p. 49) so that
43. (a) Assuming is invertible, we can multiply (on the left) each side of the equation by :
Multiply (on the left) each side by
Theorem 1.4.1(c)
Property on p. 43
(b) If is not an invertible matrix then does not generally imply as evidenced by
Example 3.
44. Invertibility of implies that is a square matrix, which is all that is required.
By repeated application of Theorem 1.4.1(m) and (l), we have
45. (a)
Theorem 1.4.1(d) and (e)
Property on p. 43
Theorem 1.4.1(a)

(b) We can multiply each side of the equality from part (a) on the left by , then on the right by
to obtain
which shows that if , , and are invertible then so is .

Furthermore, .
46. (a)
Theorem 1.4.1(f) and (g)
Property on p. 43
is idempotent so
(b)
Theorem 1.4.1(l) and (m);

Property on p. 43
is idempotent so
47. Applying Theorem 1.4.1(d) and (g), property , and the assumption we can write
48.
(a) False. and are inverses of one another if and only if .
(b) False. does not generally equal

since may not equal .
(c) False. does not generally equal since may not
equal .
(d) False. does not generally equal .
(e) False. does not generally equal .
(f) True. This follows from Theorem 1.4.5.
(g) True. This follows from Theorem 1.4.8.
(h) True. This follows from Theorem 1.4.9. (The inverse of is the transpose of .)
(i) False. .
(j) True.
If the th row vector of is then it follows from Formula (9) in Section 1.3 that
th row vector of .
inverse.
If the th column vector of is then it follows from Formula (8) in Section 1.3 that
inverse.
(k) False. E.g. and are both invertible but is not.
1.5 Elementary Matrices and a Method for Finding A-1
1. (a) Elementary matrix (corresponds to adding times the first row to the second row )
(b) Not an elementary matrix
(c) Not an elementary matrix
(d) Not an elementary matrix
2. (a) Elementary matrix (corresponds to multiplying the second row by )

(b) Elementary matrix (corresponds to interchanging the first row and the third row)
(c) Elementary matrix (corresponds to adding times the third row to the second row)
(d) Not an elementary matrix
3. (a) Add times the second row to the first row:

(b) Multiply the first row by :
(c) Add times the first row to the third row:
(d) Interchange the first and third rows:
4. (a) Add times the first row to the second row:
(b) Multiply the third row by :
(c) Interchange the first and fourth rows:
(d) Add times the third row to the first row:
5. (a) Interchange the first and second rows:
(b) Add times the second row to the third row:
(c) Add times the third row to the first row:
6. (a) Multiply the first row by :
(b) Add times the first row to the second row:
(c) Multiply the second row by :
7. (a) ( was obtained from by interchanging the first row and the third row)
1.5 Elementary Matrices and a Method for Finding A-1 59
(b) ( was obtained from by interchanging the first row and the third row)
(c) ( was obtained from by adding times the first row to the third row)
(d) ( was obtained from by adding times the first row to the third row)
8. (a) ( was obtained from by multiplying the second row by )
(b) ( was obtained from by multiplying the second row by )
(c) ( was obtained from by adding times the third row to the second row)
(d) ( was obtained from by adding times the third row to the second row)
9. (a) (Method I: using Theorem 1.4.5)

The determinant of , , is nonzero. Therefore is invertible
and its inverse is .
(Method II: using the inversion algorithm)
The identity matrix was adjoined to the given matrix.
The inverse is .
(b) (Method I: using Theorem 1.4.5)

The determinant of , . Therefore is not invertible.
A row of zeros was obtained on the left side, therefore is not invertible.
10. (a) (Method I: using Theorem 1.4.5)
The determinant of , , is nonzero. Therefore is
invertible and its inverse is .
The inverse is .
(b) (Method I: using Theorem 1.4.5)

The determinant of , . Therefore is not invertible.
A row of zeros was obtained on the left side, therefore the matrix is not invertible.
11. (a) The identity matrix was adjoined to the given matrix.
times the first row was added to the second row and
The inverse is .
(b) The identity matrix was adjoined to the given matrix.
Each row was multiplied by .

times the first row was added to the second and

The second row was multiplied by and

the third row was multiplied by .
2 times the third row was added to the first row.
The inverse is .


13. The identity matrix was adjoined to the given matrix.
times the third row was added to the second and

times the third row was added to the first row
The inverse is .
Each of the first two rows was multiplied by .
The second row was multiplied by

The inverse is .

times the second row was added to the first row
The first row was multiplied by
The inverse is .
16.
times the first row was added to each of the

remaining rows.

and to the fourth row.
times the third row was added to the fourth row

the third row was multiplied by and
the fourth row was multiplied by
The inverse is .
17. The identity matrix was adjoined to the given

matrix.
times the first row was added to the second.
The second and fourth rows were interchanged.

times the second row was added to the fourth.
The third row was multiplied by
times the third row was added

to the fourth row.
The fourth row was multiplied by
times the fourth row was added

to the second row.

to the second row and
to the first row.
times the second row was added

to the first row.
The inverse is .
times the first row was added to the fourth row

and to the fourth row.

to the fourth row.

to the fourth row.
The third row was multiplied by and

times the fourth row was added to the first row

and
times the third row was added to the second.
The inverse is .

the second row was multiplied by
The inverse is .
First row and third row were both multiplied by .

times the fourth row was added

to the third row and
to the first row.
The inverse is .
The first and fourth rows were interchanged;

the second and third rows were interchanged.

the second row was multiplied by
The inverse is .

times the first row was added

to the second row.

to the third row.

to the fourth row.
The inverse is .
21. It follows from parts (a) and (d) of Theorem 1.5.3 that a square matrix is invertible if and only if its
reduced row echelon form is identity.

If or , i.e. if or the last matrix contains at least one

row of zeros, therefore it cannot be reduced to by elementary row operations.
Otherwise (if and ), multiplying the second row by and multiplying the third row by
would result in a row echelon form with 1’s on the main dia onal. Subse uent elementar row
operations would then lead to the identity matrix.
We conclude that for any value of other than and the matrix is invertible.
22. It follows from parts (a) and (d) of Theorem 1.5.3 that a square matrix is invertible if and only if its
reduced row echelon form is identity.
times the second row was added to the third.
If , i.e. if , or the last matrix contains a row of zeros,

therefore it cannot be reduced to by elementary row operations.
Otherwise (if ), multiplying the last row by would result in a row echelon form with
1’s on the main diagonal. Subsequent elementary row operations would then lead to the identity
matrix.
We conclude that for any value of other than , and the matrix is invertible.
23. We perform a sequence of elementary row operations to reduce the given matrix to the identity
matrix. As we do so, we keep track of each corresponding elementary matrix:
times the second row was added to the first.

Since , then
and
Note that this answer is not unique since a different sequence of elementary row operations (and the
corresponding elementary matrices) could be used instead.
Since , and .

Since , we have
and .
The second and third rows were interchanged
Since , we have
and
27. Let us perform a sequence of elementary row operations to produce from . As we do so, we keep
track of each corresponding elementary matrix:
Since , the equality is satisfied by the matrix
28. Let us perform a sequence of elementary row operations to produce from . As we do so, we keep
track of each corresponding elementary matrix:
Since , the equality is satisfied by the matrix
Note that a different sequence of elementary row operations (and the corresponding elementary
matrices) could be used instead. (However, since both and in this exercise are invertible, is
uniquely determined by the formula .)
29. cannot result from interchanging two rows of (since that would create a nonzero
entry above the main diagonal).

can result from multiplying the third row of by a nonzero number

(in this case, ).
The other possibilities are that can be obtained by adding times the first row to the third
( or by adding times the second row to the third .
In all three cases, at least one entry in the third row must be zero.
30. Consider three cases:
 If then has a row of zeros (first row).

 If and then has a row of zeros (fifth row).
 If and then adding times the first row to the third, and adding times the fifth
row to the third results in the third row becoming a row of zeros.
In all three cases, the reduced row echelon form of is not . By Theorem 1.5.3, is not invertible.
(a) False. An elementary matrix results from performing a single elementary row operation on an
identity matrix; a product of two elementary matrices would correspond to a sequence of two such
operations instead, which generally is not equivalent to a single elementary operation.
(b) True. This follows from Theorem 1.5.2.
(c) True. If and are row equivalent then there exist elementary matrices such that
. Likewise, if and are row equivalent then there exist elementary matrices
such that . Combining the two equalities yields therefore and
are row equivalent.
(d) True. A homogeneous system has either one solution (the trivial solution) or infinitely many
solutions. If is not invertible, then by Theorem 1.5.3 the system cannot have just one solution.
Consequently, it must have infinitely many solutions.
(e) True. If the matrix is not invertible then by Theorem 1.5.3 its reduced row echelon form is not .
However, the matrix resulting from interchanging two rows of (an elementary row operation)
must have the same reduced row echelon form as does, so by Theorem 1.5.3 that matrix is not
invertible either.
(f) True. Adding a multiple of the first row of a matrix to its second row is an elementary row operation.
Denoting by be the corresponding elementary matrix we can write so the
resulting matrix is invertible if is.
(g) False. For instance, .

1.6 More on Linear Systems and Invertible Matrices
1. The given system can be written in matrix form as , where , , and .

We begin by inverting the coefficient matrix
The identity matrix was adjoined to the coefficient matrix.
Since , Theorem 1.6.2 states that the system has exactly one solution :
, i.e., .
2. The given system can be written in matrix form as , where , , and
. We begin by inverting the coefficient matrix
The first row was multiplied by and

the second row was multiplied by .

1.6 More on Linear Systems and Invertible Matrices 77
, i.e., .

Since , Theorem 1.6.2 states that the system has exactly one solution
: , i.e., and .

: , i.e., and .
The second row was multiplied by and

the third row was multiplied by .

and to the first row.

, i.e., and .
6. The given system can be written in matrix form as , where ,
, and . We begin by inverting the coefficient matrix
The identity matrix was adjoined to the coefficient

matrix.

and the first row was added to the fourth row.

and the second row was added to the fourth.
times the third row was added to the fourth.

times the last row was added to the third row,

times the last row was added to the second row
and times the last row was added to the first.

and
: ,
i.e., , , , and .

, i.e., , .
The third row was added to the second row and


: , i.e.,
, , and .
We augmented the coefficient matrix with two columns

9. of constants on the right hand sides of the systems
(i) and (ii) – refer to Example 2.
We conclude that the solutions of the two systems are:

(i) (ii) ,
We augmented the coefficient matrix with two columns

(i) and (ii) – refer to Example 2.


and the third row was added to the first row.
We conclude that the solutions of the two systems are:

(i) , (ii) , , .
We augmented the coefficient matrix with four columns

(i), (ii), (iii), and (iv) – refer to Example 2.
We conclude that the solutions of the four systems are:

(i) (ii) ,
(iii) (iv) ,
We augmented the coefficient matrix with three columns

(i), (ii) and (iii) – refer to Example 2.
The first row was added to the second row and


and to the second row.
We conclude that the solutions of the three systems are:

(i) ,
(ii) , ,
(iii) ,
The system is consistent for all values of and .

The system is consistent if and only if , i.e. .

The system is consistent if and only if , i.e. .

and to the third row.

The system is consistent for all values of , , and .

times the first row was added to the third row, and
The second row was added to the third row and

The system is consistent for all values of , , , and that satisfy the equations
and .
These equations form a linear system in the variables , , , and whose augmented matrix
has the reduced row echelon form . Therefore the system
is consistent if and .
18. (a) The equation can be rewritten as , which yields and
.
This is a matrix form of a homogeneous linear system - to solve it, we reduce its augmented
matrix to a row echelon form.
The augmented matrix for the homogeneous system

.

Using back-substitution, we obtain the unique solution: .

(b) As was done in part (a), the equation can be rewritten as . We solve the
latter system by Gauss-Jordan elimination
The augmented matrix for the homogeneous system

.

and the second row was added to the first row.

, , and .
19. . Let us find
The identity matrix was adjoined to the matrix.

The second row was added to the first row.
Using we obtain
20. . Let us find
The identity matrix was adjoined to the matrix.

Using we obtain
(a) True. By Theorem 1.6.1, if a system of linear equation has more than one solution then it must have
infinitely many.
(b) True. If is a square matrix such that has a unique solution then the reduced row echelon
form of must be . Consequently, must have a unique solution as well.
(c) True. Since is a square matrix then by Theorem 1.6.3(b) implies .
Therefore, .
(d) True. Since and are row equivalent matrices, it must be possible to perform a sequence of
elementary row operations on resulting in . Let be the product of the corresponding elementary
matrices, i.e., . Note that must be an invertible matrix thus .
Any solution of is also a solution of since .
Likewise, any solution of is also a solution of since .
(e) True. If then . Consequently, is a solution of .
(f) True. is equivalent to , which can be rewritten as . By Theorem
1.6.4, this homogeneous system has a unique solution (the trivial solution) if and only if its coefficient
matrix is invertible.
(g) True. If were invertible, then by Theorem 1.6.5 both and would be invertible.
1.7 Diagonal, Triangular, and Symmetric Matrices
1. (a) The matrix is upper triangular. It is invertible (its diagonal entries are both nonzero).
(b) The matrix is lower triangular. It is not invertible (its diagonal entries are zero).
(c) This is a diagonal matrix, therefore it is also both upper and lower triangular. It is invertible (its
diagonal entries are all nonzero).
(d) The matrix is upper triangular. It is not invertible (its diagonal entries include a zero).
2. (a) The matrix is lower triangular. It is invertible (its diagonal entries are both nonzero).
(b) The matrix is upper triangular. It is not invertible (its diagonal entries are zero).
(c) This is a diagonal matrix, therefore it is also both upper and lower triangular. It is invertible (its
diagonal entries are all nonzero).
(d) The matrix is lower triangular. It is not invertible (its diagonal entries include a zero).
3.
4.
5.
6.
7. , ,
8. , ,
9. , ,
10. ,
11.
12.
13.
1.7 Diagonal, Triangular, and Symmetric Matrices 93
14.
15. (a) (b)
16. (a) (b)
17. (a) (b)
18. (a) (b)
19. From part (c) of Theorem 1.7.1, a triangular matrix is invertible if and only if its diagonal entries are
all nonzero. Since this upper triangular matrix has a 0 on its diagonal, it is not invertible.
all nonzero. Since this upper triangular matrix has all three diagonal entries nonzero, it is invertible.
all nonzero. Since this lower triangular matrix has all four diagonal entries nonzero, it is invertible.
all nonzero. Since this lower triangular matrix has a 0 on its diagonal, it is not invertible.
23. . The diagonal entries of are: .
24. . The diagonal entries of are: .
25. The matrix is symmetric if and only if . In order for to be symmetric, we must have
.
26. The matrix is symmetric if and only if the following equations must be satisfied
We solve this system by Gauss-Jordan elimination


and times the first row was added to the third.
The third row was added to the second row

and times the third row was added to the first.
In order for to be symmetric, we must have , , and .

all nonzero. Therefore, the given upper triangular matrix is invertible for any real number such
that , , and .
all nonzero. Therefore, the given lower triangular matrix is invertible for any real number such
that , , and .
29. By Theorem 1.7.1, is also an upper triangular or lower triangular invertible matrix. Its diagonal
entries must all be nonzero - they are reciprocals of the corresponding diagonal entries of the matrix
.
30. By Theorem 1.4.8(e), . Therefore we have:
,
, and
since is symmetric.
31.
32. For example (there are seven other possible answers, e.g., ,
, etc.)
33.
. Since this is an upper triangular matrix, we have verified Theorem 1.7.1(b).
34. (a) Theorem 1.4.8(e) states that (if the multiplication can be performed). Therefore,
is
s mmetric
which shows that is symmetric.

(b)
h. h. and
1. . 1. . are
b-d) e) s mmetric
which shows that is symmetric.
35. (a) is symmetric, therefore we verified Theorem 1.7.4.
(b) The identity matrix was adjoined to the matrix .

Since is symmetric, we have verified Theorem 1.7.4
36. All diagonal matrices have a form .
This is a zero matrix whenever the value of , , and is either or . We conclude that the
following are all diagonal matrices that satisfy the equation:
37. (a) for all and therefore is symmetric.
(b) does not generally equal for therefore is not symmetric

(unless ).
(c) for all and therefore is symmetric.
(d) does not generally equal for therefore is not

symmetric (unless ).
38. If then is symmetric if and only if for all values of and .
39. For a general upper triangular matrix we have

Setting we obtain the equations , , .

The first and the third equations yield .
Substituting these into the second equation leads to , i.e., .
We conclude that the only upper triangular matrix such that is .
40. (a) Step 1. Solve
The first equation is .

The second equation yields .
The third equation yields
Step 2. Solve using back-substitution:
The third equation yields .

The first equation yields .
(b) Step 1. Solve

The third equation yields
Step 2. Solve using back-substitution:
The third equation yields .

41. (a) (b)
42. The condition is equivalent to the linear system

The augmented matrix has the reduced row echelon form .
If we assign the arbitrary value , the general solution is given by the formulas
, , , .
43. No. If , , and then which does
not generally equal . (The product of skew-symmetric matrices that commute is symmetric.)
44. is symmetric since and
is skew-symmetric since
therefore the result follows from the identity .
45. (a)
Theorem 1.4.9(d)
The assumption: is skew-symmetric
Theorem 1.4.7(c)
(b)
Theorem 1.4.8(a)
Theorem 1.4.8(b)
The assumption: and are skew-symmetric
Theorem 1.4.1(h)
Theorem 1.4.8(c)
The assumption: and are skew-symmetric
Theorem 1.4.1(i)
Theorem 1.4.8(d)
Theorem 1.4.1(l)
47. therefore is symmetric; thus we have .
(a) True. Every diagonal matrix is symmetric: its transpose equals to the original matrix.
(b) False. The transpose of an upper triangular matrix is a lower triangular matrix.
(c) False. E.g., is not a diagonal matrix.
(d) True. Mirror images of entries across the main diagonal must be equal - see the margin note next to
Example 4.
(e) True. All entries below the main diagonal must be zero.
(f) False. By Theorem 1.7.1(d), the inverse of an invertible lower triangular matrix is a lower triangular
matrix.
(g) False. A diagonal matrix is invertible if and only if all or its diagonal entries are nonzero (positive or
negative).
(h) True. The entries above the main diagonal are zero.
(i) True. If is upper triangular then is lower triangular. However, if is also symmetric then it
follows that must be both upper triangular and lower triangular. This requires to be a
diagonal matrix.
(j) False. For instance, neither nor is symmetric even though is.
(k) False. For instance, neither nor is upper triangular even though
is.
(l) False. For instance, is not symmetric even though is.
(m) True. By Theorem 1.4.8(d), . Since is symmetric, we also have . For

nonzero the equality of the right hand sides implies .
1.8 Matrix Transformations
1. (a) maps any vector in into a vector in .

The domain of is ; the codomain is .
(b) maps any vector in into a vector in .
(c) maps any vector in into a vector in .
(d) maps any vector in into a vector in .
2. (a) maps any vector in into a vector in .
(b) maps any vector in into a vector in .
(c) maps any vector in into a vector in .
(d) maps any vector in into a vector in .
3. (a) The transformation maps any vector in into a vector in .
Its domain is ; the codomain is .
(b) The transformation maps any vector in into a vector in .
1.8 Matrix Transformations 101

9. The transformation maps any vector in into a vector in . Its domain is ; the codomain is .
10. The transformation maps any vector in into a vector in . Its domain is ; the codomain is .
11. (a) The given equations can be expressed in matrix form as
therefore the standard matrix for this transformation is
(b) The given equations can be expressed in matrix form as
therefore the standard matrix for this transformation is .
12. (a) The given equations can be expressed in matrix form as
(b) The given equations can be expressed in matrix form as
13. (a) ; the standard matrix is

(b) ;
the standard matrix is
(c) ; the standard matrix is
(d) ; the standard matrix is
14. (a) ; the standard matrix is
(b) ; the standard matrix is
(c) ; the standard matrix is
(d) ; the standard matrix is
15. The given equations can be expressed in matrix form as therefore the
standard matrix for this operator is .
By directly substituting for into the given equation we obtain
By matrix multiplication, .
16. The given equations can be expressed in matrix form as therefore the
standard matrix for this transformation is .

By directly substituting for into the given equation we obtain
By matrix multiplication,
17. (a) ; the standard matrix is .
matches .
(b) ; the standard matrix is .
matches .
18. (a) ; the standard matrix is .
matches
(b) ; the standard matrix is .
matches .
19. (a)
(b)
20. (a)
(b)
21. (a) If and then

and .
(b) If and then
and .
22. (a) If and then
and
.
(b) If and then
and .
23. (a) The homogeneity property fails to hold since does not
generally equal . (It can be shown that the additivity property
fails to hold as well.)
(b) The homogeneity property fails to hold since
does not generally equal . (It can be shown that the
additivity property fails to hold as well.)
24. (a) The homogeneity property fails to hold since does not generally equal
. (It can be shown that the additivity property fails to hold
as well.)
(b) The homogeneity property fails to hold since does not

generally equal . (It can be shown that the
additivity property fails to hold as well.)
25. The homogeneity property fails to hold since for , does not generally equal
. (It can be shown that the additivity property fails to hold as well.)
On the other hand, both properties hold for :
and .
Consequently, is not a matrix transformation on unless
26. Both properties of Theorem 1.8.2 hold for :
On the other hand, neither property holds in general for , e.g.,

does not equal
27. By Formula (13), the standard matrix for is . Therefore
and .
and .

and .
30. For instance, satisfies the property , but the homogeneity property
fails to hold since does not generally equal
.
31. (a) , , .
(b) Since is a matrix transformation,
(c) Since is a matrix transformation, .
(a) False. The domain of is .
(b) False. The codomain of is .
(c) True. Since the statement requires the given equality to hold for some vector in , we can let .
(d) False. (Refer to Theorem 1.8.3.)
(e) True. The columns of are .
(f) False. The given equality must hold for every matrix transformation since it follows from the
homogeneity property.
(g) False. The homogeneity property fails to hold since does not generally equal
.
1.9 Applications of Linear Systems
1. There are four nodes, which we denote by , , , and (see the figure on the left).
We determine the unknown flow rates , , and assuming the counterclockwise direction (if any
of these quantities are found to be negative then the flow direction along the corresponding branch
will be reversed).
This system can be rearranged as follows
By inspection, this system has a unique solution , , . This yields the flow
rates and directions shown in the figure on the right.
2. (a) There are five nodes – each of them corresponds to an equation.
top left
top ri ht
bottom left
bottom middle
bottom ri ht
1.9 Applications of Linear Systems 107
(b) The augmented matrix of the linear system obtained in part (a) has the reduced row echelon
form . If we assign and the arbitrary values and ,
respectively, the general solution is given by the formulas

, , , , ,
(c) When and , the remaining flow rates become , , ,
and . The directions of the flow agree with the arrow orientations in the diagram.
3. (a) There are four nodes – each of them corresponds to an equation.
top left
top ri ht A)
bottom left
bottom ri ht )
(b) The augmented matrix of the linear system obtained in part (a)
has the reduced row echelon form . If we assign the arbitrary value
, the general solution is given by the formulas

, , ,
(c) In order for all values to remain positive, we must have . Therefore, to keep the traffic
flowing on all roads, the flow from A to B must exceed 500 vehicles per hour.
4. (a) There are six intersections – each of them corresponds to an equation.
top left
top middle
top ri ht
bottom left
bottom middle
bottom ri ht
We rewrite the system as follows
(b) The augmented matrix of the linear system obtained in part (a) has the reduced row echelon
form . If we assign and the arbitrary values and
, respectively, the general solution is given by the formulas

, ,
750  s  0
750  s  0
, ,
, ,
0
t
subject to the restriction that all seven 0
t
t
s
values must be nonnegative. Obviously, s

0

60
we need both and 0
60
, which in turn imply
and . Additionally
imposing the three inequalities 150 50  t  0
, 50 50  t  0
, and results in
600 750 s
the set of allowable and values
depicted in the grey region on the graph.
(c) Setting in the general solution obtained in part (b) would result in the negative value
which is not allowed (the traffic would flow in a wrong way along the street
marked as .)
5. From Kirchhoff's current law at each node, we have Kirchhoff's voltage law yields
eft oop cloc wise)

Ri ht oop cloc wise)
(An equation corresponding to the outer loop is a combination of these two equations.)
The linear system can be rewritten as
Its augmented matrix has the reduced row echelon form .
The solution is , , and .

Since is negative, this current is opposite to the direction shown in the diagram.
eft nside oop cloc wise)

Ri ht nside oop cloc wise)
(An equation corresponding to the outer loop is a combination of these two equations.)

Since is negative, this current is opposite to the direction shown in the diagram.
7. From Kirchhoff's current law, we have
op eft ode
op Ri ht ode
ottom eft ode
ottom Ri ht ode
Kirchhoff's voltage law yields
eft oop cloc wise)

iddle oop cloc wise)
Ri ht oop cloc wise)
(Equations corresponding to the other loops are combinations of these three equations.)
The solution is A, A.
op nside oop cloc wise)

ottom nside oop cloc wise)
The corresponding linear system can be rewritten as
9. We are looking for positive integers , and such that
The number of atoms of carbon, hydrogen, and oxygen on both sides must equal:
The linear system

has the augmented matrix whose reduced row echelon form is .
The general solution is , , , where is arbitrary. The smallest positive

integer values for the unknowns occur when , which yields the solution
, , , . The balanced equation is
10. We are looking for positive integers and such that
The linear system
The general solution is , , where is arbitrary. The smallest positive integer

values for the unknowns occur when , which yields the solution , , . The
balanced equation is
The number of atoms of carbon, hydrogen, oxygen, and fluorine on both sides must equal:
The linear system


integer values for the unknowns occur when , which yields the solution , ,
. The balanced equation is
The linear system

integer values for the unknowns occur when , which yields the solution , ,
, . The balanced equation is
13. We are looking for a polynomial of the form such that ,

and . We obtain a linear system
There is a unique solution , , .

The quadratic polynomial is
14. We are looking for a polynomial of the form such that ,
and . We obtain a linear system
There is a unique solution , , . The quadratic polynomial is

15. We are looking for a polynomial of the form such that
, and . We obtain a linear system
There is a unique solution , , , .

The cubic polynomial is .
16. We are looking for a polynomial of the form such that

, and . We obtain a linear system
There is a unique solution , , , .

The cubic polynomial is .
17. (a) We are looking for a polynomial of the form such that and
. We obtain a linear system

The general solution of the linear system is , , where is arbitrary.
Consequently, the family of all second-degree polynomials that pass through and
can be represented by where is an arbitrary real number.
(b)
(a) False. In general, networks may or may not satisfy the property of flow conservation at each node
(although the ones discussed in this section do).
(b) False. When a current passes through a resistor, there is a drop in the electrical potential in a circuit.
(c) True.
(d) False. A chemical equation is said to be balanced if for each type of atom in the reaction, the same
number of atoms appears on each side of the equation.
(e) False. By Theorem 1.9.1, this is true if the points have distinct -coordinates.
1.10 Leontief Input-Output Models
1. (a)
(b) The Leontief matrix is ;
the outside demand vector is .

1.10 Leontief Input-Output Models 115
The Leontief equation leads to the linear system with the augmented matrix
. Its reduced row echelon form is
To meet the consumer demand, must produce approximately $25,290.32 worth of

mechanical work and must produce approximately $22,580.65 worth of body work.
2. (a)
. Its reduced row echelon form is .
To meet the consumer demand, the economy must produce $300,000 worth of food and
$400,000 worth of housing.
3. (a)
Its reduced row echelon form is .
The production vector that will meet the given demand is .
4. (a)

Its reduced row echelon form is .
The production vector that will meet the given demand is .
5. ;
6. ;
7. (a) The Leontief matrix is .
. Its reduced row echelon form is therefore a production vector can be
found (namely, for an arbitrary nonnegative ) to meet the demand.
On the other hand, the Leontief equation leads to the linear system with the
augmented matrix . Its reduced row echelon form is ; the system is
inconsistent, therefore a production vector cannot be found to meet the demand.

1.10 Leontief Input-Output Models 117
(b) Mathematically, the linear system represented by can be rewritten as
Clearly, if the system has infinitely many solutions: ; where is an

arbitrary nonnegative number.
If the system is inconsistent. (Note that the Leontief matrix is not invertible.)
An economic explanation of the result in part (a) is that therefore the second sector
consumes all of its own output, making it impossible to meet any outside demand for its
products.
8.
If the open sector demands dollars worth from each product-producing sector, i.e. the outside
demand vector is . The Leontief equation leads to the linear system with the
augmented matrix . Its reduced row echelon form is .
We conclude that the first sector must produce the greatest dollar value to meet the specified open
sector demand.
9. From the assumption , it follows that the determinant of
is nonzero. Consequently, the Leontief matrix
is invertible; its inverse is . Since the consumption matrix

has nonnegative entries and , we conclude that all entries of are
nonnegative as well. This economy is productive (see the discussion above Theorem 1.10.1) - the
equation has a unique solution for every demand vector .
(a) False. Sectors that do not produce outputs are called open sectors.
(b) True.
(c) False. The th row vector of a consumption matrix contains the monetary values required of the th
sector by the other sectors for each of them to produce one monetary unit of output.
(d) True. This follows from Theorem 1.10.1.

(e) True.
Chapter 1 Supplementary Exercises
1. The corresponding system of linear equations is
The original augmented matrix.
This matrix is in row echelon form. It corresponds to the system of equations
formulas

Supplementary Exercises 119
This matrix is both in row echelon form and in reduced row echelon form. It corresponds to the
system of equations

This matrix is in row echelon form. It corresponds to the system of equations
then finish back-substituting to obtain the unique solution
Although this matrix is not in row echelon form yet, clearly it corresponds to an inconsistent linear
system
since the third equation is contradictory. (We could have performed additional elementary row
operations to obtain a matrix in row echelon form .)

5. The augmented matrix corresponding to the system.
6. We break up the solution into three cases:

Case I: and
The augmented matrix corresponding to the system.
times the first row was added to the second

( ).
times the second row was added to the first row

( .

Case II: which implies . The original system becomes , .
Multiplying both sides of the each equation by yields .
Case III: which implies . The original system becomes ,

. Multiplying both sides of each equation by yields ,
Notice that the solution found in case I
and .
actually applies to all three cases.
7. The original augmented matrix.
The positivity of the three variables requires that , , and . The first
inequality can be rewritten as , while the second inequality is equivalent to . All three
unknowns are positive whenever . There are three integer values of in this interval:
, , and . Of those, only yields integer values for the remaining variables: , .
8. Let and denote the number of pennies, nickels, and dimes, respectively. Since there are 13
coins, we must have
On the other hand, the total value of the coins is 83 cents so that
The resulting system of equations has the augmented matrix whose reduced row
echelon form is

However, all three unknowns must be nonnegative integers.

The nonnegativity of requires the inequality , i.e., .
Likewise for , yields .
When , all three variables are nonnegative. Of the four integer values inside this
interval ( , , , and ), only yields integer values for and .
We conclude that the box has to contain 3 pennies, 4 nickels, and 6 dimes.
(a) the system has a unique solution if and (multiplying the rows by , , and ,
respectively, yields a row echelon form of the augmented matrix ).
(b) the system has a one-parameter solution if and (multiplying the first two rows by
yields a reduced row echelon form of the augmented matrix ).
(c) the system has a two-parameter solution if and
(the reduced row echelon form of the augmented matrix is ).
(d) the system has no solution if and
(the reduced row echelon form of the augmented matrix is ).
times the second row was added to the third .

From quadratic formula we have
The system has no solutions when and (since the third row of our last matrix would
then correspond to a contradictory equation).
The system has infinitely many solutions when or .
No values of result in a system with exactly one solution.
11. For the product to be defined, must be a matrix. Letting we can write
The matrix equation can be rewritten as a system of nine linear equations
which has a unique solution , , , . (An easy way to solve this system is to first
split it into two smaller systems. The system , , involves
and only, whereas the remaining six equations involve just and .) We conclude that .
12. Substituting the values , and into the original system yields a system of three
equations in the unknowns and :
that can be rewritten as
The augmented matrix of this system has the reduced row echelon form . We
conclude that for the original system to have , , and as its solution, we must let
, and .
(Note that it can also be shown that the system with , and has , ,
and as its only solution. One way to do that would be to verify that the reduced row echelon
form of the coefficient matrix of the original system with these specific values of and is the
identity matrix.)
13. (a) must be a matrix. Letting we can write
therefore the given matrix equation can be rewritten as a system of linear equations:
so the system has a unique solution
, , , , , and .
(An alternative to dealing with this large system is to split it into two smaller systems instead:
the first three equations involve , , and only, whereas the remaining three equations involve
just , , and . Since the coefficient matrix for both systems is the same, we can follow the
procedure of Example 2 in Section 1.6; the reduced row echelon form of the matrix
is .)
Yet another way of solving this problem would be to determine the inverse
using the method introduced in Section 1.5, then multiply
both sides of the given matrix equation on the right by this inverse to determine :
(b) must be a matrix. Letting we can write
The augmented matrix of this system has the reduced row echelon form so
the system has a unique solution , , , . We conclude that .

(An alternative to dealing with this large system is to split it into two smaller systems instead:
the first three equations involve and only, whereas the remaining three equations involve
just and . Since the coefficient matrix for both systems is the same, we can follow the
procedure of Example 2 in Section 1.6; the reduced row echelon form of the matrix
is .)
(c) must be a matrix. Letting we can write
so the system has a unique solution , , , .
We conclude that .
14. (a) From Theorem 1.4.1, the properties (page 43) and the assumption , we
have
This shows that .

(b) From Theorem 1.4.1, the properties (page 43) and the assumption , we
have
15. We are looking for a polynomial of the form
such that , and . We obtain a linear system
There is a unique solution , , .

16. Since and we have the equations and .
From calculus, the derivative of is .
For the tangent to be horizontal, the derivative must equal zero. This leads to the
equation
We proceed to solve the resulting system of two equations:
The reduced row echelon form of the augmented matrix of this system is . Therefore,
the values , , and result in a polynomial that satisfies the conditions specified.
17. When multiplying the matrix by itself, each entry in the product equals . Therefore,
Property on p. 43
Theorem 1.4.1(m)
Theorem 1.4.1(j) and (k)

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1.

1 Introduction to Systems of Linear Equations 1

CHAPTER 1: SYSTEMS OF LINEAR EQUATIONS AND MATRICES

1.1 Introduction to Systems of Linear Equations

1. (a) This is a linear equation in , , and .

(e) This is not a linear equation in , , and because of the term .

(f) This is a linear equation in , , and .

(c) This is a linear equation in and .

(d) This is not a linear equation in and because of the term .

(e) This is not a linear equation in and because of the term .

4. (a) (b) (c)

7. (a) (b) (c)

8. (a) (b) (c)

where the parameter is an arbitrary real number.

If (i.e., ) then the second equation imposes no restriction on and ;

where the parameter is an arbitrary real number.

where the parameters and are arbitrary real numbers.

where the parameters , , and are arbitrary real numbers.

where the parameters , , , and are arbitrary real numbers.

where the parameter is an arbitrary real number.

where the parameters and are arbitrary real numbers.

where the parameters , , and are arbitrary real numbers.

where the parameters , , , and are arbitrary real numbers.

where the parameter is an arbitrary real number.

where the parameters and are arbitrary real numbers.

where the parameter is an arbitrary real number.

where the parameters and are arbitrary real numbers.

(b) Add the third row to the first to obtain

18. (a) Multiply the first row by to obtain .

(b) Add the third row to the first to obtain

This is a linear system in the unknowns , , and . Its augmented matrix is .

where the parameter is an arbitrary real number.

which can be rewritten as

 at least two of the three lines are parallel and distinct or

 two lines coincide and the third one intersects them or

26. We set up the linear system as discussed in Exercise 21:

1.2 Gaussian Elimination

3. (a) The linear system

and solved by back-substitution:

therefore the original linear system has a unique solution: , , .

therefore the original linear system has infinitely many solutions:

therefore the original linear system has infinitely many solutions:

where and are arbitrary values.

4. (a) A unique solution: , , .

(b) Infinitely many solutions: , , , where is an arbitrary

5. The augmented matrix for the system.

The first row was added to the second row.

times the first row was added to the third row.

The second row was multiplied by .

10 times the second row was added to the third row.

The third row was multiplied by .

and can be rewritten as

The linear system has a unique solution: , , .

6. The augmented matrix for the system.

The first row was multiplied by .

times the first row was added to the second row.

times the first row was added to the third row.

The second row was multiplied by .

7 times the second row was added to the third row.

Solve the equations for the leading variables

then substitute the second equation into the first

If we assign an arbitrary value , the general solution is given by the formulas

7. The augmented matrix for the system.