Answers to Problems in "Elementary Linear Algebra" (12th Edition) by Anton

Simplify your study of linear algebra with this comprehensive collection of answers to problems from "Elementary Linear Algebra" (12th Edition) by Anton. This guide offers step-by-step solutions for key concepts, including matrices, vector spaces, and linear transformations. Ideal for students seeking thorough and precise explanations to reinforce their learning.

1. 1.1 Introduction to Systems of Linear Equations 1

1.1 Introduction to Systems of Linear Equations
1. (a) This is a linear equation in
1
x,
2
x, and
3
x.
(b) This is not a linear equation in
1
x,
2
x, and
3
x because of the term
13
xx.
(c) We can rewrite this equation in the form 
123
730xxx therefore it is a linear equation in
1
x,
2
x, and
3
x.
(d) This is not a linear equation in
1
x,
2
x, and
3
x because of the term
2
1
x.
(e) This is not a linear equation in
1
x,
2
x, and
3
x because of the term
3/5
1
x.
(f) This is a linear equation in
1
x,
2
x, and
3
x.
2. (a) This is a linear equation in x and y.
(b) This is not a linear equation in x and y because of the terms
1/3
2x and 3y.
(c) This is a linear equation in x and y.
(d) This is not a linear equation in x and y because of the term

7
cosx.
(e) This is not a linear equation in x and y because of the term xy.
(f) We can rewrite this equation in the form 7xy thus it is a linear equation in x and y.
3. (a)


11 1 12 2 1
21 1 22 2 2
ax ax b
ax ax b

(b)



11 1 12 2 13 3 1
21 1 22 2 23 3 2
31 1 32 2 33 3 3
ax ax ax b
ax ax ax b
ax ax ax b
(c)


11 1 12 2 13 3 14 4 1
21 1 22 2 23 3 24 4 2
ax ax ax ax b
ax ax ax ax b


4. (a)




11 12 1
21 22 2
aab
aab

(b)






11 12 13 1
21 22 23 2
31 32 33 3
aaab
aaab
aaab

(c)

 
 
 
11 12 13 14 1
21 22 23 24 2
aaaab
aaaab



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2. 1.1 Introduction to Systems of Linear Equations 2

5. (a)




1
12
2
2 0
340
1
x
xx
x


(b)


 

13
12 3
23
3 2 5
743
27
xx
xxx
xx



6. (a)

 

23 4
12 4
31
52 3 6xx x
xx x


(b)


 


134
134
12 4
4
3 4 3
4 4 3
3 2 9
2
xxx
xxx
xx x
x


7. (a)






26
38
93


(b)




6134
05 11

(c)


 
 

 
 
 
020310
311001
621236


8. (a)






321
453
732


(b)

 
 

 
 
 
2021
3147
6110

(c)

 
 
 
 
 
1001
0102
0013


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
x from the second equation by adding 2 times the first equation to the second. This yields
the system



324
01
xy

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
x from the second equation by adding 2 times the first equation to the second. This yields
the system



241
00
xy

3. 1.1 Introduction to Systems of Linear Equations 3

The second equation does not impose any restriction on x and y therefore we can omit it. The lines
represented by the original system have infinitely many points of intersection. Solving the first equation for x
we obtain 
1
2
2
x y. This allows us to represent the solution using parametric equations

1
2 ,
2
x tyt
where the parameter
t is an arbitrary real number.

(c) We can eliminate
x from the second equation by adding 1 times the first equation to the second. This yields
the system



20
28
xy
y

From the second equation we obtain 4y. Substituting 4 for y into the first equation results in 8x.
Therefore, the original system has the unique solution
8, 4xy
The represented by the equations in that system have one point of intersection:

8, 4.
12. We can eliminate x from the second equation by adding 2 times the first equation to the second. This yields
the system



23
02
xya
ba

If
20ba (i.e., 2ba) then the second equation imposes no restriction on
x and y; consequently, the
system has infinitely many solutions.
If
20ba (i.e., 2ba) then the second equation becomes contradictory thus the system has no solutions.
There are no values of
a and b for which the system has one solution.
13. (a) Solving the equation for
x we obtain 
35
77
x y therefore the solution set of the original equation can be
described by the parametric equations

35
,
77
x tyt
where the parameter
t is an arbitrary real number.

(b) Solving the equation for
1
x we obtain  
75 4
12333 3
x xx therefore the solution set of the original equation can
be described by the parametric equations
   
123
75 4
, ,
33 3
x r s xr xs
where the parameters
r and
s are arbitrary real numbers.

4. 1.1 Introduction to Systems of Linear Equations 4

(c) Solving the equation for
1
x we obtain    
5311
123484 8 4
x xxx therefore the solution set of the original
equation can be described by the parametric equations
      
1234
11 5 3
, , ,
84 8 4
x rstxrxsxt
where the parameters
r,
s, and t are arbitrary real numbers.

(d) Solving the equation for v we obtain

8 214
3 333
vwxyz therefore the solution set of the original equation
can be described by the parametric equations

    
123 4 1 2 3 4
8214
, , , ,
3333
vt t t t wtxt ytzt
where the parameters
1
t,
2
t,
3
t, and
4
t are arbitrary real numbers.
14. (a) Solving the equation for
x we obtain 210xy therefore the solution set of the original equation can be
described by the parametric equations
2 10 , x tyt
where the parameter
t is an arbitrary real number.

(b) Solving the equation for
1
x we obtain  
123
33 12x xx therefore the solution set of the original equation can
be described by the parametric equations

   
123
3 3 12 , ,
x rsxrxs
where the parameters
r and
s are arbitrary real numbers.

(c) Solving the equation for
1
x we obtain   
311
1234 244
5
x xxx therefore the solution set of the original
equation can be described by the parametric equations
     
12
131
5 , , ,
244
x rstxr yszt
where the parameters
r,
s, and t are arbitrary real numbers.
(d) Solving the equation for v we obtain  57vwx yz therefore the solution set of the original equation
can be described by the parametric equations

       
1234 1 2 3 4
5 7 , , , , vtt t t wt xt yt zt
where the parameters
1
t,
2
t,
3
t, and
4
t are arbitrary real numbers.
15. (a) We can eliminate
x from the second equation by adding 3 times the first equation to the second. This yields
the system



231
00
xy

5. 1.1 Introduction to Systems of Linear Equations 5

The second equation does not impose any restriction on x and y therefore we can omit it. Solving the first
equation for x we obtain 
31
22
x y. This allows us to represent the solution using parametric equations

13
,
22
x tyt
where the parameter
t is an arbitrary real number.

(b) We can see that the second and the third equation are multiples of the first: adding 3 times the first equation
to the second, then adding the first equation to the third yields the system



123
34
00
00
xxx

The last two equations do not impose any restriction on the unknowns therefore we can omit them. Solving the
first equation for
1
x we obtain   
123
43x xx. This allows us to represent the solution using parametric
equations

    
123
4 3 , ,
x rs x r x s
where the parameters
r and
s are arbitrary real numbers.
16. (a) We can eliminate
1
x from the first equation by adding 2 times the second equation to the first. This yields
the system

00



12
34xx
The first equation does not impose any restriction on
1
x and
2
x therefore we can omit it. Solving the second
equation for
1
x we obtain  
41
12 33
x x. This allows us to represent the solution using parametric equations
 
12
41
,
33
x txt
where the parameter
t is an arbitrary real number.

(b) We can see that the second and the third equation are multiples of the first: adding 3 times the first equation
to the second, then adding
2 times the first equation to the third yields the system
224xy z


00


00
The last two equations do not impose any restriction on the unknowns therefore we can omit them. Solving the
first equation for x we obtain   
1
2
2
x yz. This allows us to represent the solution using parametric
equations

6. 1.1 Introduction to Systems of Linear Equations 6

    
1
2 , ,
2
x rs yr zs
where the parameters
r and
s are arbitrary real numbers.
17. (a) Add 2 times the second row to the first to obtain
 
 

 
 
 
1788
2332
0231
.

(b) Add the third row to the first to obtain

 
 

 
 
 
1383
2932
1433

(another solution: interchange the first row and the third row to obtain

 
 

 
 
 
14 33
2932
0150
).
18. (a) Multiply the first row by
1
2
to obtain

 
 
 
 
 
12 34
71 43
54 27
.

(b) Add the third row to the first to obtain

 
 

 
 
 
1136
3181
6314

(another solution: add
2 times the second row to the first to obtain

 
 

 
 
 
12180
31 81
63 14
).
19. (a) Add 4 times the first row to the second to obtain
 
 
 
14
084 18
k
k
which corresponds to the system
  4xky

  84 18ky


If
2k then the second equation becomes
018, which is contradictory thus the system becomes
inconsistent.
If
2k then we can solve the second equation for y and proceed to substitute this value into the first equation
and solve for
x.
Consequently, for all values of
2k the given augmented matrix corresponds to a consistent linear system.

(b) Add 4 times the first row to the second to obtain
 
 
 
11
084 0
k
k
which corresponds to the system

7. 1.1 Introduction to Systems of Linear Equations 7


  1xky

  84 0ky


If
2k then the second equation becomes
00, which does not impose any restriction on x and y 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 2
k then the second equation yields
0y and the first equation becomes 1 x .
Consequently, for all values of
k the given augmented matrix corresponds to a consistent linear system.
20. (a) Add 2 times the first row to the second to obtain

 
 
 
34
0025
k
k
which corresponds to the system
34xyk


02 5k


If 
5
2
k then the second equation becomes
00, which does not impose any restriction on x and y
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 
5
2
k then the second equation is contradictory thus the system becomes inconsistent.
Consequently, the given augmented matrix corresponds to a consistent linear system only when

5
2
k .

(b) Add the first row to the second to obtain
 
 
 
12
400
k
k
which corresponds to the system


 

2
4 0
kx y
kx

If
4k then the second equation becomes
00, which does not impose any restriction on x and y
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
4k then the second equation yields
0x and the first equation becomes 2y.
Consequently, for all values of
k 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


2
11 1
yaxbxc
Repeating this for the other two points and rearranging the three equations yields

2
11 1
xaxbcy


2
22 2
xaxbcy


2
33 3
xaxbcy

8. 1.1 Introduction to Systems of Linear Equations 8

This is a linear system in the unknowns a, b, and c. Its augmented matrix is
 
 
 
 
 
2
11 1
2
22 2
2
33 3
1
1
1
xxy
xxy
xxy
.
23. Solving the first equation for
1
x we obtain 
12
xckx therefore the solution set of the original equation can be
described by the parametric equations

12
, xckt x t
where the parameter
t is an arbitrary real number.
Substituting these into the second equation yields

cktltd
which can be rewritten as
cktdlt
This equation must hold true for all real values
t, which requires that the coefficients associated with the same power
of
t on both sides must be equal. Consequently,
cd and kl.
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. 


23 7
239
42516
xyz
xy z
xyz
26. We set up the linear system as discussed in Exercise 21:





2
2
2
11 1
22 4
11 1
abc
abc
abc
i.e.



1
42 4
1
abc
abc
abc

One solution is expected, since exactly one parabola passes through any three given points

11
,
xy,  
22
,xy,  
33
,xy
if
1
x,
2
x, and
3
x are distinct.
27.  
 

12
225
1
xy z
xy z
xz

9. 1.1 Introduction to Systems of Linear Equations 9


True-False Exercises
(a) True. 0,0, ,0 is a solution.
(b) False. Only multiplication by a nonzero constant is a valid elementary row operation.
(c) True. If 6k then the system has infinitely many solutions; otherwise the system is inconsistent.
(d) True. According to the definition, 
11 2 2 nn
ax ax ax b is a linear equation if the a's are not all zero. Let us
assume
0
j
a . The values of all x's except for
j
x can be set to be arbitrary parameters, and the equation can be used
to express
j
x 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 0c then the new system has the same solution set as the original one.
(g) True. Adding 1 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
01, which is contradictory.

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 (
7) 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 leading 1 and a
nonzero number (4) above it. The matrix is in row echelon form but not reduced row echelon form.

(d) This matrix has properties 1-3 but does not have property 4: the second column contains a leading 1 and a
nonzero number (5) above it. The matrix is in row echelon form but not reduced row echelon form.

10. 1.2 Gaussian Elimination 10


(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 first three columns are pivot columns and all three rows are pivot rows. The linear system



347
22
5
xyz
yz
z
can be rewritten as



73 4
22
5
x yz
yz
z

and solved by back-substitution:

 

 
 
5
225 8
738 45 37
z
y
x

therefore the original linear system has a unique solution:
37x, 8y, 5z.

(b) The first three columns are pivot columns and all three rows are pivot rows. The linear system



856
493
2
wyz
xyz
yz
can be rewritten as



68 5
34 9
2
wyz
x yz
yz


Let zt. Then



   
    
2
342 9 513
6 8 2 5 10 13
yt
x tt t
wttt


therefore the original linear system has infinitely many solutions:

 10 13wt ,
513x t, 2yt, zt
where
t is an arbitrary value.

(c) Columns 1, 3, and 4 are pivot columns. The first three rows are pivot rows. The linear system
 
 


12 3 5
34 5
45
72 8 3
65
3 9
0 0
xxx x
xx x
xx

can be rewritten:
   
1235
37 2 8
x xxx ,  
345
56x xx , 
45
93x x.
Let

2
xs and 
5
xt. Then

11. 1.2 Gaussian Elimination 11




   
        
4
3
1
93
593 6 43
37 2 43 8 117 2
xt
xttt
x sttst

therefore the original linear system has infinitely many solutions:

        
12345
11 7 2 , , 4 3 , 9 3 ,
x stxsx tx txt
where s and t are arbitrary values.

(d) The first two columns are pivot columns and the first two rows are pivot rows. The system is inconsistent since
the third row of the augmented matrix corresponds to the equation 0001.xyz
4. (a) The first three columns are pivot columns and all three rows are pivot rows. A unique solution: 3x, 0y,
7z.

(b) The first three columns are pivot columns and all three rows are pivot rows. Infinitely many solutions:
87wt , 23
x t,  5yt , zt where t is an arbitrary value.

(c) Columns 1, 3, and 4 are pivot columns. The first three rows are pivot rows. Infinitely many solutions:
  26 3vst , ws, 74
x t, 85yt, zt where s and t are arbitrary values.

(d) Columns 1 and 3 are pivot columns. The first two rows are pivot rows. The system is inconsistent since the
third row of the augmented matrix corresponds to the equation 0001.xyz
5.






1128
1231
37410


 

The augmented matrix for the system.
 







1128
0159
37410


 
The first row was added to the second row. 







1128
0159
010214


 
3 times the first row was added to the third row. 







1128
0159
010214


 
The second row was multiplied by 1. 





 

11 2 8
01 5 9
0 0 52 104


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

12. 1.2 Gaussian Elimination 12








11 2 8
01 5 9
00 1 2
 

 
The third row was multiplied by

1
52
.

The system of equations corresponding to this augmented matrix in row echelon form is

 


12 3
23
3
28
59
2
xx x
xx
x
and can be rewritten as



123
23
3
82
95
2
x xx
xx
x

Back-substitution yields



 
 
3
2
1
2
952 1
8122 3
x
x
x

The linear system has a unique solution:

1
3x, 
2
1x, 
3
2x.

6.




 

222 0
252 1
814 1


 

The augmented matrix for the system.
 





 

111 0
252 1
814 1


 
The first row was multiplied by
1
2
. 




 

111 0
074 1
814 1


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






1110
0741
0741


 
8 times the first row was added to the third row. 






41
77
1110
01
0741


 
The second row was multiplied by
1
7
. 






41
77
1110
01
0000
 

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

The system of equations corresponding to this augmented matrix in row echelon form is

13. 1.2 Gaussian Elimination 13


 


12 3
23
0
41

77
0 0
xx x
xx
Solve the equations for the leading variables

123
xxx


23
14
77
x x
then substitute the second equation into the first
 

13
23
13
77
14
77
x x
x x

If we assign
3
x an arbitrary value t, the general solution is given by the formulas
    
12 3
13 14
, ,
77 77
x tx t x t
7.







112 11
21222
12 4 1 1
300 33


 

The augmented matrix for the system.
 








11211
03600
12 4 11
30033


 
2 times the first row was added to the second row. 





 


11211
03600
01200
30033


 
The first row was added to the third row. 





 


11211
03600
01200
03600


 
3 times the first row was added to the fourth row. 

14. 1.2 Gaussian Elimination 14






 


11211
01200
01200
03600


 
The second row was multiplied by
1
3
. 








11211
01200
00000
03600
 

 
1 times the second row was added to the third row. 








11211
01200
00000
00000
 

 
3 times the second row was added to the fourth row. 

The system of equations corresponding to this augmented matrix in row echelon form is

  



21
2 0
0 0
0 0
xy zw
yz

Solve the equations for the leading variables
  

12
2
x yzw
yz

then substitute the second equation into the first
   

12 2 1
2
x zzw w
yz

If we assign z and w the arbitrary values
s and t, respectively, the general solution is given by the formulas
    1 , 2 , , x ty s zs wt
8.






0231
3632
6635


 

The augmented matrix for the system.
 







3632
0231
6635


 
The first and second rows were interchanged. 

15. 1.2 Gaussian Elimination 15








2
3
121
0231
6635


 
The first row was multiplied by
1
3
. 







2
3
121
0231
0699


 
6 times the first row was added to the third row. 







2
3
3 1
22
12 1
01
0699


 
The second row was multiplied by

1
2
. 







2
3
3 1
22
12 1
01
0006
 

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







2
3
3 1
22
12 1
01
00 0 1
 

 
The third row was multiplied by
1
6
.

The system of equations corresponding to this augmented matrix in row echelon form

 
 

2
2
3
31

22
01
ab c
bc
is clearly inconsistent.
9.






1128
1231
37410


 

The augmented matrix for the system.
 







1128
0159
37410


 
The first row was added to the second row. 







1128
0159
010214


 
3 times the first row was added to the third row. 







1128
0159
010214


 
The second row was multiplied by 1. 

16. 1.2 Gaussian Elimination 16






 

11 2 8
01 5 9
0 0 52 104


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







11 2 8
01 5 9
00 1 2
 

 
The third row was multiplied by

1
52
. 






1128
0101
0012
 

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






1104
0101
0012
 

 
2 times the third row was added to the first row. 






1003
0101
0012
 

 
1 times the second row was added to the first row.

The linear system has a unique solution: 
1
3x, 
2
1x, 
3
2x.

10.




 

222 0
252 1
814 1


 

The augmented matrix for the system.
 





 

111 0
252 1
814 1


 
The first row was multiplied by
1
2
. 




 

111 0
074 1
814 1


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






1110
0741
0741


 
8 times the first row was added to the third row. 






41
77
1110
01
0741


 
The second row was multiplied by
1
7
. 

17. 1.2 Gaussian Elimination 17







41
77
1110
01
0000
 

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






3 1
77
41
77
10
01
000 0
 

 
1 times the second row was added to the first row.

Infinitely many solutions:  
31
1 77
x t, 
14
2 77
x t, 
3
xt where t is an arbitrary value.
11.







112 11
21222
12 4 1 1
300 33


 

The augmented matrix for the system.
 








11211
03600
12 4 11
30033


 
2 times the first row was added to the second row. 





 


11211
03600
01200
30033


 
the first row was added to the third row. 





 


11211
03600
01200
03600


 
3 times the first row was added to the fourth row. 





 


11211
01200
01200
03600


 
The second row was multiplied by
1
3
. 








11211
01200
00000
03600
 

 
1 times the second row was added to the third row. 








11211
01200
00000
00000
 

 
3 times the second row was added to the fourth row. 

18. 1.2 Gaussian Elimination 18









10011
01200
00000
00000
 

 
the second row was added to the first row. 

The system of equations corresponding to this augmented matrix in row echelon form is

 



1
2 0
0 0
0 0
xw
yz

Solve the equations for the leading variables
1x w

2yz


If we assign
z and w the arbitrary values
s and t, respectively, the general solution is given by the formulas
    1 , 2 , , x tyszswt
12.






0231
3632
6635


 

The augmented matrix for the system.
 







3632
0231
6635


 
The first and second rows were interchanged. 







2
3
121
0231
6635


 
The first row was multiplied by
1
3
. 







2
3
121
0231
0699


 
6 times the first row was added to the third row. 







2
3
3 1
22
12 1
01
0699


 
The second row was multiplied by

1
2
. 







2
3
3 1
22
12 1
01
0006
 

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

19. 1.2 Gaussian Elimination 19








2
3
3 1
22
12 1
01
00 0 1
 

 
The third row was multiplied by
1
6
. 







2
3
3
2
12 1
01 0
00 0 1
  

 
1
2
times the third row was added to the second row. 







3
2
12 10
01 0
00 01
  

 
2
3
times the third row was added to the first row. 







3
2
10 20
01 0
00 01
  

 
2 times the second row was added to the first row.

The last row corresponds to the equation
0001abc
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

3
0x, which substituted into the second equation yields 
2
0.x Both of these
substituted into the first equation result in

1
0x.)
15. We present two different solutions.
Solution I uses Gauss-Jordan elimination






2130
1200
0110




The augmented matrix for the system.







31
22
10
1200
0110



The first row was multiplied by
1
2
.







31
22
33
22
10
00
01 10



1 times the first row was added to the second row.

20. 1.2 Gaussian Elimination 20








31
22
10
01 10
01 10



The second row was multiplied by
2
3
.







31
22
10
01 10
00 20



1 times the second row was added to the third row.







31
22
10
01 10
00 10



The third row was multiplied by
1
2
.






1
2
100
0100
0010





The third row was added to the second row
and

3
2
times the third row was added to the first row






1000
0100
0010




1
2
times the second row was added to the first row.

Unique solution:

1
0x, 
2
0x, 
3
0x.
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” in Section 1.2)






2130
1200
0110




The augmented matrix for the system.







1200
2130
0110




The first and second rows were interchanged
(to avoid introducing fractions into the first row).







1 200
0330
0110



2 times the first row was added to the second row.







12 00
01 10
01 10



The second row was multiplied by 
1
3
.

21. 1.2 Gaussian Elimination 21








12 00
01 10
00 20



1 times the second row was added to the third row.







12 00
01 10
00 10



The third row was multiplied by
1
2
.






1200
0100
0010



The third row was added to the second row.






1000
0100
0010



2 times the second row was added to the first row.

Unique solution:

1
0x, 
2
0x, 
3
0x.
16. We present two different solutions.
Solution I uses Gauss-Jordan elimination







2130
12 30
1140




The augmented matrix for the system.








31
22
10
1230
1140



The first row was multiplied by
1
2
.







31
22
39
22
10
00
1140



The first row was added to the second row.







31
22
39
22
3 11
22
10
00
00



1 times the first row was added to the third row.







31
22
3 11
22
10
0130
00



The second row was multiplied by
2
3
.







31
22
10
0130
00100




3
2
times the second row was added to the third row.

22. 1.2 Gaussian Elimination 22








31
22
10
0130
0010



The third row was multiplied by
1
10
.






1
2
100
0100
0010




3 times the third row was added to the second row
and
3
2
times the third row was added to the first row






1000
0100
0010



1
2
times the second row was added to the first row.

Unique solution:
0x, 0y, 0z.
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” in Section 1.2)







2130
12 30
1140




The augmented matrix for the system.








1140
12 30
2130




The first and third rows were interchanged
(to avoid introducing fractions into the first row).






1140
0310
2130



The first row was added to the second row.






11 40
03 10
03110



2 times the first row was added to the third row.




 

11 40
03 10
00 100



The second row was added to the third row.






1140
0310
0010



The third row was multiplied by

1
10
.

23. 1.2 Gaussian Elimination 23







1140
0300
0010



1 times the third row was added to the second row.






1100
0300
0010



4 times the third row was added to the first row.






1100
0100
0010



The second row was multiplied by
1
3
.






1000
0100
0010



1 times the second row was added to the first row.

Unique solution: 0
x, 0y, 0z.
17.



31110
51110




The augmented matrix for the system.





11 1
33 3
10
51110



The first row was multiplied by
1
3
.





111
333
88 2
333
10
00



5 times the first row was added to the second row.




111
333
1
4
10
01 10



The second row was multiplied by

3
8
.




1
4
1
4
10 00
01 10




1
3
times the second row was added to the first row.

If we assign
3
x and
4
x the arbitrary values s and t, respectively, the general solution is given by the formulas
    
12 34
11
, , ,
44
x sx stxsxt .
(Note that fractions in the solution could be avoided if we assigned 
3
4xs instead, which along with 
4
xt would
yield

1
xs,  
2
xst, 
3
4xs, 
4
xt.)

24. 1.2 Gaussian Elimination 24

18.




 

 
01320
21430
23210
43540




The augmented matrix for the system.






 

 
21430
01320
23210
43540



The first and second rows were interchanged.





 

 
31
22
120
01320
23210
43540



The first row was multiplied by
1
2
.





 


31
22
120
01320
02640
01320




2 times the first row was added to the third row
and
4 times the first row was added to the fourth row.








31
22
120
01 3 20
00000
00000




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








75
22
10 0
01 3 20
00 0 00
00 0 00




1
2
times the second row was added to the first row.

If we assign
w and
x the arbitrary values s and t, respectively, the general solution is given by the formulas
   
75
, 3 2 , ,
22
ustv stwsxt .

19.







02 2 40
10 1 30
23 1 10
21 3 20




The augmented matrix for the system.

25. 1.2 Gaussian Elimination 25








10 1 30
02 2 40
23 1 10
21 3 20



The first and second rows were interchanged.







10 1 30
02 2 40
03 3 70
01180




2 times the first row was added to the third row
and
2 times the first row was added to the fourth row.







10 1 30
01 1 20
03 3 70
01180



The second row was multiplied by 1
2
.







10 1 30
01 1 20
00 0 10
00 0 100




3 times the second row was added to the third and
1 times the second row was added to the fourth row.







10 1 30
01 1 20
00 0 10
00 0 00



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







10 100
01 100
00 010
00 000




2 times the third row was added to the second and
3 times the third row was added to the first row.

If we assign
y an arbitrary value t the general solution is given by the formulas
   , , , 0wt x t yt z .

20.








13010
14200
02210
24110
12 110




The augmented matrix for the system.

26. 1.2 Gaussian Elimination 26











13010
01210
0 2210
010110
05100





1 times the first row was added to the second row,
2 times the first row was added to the fourth row,
and 1 times the first row was added to the fifth row.





 



 
13 0 10
01 2 10
00 2 30
0021 110
00 9 50





2 times the second row was added to the third row,
10 times the second row was added to the fourth row,
and
5 times the second row was added to the fifth row.





 



 
3
2
13 0 10
01 2 10
00 1 0
0021 110
00 9 50



The third row was multiplied by
1
2
.





 




3
2
41
2
17
2
130 10
012 10
001 0
000 0
000 0





21 times the third row was added to the fourth row
and 9 times the third row was added to the fifth row.






 




3
2
17
2
130 10
012 10
001 0
000 10
000 0



The fourth row was multiplied by
2
41
.





 




3
2
130 10
012 10
001 0
000 10
000 00




17
2
times the fourth row was added to the fifth row.

The augmented matrix in row echelon form corresponds to the system


 


12 4
23 4
34
4
3 0
20
3
0
2
0
xx x
xx x
xx
x

27. 1.2 Gaussian Elimination 27

Using back-substitution, we obtain the unique solution of this system

1234
0, 0, 0, 0xxxx .
21.







21349
102711
33158
214410




The augmented matrix for the system.









102711
21349
33158
214410




The first and second rows were interchanged
(to avoid introducing fractions into the first row).








102 7 11
0171013
0371625
0181012





2 times the first row was added to the second row,
3 times the first row was added to the third row,
and 2 times the first row was added to the fourth.








102 7 11
0171013
0371625
0181012



The second row was multiplied by
1.





 


10 2 7 11
01 7 10 13
0 0 14 14 14
0 0 15 20 25





3 times the second row was added to the third row and
1 times the second row was added to the fourth row.






 


10 2 7 11
01 7 10 13
00 1 1 1
0 0 15 20 25



The third row was multiplied by

1
14
.





 


10 2 7 11
01 710 13
00 1 1 1
00 0 5 10



15 times the third row was added to the fourth row.





 


10 2 711
01 71013
00 111
00012



The fourth row was multiplied by

1
5
.

28. 1.2 Gaussian Elimination 28









10 20 3
01 70 7
00 10 1
00 01 2





The fourth row was added to the third row,
10 times the fourth row was added to the second,
and 7 times the fourth row was added to the first.







1000 1
0100 0
0010 1
0001 2




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

Unique solution:

1
1I, 
2
0I, 
3
1I,

4
2I.
22.


 

 


00 1110
112310
112010
22 1010

The augmented matrix for the system.




 




112010
112310
00 1110
22 1010



The first and third rows were interchanged.








11 2 0 10
000300
00 1 1 10
00 3 0 30




The first row was added to the second row
and
2 times the first row was added to the last row.




 


11 2 0 10
00 1 1 10
000300
00 3 0 30



The second and third rows were interchanged.




 


11 2 0 10
00 1 1 10
000300
000300



3 times the second row was added to the fourth row.







11 2 0 10
00 1 1 10
00 0 1 00
000300



The third row was multiplied by

1
3
.

29. 1.2 Gaussian Elimination 29








11 20 10
00 11 10
00 01 00
00 00 00



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







11 20 10
00 10 10
00 01 00
00 00 00



1 times the third row was added to the second row.







110010
001010
000100
000000



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

If we assign
2
Z and
5
Z the arbitrary values s and t, respectively, the general solution is given by the formulas
     
12345
, , , 0, Z st Z s Z t Z Z t .
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
t, then back-substitution can be used to solve for the first two unknowns).

(c) The system is inconsistent since the third equation
01 is contradictory.

(d) There is insufficient information to decide whether the system is consistent as illustrated by these examples:

For



 

1
0000
001
the system is consistent with infinitely many solutions.

For





1
0010
0011
the system is inconsistent (the matrix can be reduced to





1
0010
0001
).
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 1 times the first row to the second yields





1000
0001
1
; the second
equation
01 is contradictory).

(d) There is insufficient information to decide whether the system is consistent as illustrated by these examples:

30. 1.2 Gaussian Elimination 30


For





1001
1001
1001
the system is consistent with infinitely many solutions.

For





1002
1001
1001
the system is inconsistent (the matrix can be reduced to





1002
0001
0000
).
25.




 

2
12 3 4
31 5 2
41 14 2
aa




The augmented matrix for the system.








2
12 3 4
07 1 410
07 2 14
aa




3 times the first row was added to the second row
and 4 times the first row was added to the third row.





 

2
12 3 4
07 1 410
00 16 4
aa



1 times the second row was added to the third row.





 

10
7
2
12 3 4
01 2
00 16 4
aa



The second row was multiplied by

1
7
.

The system has no solutions when
4a (since the third row of our last matrix would then correspond to a
contradictory equation
08).
The system has infinitely many solutions when
4a (since the third row of our last matrix would then correspond
to the equation
00).
For all remaining values of
a (i.e., 4a and
4a) the system has exactly one solution.

26.




 

2
12 1 2
22 3 1
12( 3)
aa




The augmented matrix for the system.






  

2
12 1 2
06 1 3
00 2 2
aa




2 times the first row was added to the second row
and 1 times the first row was added to the third row.





  

11
62
2
12 1 2
01
00 2 2
aa



The second row was multiplied by

1
6
.

31. 1.2 Gaussian Elimination 31

The system has no solutions when 2a or 2a (since the third row of our last matrix would then correspond
to a contradictory equation).
For all remaining values of
a (i.e.,
2a and 2a ) the system has exactly one solution.
There is no value of
a for which this system has infinitely many solutions.
27.



 

13 1
11 2
02 3a
b
c




The augmented matrix for the system.






 

131
023
023 a
ab
c



1 times the first row was added to the second row.





  

131
023
000 a
ab
abc



The second row was added to the third row.





  

3
222
13 1
01
00 0
ab
a
abc



The second row was multiplied by

1
2
.

If
  0abc then the linear system is consistent. Otherwise (if
 0abc ) it is inconsistent.
28.




 

131
121
371 a
b
c




The augmented matrix for the system.






 

131
0 1 2
024 3 a
ab
ac




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





  

131
012
000 2 a
ab
abc



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

If
 20abc then the linear system is consistent. Otherwise (if
20abc ) it is inconsistent.
29.


21
36
a
b



The augmented matrix for the system.




11
22
1
36 a
b



The first row was multiplied by
1
2
.

32. 1.2 Gaussian Elimination 32






11
22
93
22
1
0 a
ab



3 times the first row was added to the second row.





11
22
12
39
1
01
a
ab



The third row was multiplied by
2
9
.





21
39
12
39
10
01
ab
ab




1
2
times the second row was added to the first row.

The system has exactly one solution:

21
39
xab and  
12
39
yab .
30.





111
202
033 a
b
c




The augmented matrix for the system.








111
0202
033 a
ab
c



2 times the first row was added to the second row.







2
111
010
033
b
a
a
c



The second row was multiplied by

1
2
.





  

2
3
2
111
010
003 3
b
a
a
abc



3 times the second row was added to the third row.





 

2
23
111
010
001
b
bc
a
a
a



The third row was multiplied by
1
3
.





 

23
2
23
1102
010
001
bc
b
bc
a
a
a



1 times the third row was added to the first row.





 

3
2
23
100
010
001
c
b
bc
a
a
a



1 times the second row was added to the first row.

The system has exactly one solution: 
1 3
c
xa , 
2 2
b
xa , and 
3 23
bc
xa .
31. Adding 2 times the first row to the second yields a matrix in row echelon form



13
01
.

33. 1.2 Gaussian Elimination 33

Adding
3 times its second row to the first results in



10
01
, which is also in row echelon form.
32.






21 3
0229
34 5












21 3
0229
13 2



1 times the first row was added to the third row.







13 2
0229
21 3



The first and third rows were interchanged.







13 2
0229
05 1



2 times the first row was added to the third row.







13 2
0229
0186



3 times the second row was added to the third row.






13 2
0186
0229



The second and third rows were interchanged.






13 2
01 86
0 0 143



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






13 2
0186
00 1



The third row was multiplied by
1
143
.






130
010
001




86 times the third row was added to the second row
and 2 times the third row was added to the first row.






100
010
001



3 times the second row was added to the first row.

33. We begin by substituting sinx , cosy , and tanz so that the system becomes

34. 1.2 Gaussian Elimination 34




  
230
2530
550xyz
xyz
xyz







1230
2 530
1 550




The augmented matrix for the system.








1230
0130
0380




2 times the first row was added to the second row
and the first row was added to the third row.





 

12 30
01 30
00 10



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







12 30
01 30
00 10



The third row was multiplied by
1.






1200
0100
0010




3 times the third row was added to the second row and
3 times the third row was added to the first row.






1000
0100
0010



2 times the second row was added to the first row.

This system has exactly one solution
0, 0, 0.xyz
On the interval
02 , the equation sin 0 has three solutions: 0, , and 2.
On the interval 02 , the equation cos 0 has two solutions:


2
and


3
2
.
On the interval
02 , the equation tan 0 has three solutions: 0, , and 2.
Overall,
323 18 solutions 
, , can be obtained by combining the values of , , and  listed above:




0, ,0 , , ,0
22
, etc.
34. We begin by substituting
sinx , cosy , and tanz so that the system becomes



233
4222
63 9xyz
xyz
xyz

35. 1.2 Gaussian Elimination 35








2133
4222
6319




The augmented matrix for the system.






 

2133
0484
0080




2 times the first row was added to the second row
and 3 times the first row was added to the third row.







2133
0484
00 10



The third row was multiplied by 
1
8
.







2103
0404
0010




8 times the third row was added to the second row
and
3 times the third row was added to the first row.







2103
0101
0010



The second row was multiplied by
1
4
.







200 2
010 1
001 0



The second row was added to the first row.







100 1
010 1
001 0



The first row was multiplied by
1
2
.

This system has exactly one solution
1, 1, 0.xy z
The only angles
, , and  that satisfy the inequalities 02, 02, 0 and the equations
 sin1, cos 1, tan0
are


2
, , and 0.
35. We begin by substituting 
2
Xx, 
2
Yy, and 
2
Zz so that the system becomes

 
 
 
6
22
23XY Z
XY Z
XY Z






 

1116
1122
2113




The augmented matrix for the system.

36. 1.2 Gaussian Elimination 36








1116
0214
0139




1 times the first row was added to the second row
and 2 times the first row was added to the third row.







1116
0139
0214




The second and third rows were interchanged
(to avoid introducing fractions into the second row).






1116
0139
0214



The second row was multiplied by
1.






111 6
013 9
00714



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






1116
0139
0012



The third row was multiplied by
1
7
.






1104
0103
0012




3 times the third row was added to the second row
and 1 times the third row was added to the first row.






1001
0103
0012



1 times the second row was added to the first row.

We obtain




11
33
22
Xx
Yy
Zz

36. We begin by substituting 
1
x
a, 
1
y
b, and 
1
z
c so that the system becomes

 
 
  
241
23 80
9105ab c
ab c
ab c







12 4 1
23 80
19105




The augmented matrix for the system.

37. 1.2 Gaussian Elimination 37








12 4 1
01162
011 6 6




2 times the first row was added to the second row
and the first row was added to the third row.







12 41
01162
011 66



The second row was multiplied by
1.





 

12 4 1
01 16 2
0 0 182 16



11 times the second row was added to the third row.





 

8
91
12 4 1
01 16 2
00 1



The third row was multiplied by
1
182
.

Using back-substitution, we obtain

   
    
     
8191
91 8
54 1 91
216
91 54
7113
12 4
13 7
cz
c
bc y
b
abc x
a

37. Each point on the curve yields an equation, therefore we have a system of four equations










equation corresponding to 1,7 : 7
equation corresponding to 3, 11 : 27 9 3 11
equation corresponding to 4, 14 : 64 16 4 14
equation corresponding to 0,10 : 10
abcd
abcd
abcd
d






 


1111 7
27 9 3 1 11
64 16 4 1 14
000110




The augmented matrix for the system.









1111 7
0182426200
0 48 60 63 462
000110




27 times the first row was added to the second row
and 64 times the first row was added to the third.

38. 1.2 Gaussian Elimination 38








13 1004
39 9
1111 7
01
0486063462
000110



The second row was multiplied by

1
18
.







13 1004
39 9
19214
33
111 1 7
01
004
000 110



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







13 1004
39 9
19 107
12 6
111 1 7
01
001
000 110



The third row was multiplied by
1
4
.








104
33
1110 3
01 0
0010 2
0001 10






19
12
times the fourth row was added to the third row,

13
9
times the fourth row was added to the second row,
and
1 times the fourth row was added to the first.








1100 5
0100 6
0010 2
000110





4
3
times the third row was added to the second row and
1 times the third row was added to the first row.








1000 1
0100 6
0010 2
000110



1 times the second row was added to the first row.

The linear system has a unique solution: 1a, 6b , 2c, 10d . These are the coefficient values required
for the curve

32
yax bx cxd to pass through the four given points.
38. Each point on the curve yields an equation, therefore we have a system of three equations








equation corresponding to 2,7 : 53 2 7 0
equation corresponding to 4,5 : 41 4 5 0
equation corresponding to 4, 3 : 25 4 3 0abcd
abcd
abcd

The augmented matrix of this system




 

53 2 7 1 0
41 4 5 1 0
25 4 3 1 0
has the reduced row echelon form

39. 1.2 Gaussian Elimination 39

1
29
2
29
4
29
100 0
010 0
001 0
 
 

 
 
 

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

   
124
, , ,
29 29 29
atbtctdt
(For instance, letting the free variable
d have the value
29 yields 1a, 2b , and 4c .)
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
 
 
 
 
 
1000
0100
0010
.
Applying the
same sequence of elementary row operations to the augmented matrix of the nonhomogeneous system
yields the reduced row echelon form





100
010
001 r
s
t
where r, s, and t 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 B 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:






100
010
001
,





10
01
000r
s,





10
001
000r
,
 
 
 
 
 
1
000
000
rs
,
 
 
 
 
 
010
001
000
,
 
 
 
 
 
01
000
000
r
,
 
 
 
 
 
001
000
000
, and
 
 
 
 
 
000
000
000

where r and
s can be any real numbers.

(b) There are sixteen possible reduced row echelon forms:

40. 1.2 Gaussian Elimination 40








1000
0100
0010
0001
,






100
010
001
0000 r
s
t
,






10 0
01 0
0001
0000r
s
,
 
 
 
 
 
 
10
01
0000
0000
rt
su
,
 
 
 
 
 
 
100
0010
0001
0000
r
,






10
001
0000
0000rs
t
,






10
0001
0000
0000rs
,






1
0000
0000
0000rst
,






0100
0010
0001
0000
,
 
 
 
 
 
 
010
001
0000
0000
r
s
,
 
 
 
 
 
 
01 0
0001
0000
0000
r
,






01
0000
0000
0000rs
,






0010
0001
0000
0000
,






001
0000
0000
0000
r
,






0001
0000
0000
0000
, and
 
 
 
 
 
 
0000
0000
0000
0000
.
where r,
s, t, and u 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)
0a, and (ii) 0a.
(i) If 0a then the assumption 0ad bc implies that 0b and 0c. Gauss-Jordan elimination yields



0b
cd




We assumed
0a





0
cd
b



The rows were interchanged.




1
01
d
c




The first row was multiplied by
1
c
and
the second row was multiplied by
1
.
b
(Note that ,0.)bc



10
01




d
c
times the second row was added to the first row.

(ii) If
0a then we perform Gauss-Jordan elimination as follows:



ab
cd









1
b
a
cd



The first row was multiplied by
1
a
.





1
0
b
a
ad bc
a



c times the first row was added to the second row.

41. 1.2 Gaussian Elimination 41





1
01
b
a




The second row was multiplied by

a
ad bc
.
(Note that both
a and
ad bc are nonzero.)



10
01




b
a
times the second row was added to the first row.

In both cases (
0a as well as 0a) we established that the reduced row echelon form of
 
 
 
ab
cd
is



10 01

provided that
0ad bc .

(b) Applying the same elementary row operation steps as in part (a) the augmented matrix


abk
cdl
will be
transformed to a matrix in reduced row echelon form
 
 
 
10
01
p
q
where
p and q are some real numbers. We
conclude that the given linear system has exactly one solution: xp, yq.
True-False Exercises
(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



10
01
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 Section 1.2).
(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
0nn 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
2
xyz
xyz

1.3 Matrices and Matrix Operations
1. (a) Undefined (the number of columns in B does not match the number of rows in
A)

(b) Defined; 44 matrix

42. 1.3 Matrices and Matrix Operations 42


(c) Defined; 42 matrix

(d) Defined; 52 matrix

(e) Defined; 45 matrix

(f) Defined; 55 matrix
2. (a) Defined; 54 matrix

(b) Undefined (the number of columns in
D does not match the number of rows in C)

(c) Defined; 42 matrix

(d) Defined; 24 matrix

(e) Defined; 52 matrix

(f) Undefined (
T
BA is a 44 matrix, which cannot be added to a 42 matrix D)
3. (a) 


   



16 5123 765
1 1 0112 2 13
34 2143 737


(b) 


     



16 5123 5 4 1
1 1 0112 0 1 1
34 2143 1 1 1


(c) 


   



53 50 15 0
5152 510
51 51 5 5


(d)
     


     
71 74 72 7 28 14
73 71 75 21 7 35


(e) Undefined (a 23 matrix C cannot be subtracted from a
22 matrix 2B)

(f)   
  
 
           
 
  
 
4 6 4 1 4 3 2 1 2 5 2 2 24 2 4 10 12 4
414142 212021 4 24082
44 41 43 23 22 24 16 6 4 4 12 8








22 6 8
246
10 0 4


(g)  
    
  
      
  
    
  
152 26 2123 112 5226
3101 212122 31 20214
324 24 2123 38 2246

     

     

     

313 37 38 39 21 24
33 3235 9 615
3 11 3 4 3 10 33 12 30

43. 1.3 Matrices and Matrix Operations 43

(h) 


  



33 00 00
112200
11 11 0 0


(i) 1045

(j)  
    
  

  
    
  
152 36 3133 118 5329
tr 1 0 1 3 1 3 1 3 2 tr 1 3 0 3 1 6
324 34 3133 312 2349








17 2 7
tr 2 3 5 1735 25
915


(k)



  
  
  
28 774 7 1
4tr 4tr 4 28 14 4 42 168
01470 72


(l) Undefined (trace is only defined for square matrices)
4.
(a)
     
 
 
    
311 142 724 23 1 2 1 4 21 2
2
021 315 357 20 3 22 1 21 5


(b)
  
  

  
   
   
113 614 161 134 501
5 02 1 11 51 01 21 4 1 1
2 14 3 23 23 1243 1 1 1


(c) 
  
 
  
 
   
 
16 5123 5 4 1 5 0 1
1 1 0112 0 1 1 4 1 1
34 2143 111 111
TT


(d) Undefined (a 22 matrix
T
B cannot be added to a 32 matrix 5
T
C)

(e)
      
 

 
    
  
33 311 1 1 1 1
22 4 4 242 42
911 1 1 111
22 4 4 422 4
53911 1 1 1 1
22 4 4 424 44
13 3 0 0
41 (1)2 2 0
25 1 1 1


(f)

 
   
   
44 1041 40 01
012202 12 10


(g)
 



    
 
614 113 262124 313133
2 1 1 1 35 0 2 21 21 21 35 30 32
3 2 3 2 1 4 23 22 23 32 31 34

   
 
   
 
   

12 3 2 3 8 9 9 1 1
21520261324
66 43 612 01 6

44. 1.3 Matrices and Matrix Operations 44


(h)
  

  

   
       
 
614 113 262124 313133
21 11 35 02 2121 21 3530 32
3 2 3 2 1 4 23 22 23 32 31 34
TT


   
  
    
     
    
     
  
1232 3 89 911 9130
215 20 26 132 4 1 2 1
66 43 612 01 6 1 4 6
T T


(i)
   

 
          
 

 
       

 
 
613
11 41 23 15 40 22 12 41 24
112
31 11 53 35 10 52 32 11 54
413


 

 



613
3914
112
17 25 27
413


  

       

        

36 91 144 31 91 141 33 92 143
17 6 25 1 27 4 17 1 25 1 27 1 17 3 25 2 27 3







65 26 69
185 69 182


(j) Undefined (a 22 matrix B cannot be multiplied by a
32 matrix A)
(k)
 





1526 14
tr 1 0 1 1 1 1
3243 23




 
  
       

   

          

16 51 23 11 51 22 14 51 23
tr 16 01 13 11 01 12 14 01 13
36 21 43 31 21 42 34 21 43




 



17 8 15
tr 3 3 1 17 3 26
32 7 26
46

(l) Undefined (
BC is a 23 matrix; trace is only defined for square matrices)
5. (a)


   
 
  
     
  
   
 
34 00 31 02 12 3
14 20 11 22 4 5
14 10 11 12 4 1


(b) Undefined (the number of columns of B does not match the number of rows in
A)

45. 1.3 Matrices and Matrix Operations 45


(c) 


   



36 31 33 1 5 2
313132 101
34 31 33 3 2 4




     
 
       

  

       

1813193 1853092 1823194
31 31 63 35 30 62 32 31 64
12 1 3 1 9 3 12 5 3 0 9 2 12 2 3 1 9 4








42 108 75
12 3 21
36 78 63


(d)


    
  
  
  
 
  

34 00 31 02 12 3
142 142
14 20 11 22 4 5
315 315
14 10 11 12 4 1

 

  
        

           

    

12 1 3 3 12 4 3 1 12 2 3 5
41 53 44 51 42 55
41 13 44 11 42 15








3459
11 11 17
71713


(e)
  
  
 
    
  
   
       
 
 
 
30 30
41 13 44 11 42 15 115 3
12 12
01 23 04 21 02 25 6210
11 11


   

      
            

           

   

31 06 315 02 33 010
11 2 6 115 2 2 13 210
11 16 115 12 13 110

 
 

 
 
 
3459
11 11 17
71713


(f)
    
  

       
 
  
  


13
11 44 22 13 41 25142 2117
41
31 14 52 33 11 55315 1735
25


(g)

 
  

 
       
  
  

13 51 21 10 52 21
0211
13 01 11 10 02 11
12 1 8
33 21 41 30 22 41
T