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CHAPTER 1: SYSTEMS OF LINEAR EQUATIONS AND MATRICES
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) This can be rewritten as , so it is linear in and .
(d) This is not a linear equation in and since appears to a negative power.
(e) This is not a linear equation in and since appears to a power that is not
a positive integer.
(f) This is a linear equation in and .
2. Equations in part (a) do not form a linear system since the second one is not a linear equation.
Equations in parts (b), (c), and (d) form linear systems.
3. (a) Not linear because of the term . (b) Linear
4. (b) We can solve this system by inspection since the second equation is a multiple of the first.
Letting solves both equations, therefore the system is consistent.
(c) An easy way to find an example of a solution of this system would be by letting first,
then look for values of and that solve the resulting system
4x + 2z –1
–x – 3z 0
Adding 4 times the second equation to the first will yield a simplified system
–10z –1
–x – 3z 0
which is solved by
and
. Together with , this is a solution of the original
system, showing it to be consistent.
(d) We add times the first equation to the third, then proceed to add the first equation to
the fourth, and the second one to the fourth, obtaining a simplified system
Chapter 1: Systems of Linear Equations and Matrices 2
Clearly, the last two equations are contradictory, thus the system is inconsistent.
5. (b) Subtracting the first equation from the second gives and the first
equation is Parameterizing we get the solution
((c) and (d) unchanged)
6. For example,
(a)
(b)
(c)
7. (a) No, the first equation is not satisfied by the vector: 3(5) + 2( 4) 2(0) = 7 1.
(b) Yes, all the equations are satisfied by the vector. For example, the first equation becomes
5 15 847 7 7
3 +2 2 0 = =1. Verify the other equations also are satisfied.
(c) No, the second equation is not satisfied by the vector.
(d) Yes, all the equations are satisfied by the vector.
(e) No, the second equation is not satisfied by the vector.
8. The values in (b), (d), and (e) satisfy all three equations – these vectors are solutions of the
system.
The vectors in (a) and (c) are not solutions of the system.
9. (a) Solving the equation for y, we get y x,3 14 2
= so parameterizing gives us x = t, 3 14 2
=y t.
(b) Parameterize the last three variables, and solve for x1 in terms of those parameters:
5 1 41 3 3 3= 3 + x r s t, x2 = r, x3 = s, x4 = t.
10. (a)
where and are arbitrary values
(b)
where , , , and are arbitrary values
Chapter 1: Systems of Linear Equations and Matrices 3
11. (a) 2 +5 =6x y (c) x1 + 5x2 + 7x3 x4 = 3
y = 2 2x1 + 2x2 + x3 + x4 = 0 x = 0
12. (a)
(b)
(c)
(d)
13. (b)
3 0 1 6 0
0 2 1 5 2
(d)
1 0 1 0 4
0 1 0 1 9
14. (a)
(b)
(c)
(d)
Gaussian Elimination
16. (a) Row echelon form only; Property 4 is not satisfied.
(b) Neither; Property 3 is not satisfied.
(c) Reduced row echelon form (and so also row echelon form).
(d) Neither; Property 2 is not satisfied.
(e) Reduced row echelon form (and so also row echelon form).
(f) Neither; Property 1 is not satisfied.
(g) Reduced row echelon form (and so also row echelon form).
18. (a) x1 = 0, x2 = 2, x3 = 4.
(b) Because the column representing x4 does not have a leading 1, we parameterize it, x4 = t.
The last equation is then x3 4t = 1, or x3 = 1 + 4t. The second equation becomes x2 = t,
and the first equation is x1 = 2 3t. Thus, the parameterized solution is x1 = 2 3t, x2 = t,
x3 = 1 + 4t, x4 = t.
Chapter 1: Systems of Linear Equations and Matrices 4
(c) Variables x3 and x5 are parameterized because the third and fifth columns do not have
leading ones. The remaining variables are solved for in terms of x3 and x5. Let x3 = s, and
x5 = t. Then x4 = 7t, x2 = 1 + 2s, and x1 = 2 2t. Thus the parameterized solution is
x1 = 2 2t, x2 = 1 + 2s, x3 = s, x4 = 7t, and x5 = t.
(d) The bottom row corresponds to the equation 0 = 1, so there are no solutions.
19. The augmented matrix for this system is
1 2 3 6
2 1 4 1
1 1 1 3
.
Replace R2 with R2 2R1 and replace R3 with R3 R1.
1 2 3 6
0 5 10 11
0 3 4 3
Replace R2 with 15 R2.
115
1 2 3 6
0 1 2
0 3 4 3
Replace R3 with R3 + 3R2.
115
185
1 2 3 6
0 1 2
0 0 2
Replace R3 with 12 R3.
115
95
1 2 3 6
0 1 2
0 0 1
The matrix is now in Reduced Echelon form. To get it into Reduced Row Echelon form, we
replace R2 with R2 + 2R3 and replace R1 with R1 + 3R3.
35
75
95
1 2 0
0 1 0
0 0 1
Finally, replace R1 with R1 2R2.
175
75
95
1 0 0
0 1 0
0 0 1
.
The solution is x1 = 175 , x2 = 7
5 , x3 = 95 .
Chapter 1: Systems of Linear Equations and Matrices 5
20.
11
2 2 2 4
2 5 2 1
8 1 4
The augmented matrix for the system
11
1 1 1 2
2 5 2 1
8 1 4
The first row was multiplied by
11
1 1 1 2
0 7 4 5
8 1 4
2 times the first row was added to the second row
5
1 1 1 2
0 7 4 5
0 7 4
times the first row was added to the third row
547 7
1 1 1 2
0 1
0 7 4 5
The second row was multiplied by
547 7
1 1 1 2
0 1
0 0 0 0
7 times the second row was added to the third row
3 97 7
547 7
1 0
0 1
0 0 0 0
times the second row was added to the first row
There are infinitely many solutions:
,
, .
21. The augmented matrix is
3 1 1 7 13
2 1 1 3 9
2 1 0 7 8
.
To get a leading one in the first row, we replace R1 with R1 + R2.
1 0 0 4 4
2 1 1 3 9
2 1 0 7 8
Replace R2 with R2 + 2R1 and replace R3 with R3 + 2R3.
1 0 0 4 4
0 1 1 5 1
0 1 0 1 0
Chapter 1: Systems of Linear Equations and Matrices 6
Replace R3 with R3 R2.
1 0 0 4 4
0 1 1 5 1
0 0 1 4 1
The matrix is now row reduced. To get it in
reduced row echelon form replace R2 with R2 + R3.
1 0 0 4 4
0 1 0 1 0
0 0 1 4 1
Since the fourth column does not have a leading one, the variable corresponding to the column,
w, will be parameterized, w = t. The solution is x = 4 4t, y = t, z = 1 + 4t, w = t.
22.
0 2 3 3
3 6 3 2
6 6 3 4
The augmented matrix for the system
3 6 3 2
0 2 3 3
6 6 3 4
The first and second rows were interchanged
23
1 2 1
0 2 3 3
6 6 3 4
The first row was multiplied by
23
1 2 1
0 2 3 3
0 6 9 8
–6 times the first row was added to the third row
23
3 32 2
1 2 1
0 1
0 6 9 8
The second row was multiplied by
23
3 32 2
1 2 1
0 1
0 0 0 1
6 times the second row was added to the third row
23
3 32 2
1 2 1
0 1
0 0 0 1
The third row was multiplied by –1
6
5
3 32 2
1 0 2
0 1
0 0 0 1
–2 times the second row was added to the first row
3 32 2
1 0 2 0
0 1
0 0 0 1
–
times the third row was added to the first
row
Chapter 1: Systems of Linear Equations and Matrices 7
32
1 0 2 0
0 1 0
0 0 0 1
times the third row was added to the first row
The third row represents the equation which has no solutions. Thus the
original system is inconsistent. Note that this was already evident four steps earlier; there was
no need to continue reducing.
23. The reduced echelon form of the matrix from Exercise 19 is
115
95
1 2 3 6
0 1 2
0 0 1
. To solve using
back-substitution, we let 93 5
=x . Row 2 becomes 9 112 5 5
2 =x and solving for x2 yields
1811 72 5 5 5
= =x . Substituting for x3 and x2 in Row 1 yields 971 5 5
+2 3 =6x , and thus,
14 27 171 5 5 5=6 + =x . The solution is x1 = 17
5 , x2 = 75 , x3 = 9
5 .
24.
11
2 2 2 4
2 5 2 1
8 1 4
The augmented matrix for the system
11
1 1 1 2
2 5 2 1
8 1 4
The first row was multiplied by
11
1 1 1 2
0 7 4 5
8 1 4
2 times the first row was added to the second row
1 1 1 2
0 7 4 5
0 7 4 5
–8 times the first row was added to the third row
547 7
1 1 1 2
0 1
0 7 4 5
The second row was multiplied by
547 7
1 1 1 2
0 1
0 0 0 0
7 times the second row was added to the third row
The system of equations corresponding to this augmented matrix is
Chapter 1: Systems of Linear Equations and Matrices 8
Solve these equations for the leading variables, giving
then substitute the second equation into the first, giving
If we assign an arbitrary value t, the general solution is given by the formulas
25. The reduced echelon form of the matrix from Exercise 21 is
1 0 0 4 4
0 1 1 5 1
0 0 1 4 1
. The fourth
w is parameterized, w = t. Then back-substitution gives z = 1 + 4t. Row 2 becomes y (1 + 4t) + 5t = 1, so y = t.
Solving row 1 for x yields x = 4 4t. The solution is x = 4 4t, y = t, z = 1 + 4t, w = t.
26.
0 2 3 3
3 6 3 2
6 6 3 4
The augmented matrix for the system
3 6 3 2
0 2 3 3
6 6 3 4
The first row and second rows were interchanged
23
1 2 1
0 2 3 3
6 6 3 4
The first row was multiplied by
23
1 2 1
0 2 3 3
0 6 9 8
–6 times the first row was added to the third row
23
3 32 2
1 2 1
0 1
0 6 9 8
The second row was multiplied by
Chapter 1: Systems of Linear Equations and Matrices 9
23
3 32 2
1 2 1
0 1
0 0 0 1
6 times the second row was added to the third row
23
3 32 2
1 2 1
0 1
0 0 0 1
The third row was multiplied by –1
The system of equations corresponding to this augmented matrix is
This system is clearly inconsistent.
28. The third equation forces then substituting into the second equation forces
Then the first equation implies . So the only solution is the trivial solution.
30. The second equation is a multiple of the first one.
This system has nontrivial solutions (e.g., and ).
31. The augmented matrix is
2 1 4 0
3 1 6 0
4 1 9 0
. Replace R1 with R2 R1
1 0 2 0
3 1 6 0
4 1 9 0
Replace R2 with R2 3R1 and replace R3 with R3 4R1.
1 0 2 0
0 1 0 0
0 1 1 0
Replace R3 with R3 R2.
1 0 2 0
0 1 0 0
0 0 1 0
For reduced row echelon form replace R1 with R1 2R3.
1 0 0 0
0 1 0 0
0 0 1 0
The only solution is the trivial solution x1 = 0, x2 = 0, x3 = 0.
32.
1 1
1
3 0
1 2 0
1 1 1 0
The augmented matrix for the system
Chapter 1: Systems of Linear Equations and Matrices 10
2 1
3 1
1 0
1 0
1 1 1 0
The first and second rows were reversed
2 1
1
1 0
3 1 0
1 1 1 0
The first row was multiplied by –1
2 1
4
1 0
0 7 0
1 1 1 0
–3 times the first row was added to the second row
2 1
4
1 0
0 7 0
0 3 2 0
–1 times the first row was added to the third row
2 1
47
1 0
0 1 0
0 3 2 0
The second row was multiplied by
1
0
47
27
1 2 0
0 1 0
0 0
–3 times the second row was added to the third row
1
47
1 2 1 0
0 1 0
0 0 0
The third row was multiplied by
17
47
1 0 0
0 1 0
0 0 1 0
times the second row was added to the first row.
0
47
1 0 0
0 1 0
0 0 1 0
times the third row was added to the first
row
0
0
1 0 0
0 1 0
0 0 1 0
times the third row was added to the
second row
The only solution is , , .
Chapter 1: Systems of Linear Equations and Matrices 11
33. The augmented matrix is
1 1 7 4 0
1 2 6 1 0.
Replace R2 with R2 R1
1 1 7 4 0
0 3 13 5 0.
Replace R2 with 13
R2.
13 53 3
1 1 7 4 0
0 1 0.
For row reduced echelon form replace R1 with R1 + R2.
8 73 3
13 53 3
1 0 0
0 1 0.
Parameterize x3 = s and x4 = t. Then 13 52 3 3
= +x s t and .8 71 3 3=x s t
The solution is ,8 71 3 3=x s t ,13 5
2 3 3=x s t x3 = s, x4 = t. (Or, let x3 = 3S, x4 = 3T, and then
x2 = 13S + 5T and x1 = 8S 7T.
34.
0 1 2 2 0
2 1 4 3 0
4 1 6 4 0
2 2 6 5 0
The augmented matrix for the system
2
2 5
2 1 4 3 0
0 1 2 0
4 1 6 4 0
2 6 0
The first and second rows were interchanged
2
1
2 6 5
312 2
1 0
0 1 2 2 0
4 6 4 0
2 0
The first row was multiplied by
2
1
312 21 2 0
0 1 2 2 0
0 1 2 0
0 2 2 0
2 times the first row was added to the fourth row, and –4 times the first row was added to the third row
312 2
1 2 0
0 1 2 2 0
0 0 0 0 0
0 0 0 0 0
–1 times the second row was added to each of the third and fourth rows
1
2 2
12
1 0 0
0 1 0
0 0 0 0 0
0 0 0 0 0
times the second row was added to the first
row
Chapter 1: Systems of Linear Equations and Matrices 12
The general solution is then
36.
1 3 0 1 0
1 4 2 0 0
0 1 1 1 0
2 4 1 1 0
1 2 1 1 0
The augmented matrix for the system
1
1 3 0 1 0
0 7 2 1 0
0 1 1 1 0
2 4 1 0
1 2 1 1 0
The first row was added to the second row
3
2 1
1 3 0 1 0
0 7 2 1 0
0 1 1 1 0
0 10 1 0
1 1 0
–2 times the first row was added to the fourth row
0 3
5 2
1 3 0 1 0
0 7 2 1 0
0 1 1 1 0
0 1 1 0
0 1 0
–1 times the first row was added to the fifth row
7 2
0
5
1 3 0 1 0
0 1 1 1 0
0 1 0
0 1 1 3 0
0 1 2 0
The second and third rows were interchanged
0
5
1 3 0 1 0
0 1 1 1 0
0 7 2 1 0
0 1 1 3 0
0 1 2 0
The second row was multiplied by –1
1
5
0
5
1 3 0 1 0
0 1 1 0
0 0 8 0
0 1 1 3 0
0 1 2 0
–7 times the second row was added to the third row
Chapter 1: Systems of Linear Equations and Matrices 13
1
5
0
5
1 3 0 1 0
0 1 1 0
0 0 8 0
0 11 13 0
0 1 2 0
10 times the second row was added to the fourth row
1
5
0
0 4
1 3 0 1 0
0 1 1 0
0 0 8 0
0 11 13 0
0 7 0
5 times the second row was added to the fifth row
1
1
0
0 4
85
1 3 0 1 0
0 1 1 0
0 0 0
0 11 13 0
0 7 0
The third row was multiplied by
1
1
0
0 4
85
235
1 3 0 1 0
0 1 1 0
0 0 0
0 0 0
0 7 0
–11 times the third row was added to the fourth row
3
1
1
0
0 0
85
235
5
1 3 0 1 0
0 1 1 0
0 0 0
0 0 0
0 0
–4 times the third row was added to the fifth row
3
1
1
0 1
0 0
85
5
1 3 0 1 0
0 1 1 0
0 0 0
0 0 0
0 0
The fourth row was multiplied by
1
1
0 1
0 0 0
85
1 3 0 1 0
0 1 1 0
0 0 0
0 0 0
0 0
times the fourth row was added to the
fifth row
Chapter 1: Systems of Linear Equations and Matrices 14
We could continue, placing the matrix into reduced row echelon form, but it is clear from this row
echelon form that from the fourth row, which implies that from the third row.
Then the second and first rows imply that and Thus the only
solution is
38. The augmented matrix is
1 2 1 2
2 2 3 1
1 2 a a
.
Replace R2 with R2 2R1 and replace R3 with R3 R1.
1 2 1 2
0 6 1 3
0 0 1 2a a
.
The system will have no solutions if a = 1, as this would make the equation for R3: 0 = 3.
If a 1, then replace R3 with 1
1
aR3. .
aa
2+1
1 2 1 2
0 6 1 3
0 0 1
At this point, one can see that there
will be a unique solution to the system for all a 1.
39.
( )
2
1 2 1 2
2 2 3 1
1 2 3a a
The augmented matrix for the system.
2
1 2 1 2
0 6 1 3
0 0 +2 2a a
–2 times the first row was added to the second row and –1 times the first row was added to the third row.
1 16 2
2
1 2 1 2
0 1
0 0 2 2
a a
The second row was multiplied by
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.
40.
( )
2
1 2 3 4
3 1 5 2
4 1 2 + 4a a
Chapter 1: Systems of Linear Equations and Matrices 15
Replace R2 with R2 3R1 and replace R3 with R3 4R1.
( )
2
1 2 3 4
0 7 14 10
0 7 +10 12a a
Replace R3 with R3 R2.
( )
2
1 2 3 4
0 7 14 10
0 0 4 2a a
Replace R2 with 17
R2. (This step is not actually necessary, but it is the usual next step in putting
the matrix in RRE form.)
( )
107
2
1 2 3 4
0 1 2
0 0 4 2a a
.
If a = 2, R3 becomes 0 = 4, so the system is inconsistent.
If a = 2, R3 becomes 0 = 0 so the third variable would be parameterized, and the system has
infinitely many solutions.
For any other value of a, the system will have a unique solution.
41. 11
2
1 1 7 7
2 3 17
1 2 +1 6a a
The augmented matrix for the system.
25
a 6a
2
1 1 7 7
0 1 3
1 2 +1
–2 times the first row was added to the second row
25
2
1 1 7 7
0 1 3
0 1 6 6 +7a a
–1 times the first row was added to the third row.
25
9 18
2
1 1 7 7
0 1 3
0 1 6a a
–1 times the second row was added to the third row
The system has no solutions if because then the third row of this matrix corresponds to an equation with no solutions. If then the third row of the matrix is zero, and the equation has an infinite number of solutions. Finally, if and then the equation has a unique solution.
Chapter 1: Systems of Linear Equations and Matrices 16
43. Compute the reduced row echelon form:
a
b
c
1 1 2
2 0 1
0 1 3
The augmented matrix for the system.
a
a
c
b
1 1 2
0 2 3 2
0 1 3
–2 times the first row was added to the second row.
a
a
c
b3 12 2
1 1 2
0 1
0 1 3
The second row was multiplied by
.
a
a
a b
b
c
3 12 2
3 12 2
1 1 2
0 1
0 0 +
–1 times the second row was added to the third row.
3
2
a
a
a
b
c b
12 2
2 13 3 3
1 1 2
0 1
0 0 1 +
The third row was multiplied by
.
2 2 13 3 3
2
a
a
b c
c a b
1 1 2
0 1 0
0 0 1
–
times the third row was added to the
second row
7 2 43 3 3
2 2 13 3 3
2
a
b c
a b c
c a b
1 1 0
0 1 0
0 0 1
–2 times the third row was added to the first row.
1 1 13 3 3
2 2 13 3 3
2
a
b c
a b c
c a b
1 1 0
0 1 0
0 0 1
–1 times the second row was added to the first row
This system has exactly one solution for each set of values of a, b, and c:
46.
2
1 1 1 4
0 0 1 2
0 0 4 2a a
The augmented matrix for the system.
Chapter 1: Systems of Linear Equations and Matrices 17
2
1 1 1 4
0 0 1 2
0 0 0 2 + + 6a a
times the second row was added to the third row.
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.
47. Begin by substituting x = sin α, y = cos , and z = tan in the system, giving
Now compute the reduced row echelon form:
1 1 1
1 1
3
2
4 0
2 2 4
The augmented matrix for the system
1 1 1
1 2
2
2
0 3
2 3 4
Twice the first row was added to the second row
1 1 1
1 2
1
2
0 3
0 4 5
The first row was added to the third row
7 7
3 3
1 1 1
1 2
2
0 3
0 0
times the second row was added to the
third row
2
2 1 1 1
0 3 1
0 0 1 1
The third row was multiplied by
3
2 1 1 1
0 3 0
0 0 1 1
The third row was added to the second row
Chapter 1: Systems of Linear Equations and Matrices 18
1 1
1 1
1 1
2 1
0 0
0 0
The second row was multiplied by
2 1 0
1 1
1 1
0
0 0
0 0
−1 times the second row was added to the first row
2 0 1
1 1
1 1
0
0 0
0 0
The third row was added to the first row
11 0
2
1 1
1 1
0
0 0
0 0
The first row was multiplied by
This system has the unique solution
Thus we have
,
,
,
The only angles satisfying these requirements are
,
52. We break up the solution into three cases:
Case I: and
cos sin
sin cos
θ θ x
θ θ y
The augmented matrix corresponding to the system.
sincos cos1
sin cos
θ xθ θ
θ θ y
The first row was multiplied by
.
sincos cos
sin1cos cos
1
0
θ xθ θ
θθ θy x
times the first row was added to the
second row
.
sincos cos
1
0 1 cos sin
θ xθ θ
y θ x θ
The second row was multiplied by .
Chapter 1: Systems of Linear Equations and Matrices 19
1 0 cos + sin
0 1 cos sin
x θ y θ
y θ x θ
times the second row was added to the first
row
.
The system has exactly one solution: and .
Case II: which implies . The original system becomes ,
. Multiplying both sides of the each equation by yields
sin .
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.
53. (a) If the reduced row echelon form has no rows of zeros, then each of the four rows will have a leading 1, so the maximum possible number is 4.
(b) If all entries of B are zero, then any values of the variables will satisfy the system, so there are a maximum of 5 parameters, one for each variable.
(c) If each column contains a leading 1, there will still be three rows with no leading 1, so those rows are zero. The minimum possible number is therefore 3.
54. 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.
55. (a) Substitute y for t in the first and third equations, giving the system x = 3 + y and z = 7 − 2y. This obviously has the parametric solution given.
(b) In the parametrization in part (a), if we set x = 3 + t = r, then t = r − 3. Then that parametrization becomes x = r, y = r − 3, and z = 7 − 2t = 7 − 2(r − 3) = 28 − 2r. Note that this parametrization does in fact satisfy the system from part (a):
x = r = 3 + y = 3 + (r – 3) z = 28 – 2r = 7 – 2y = 7 – 2(r − 3)
Chapter 1: Systems of Linear Equations and Matrices 20
56. Clearly the second row is unaffected. Let the first row originally be R1 and the third be R3. Then
the effect of the operations is
First row Third row
R1 R3 + R1
R1 – (R3 + R1) = − R3 R3 + R1
− R3 R3 + R1 + (− R3) =R1
R3 R1
The effect is to interchange the first and third rows.
Matrices and Matrix Operations
58. (a) Defined; 4 4 matrix.
(b) Undefined; AT is 1 4, which cannot be multiplied by the 1 4 matrix B.
(c) Defined; BT is 5 4, and ET + A is 4 1, so the product is 5 1.
(d) Undefined; 3B is 4 5 while D is 3 5.
(e) Defined; B is 4 5, and C + DT is 5 3, so the product is 4 3.
(f) Undefined; EB is 1 5, so (EB)T is 5 1, while CD is 5 5.
(g) Defined; DBT is 3 4, so C(DBT ) is 5 4.
(h) Undefined; EA is 1 1, while DC is 3 3.
59. (a)
2 4 8
2 1 3
11 0 5
(b)
2 2 8
8 1 1
3 12 1
(c) 10 0
20 30
(d)
18 9 72
27 0 18
36 54 27
(e) Not possible since B and C are not the same size.
(f)
6 18 24
44 7 1
37 60 5
(g)
4 32 16
44 10 14
78 48 26
(h)
0 0 0
0 0 0 (i) 2 + 0 + 3 = 1
(j) 2 (k) Undefined since B is not square. (l) 8
60. (a) 2 14 4
26 46 8
(b) Not possible since B has more columns than A has rows.
(c)
27 0 18
51 33 105
36 15 222
Chapter 1: Systems of Linear Equations and Matrices 21
(d)
4 92 14 4 58 22
3 0 =26 46 8 50 226
2 1
(e) 2 0 29 11 58 22
=4 6 11 45 50 226
(so (AB)C = A(BC))
(f)
4 9 97 12 174 3 2
3 0 = 12 9 69 0 1
2 1 17 6 5
(g) 5 16 40
10 29 39
(h) Since CT is a 2 by 3 matrix, multiplication with the 2 by 3 matrix B is not defined.
(i) Trace
69 10 10
10 13 18
10 18 61
= 143 (j) 11
(k) The addition ATCT + 2ET is not defined.
(l) Trace
64 182 58
186 48 42
11 1 2
= 18.
62.
1 4 1
2 8 2
3 12 3
0 0 0
The original augmented matrix.
1 4 1
0 0 0
0 0 0
0 0 0
2 times the first row was added to the second row and times the first row was added to the third row.
This matrix is both in row echelon form and in reduced row echelon form. It corresponds to the
system of equations
If we assign an arbitrary value , the general solution is given by the formulas
Chapter 1: Systems of Linear Equations and Matrices 22
64.
3 1 2
9 3 6
6 2 1
The original augmented matrix.
3 1 2
0 0 0
0 0 5
3 times the first row was added to the second row and times the first row was added to the third row.
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
.)
66.
(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 ]
68.
(a) first column of
second column of
Chapter 1: Systems of Linear Equations and Matrices 23
third column of
69. (a)
5 1 1 2
= 2 0 3 , = = 1 ,
1 2 0 0
x
A x y b
z
so the equation is
5 1 1 2
2 0 3 = 1 .
1 2 0 0
x
y
z
(b)
1
2
3
4
1 1 1 7 6
0 1 4 1 = 1
4 2 1 8 0
x
x
x
x
70. (a)
,
,
; The matrix equation Ax = b is
(b)
,
,
. The matrix equation is
72. (a)
(b)
74.
Thus values of that satisfy the equation are and .
75. Setting the components of each matrix equal and moving the constants to the right hand side
gives
Chapter 1: Systems of Linear Equations and Matrices 24
= 3
+2 = 0
+2 =1
+ = 0
b
a c d
c d
a b
This is a system of 4 equations in 4 unknowns so can be solved by inspection or by writing it as a
linear system:
0 1 0 0 3
1 0 1 2 0
0 0 1 2 1
1 1 0 0 0
which row reduces to
1 0 0 0 3
0 1 0 0 3
0 0 1 0 1
0 0 0 1 1
. Thus, the solution is
76. The given matrix is equivalent to the linear system
Adding the second equation to the first gives so that a = 6; back-substituting then gives b =
−3. Adding twice the third equation to the fourth gives so that ; back-substituting
then gives . So the solution is and
80. (a) Suppose the jth column of B is a column of zeros. Using the column view of matrix
multiplication, the jth column vector of AB = A [ jth column vector of B] = A 0 = 0. Thus, AB has a
column of zeros.
(b) If A has a row of all zeros, then AB also has a row of all zeros.
81. (a) If aij = i j, the 4 4 matrix is
0 1 2 3
1 0 1 2
2 1 0 1
3 2 1 0
.
(b) If aij = ( 1)i ij, the 4 4 matrix is
1 2 3 4
2 4 6 8
3 6 9 12
4 8 12 16
.
Chapter 1: Systems of Linear Equations and Matrices 25
(c) If 0 1
=1 <1ij
i ja
i j
, the only non-zero entries come from | i j | = 0 < 1, so the 4 4
matrix is
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
.
84. The second row of A must be [0 1 0], and the third row must be [0 0 1], to get y and z
as the second and third components of the result. If the first row is [a b c], then
for all choices of x, y, and z. Hence c = 0 and a = ry, b = (1 r)x give
a solution for any real number r. Thus there are infinitely many choices:
85. Suppose that the entries of A are aij and those of B are bij .
(a) We have
(b) By formula (5), the ii component of AB is
(AB)ii = ai1 b1i + ai2 b2i + + ain bni,
so that
Chapter 1: Systems of Linear Equations and Matrices 26
86. Since tr(AB BA) = tr(AB) + tr( BA) = tr(AB) tr(BA) = 0 by the previous exercise, but
tr(I) = 1 + 1 = 2, the equation AB BA = I cannot hold.
Inverses; Algebraic Properties of Matrices
88. (a)
(b)
(c)
(d)
89. (a)
8 0 4
= 3 1 7
5 2 6
TB
, and .
8 3 5
= 0 1 2 =
4 7 6
TTB B
(b)
2 1 3 0 2 3 2 3 6 2 1 1
+ = 0 4 5 + 1 7 4 = 1 11 9 = 3 11 6 ;.
2 1 4 3 5 9 1 6 13 6 9 13
T
TA C
T
2 0 2 0 1 3 2 1 1
+ = 1 4 1 + 2 7 5 = 3 11 6
3 5 4 3 4 9 6 9 13
T TA C
So we see that (A + C)T = AT + CT.
(c)
14 7 21 14 0 14
= 0 28 35 = 7 28 7
14 7 28 21 35 28
T
TbA
2 0 2 14 0 14
= 7 1 4 1 = 7 28 7 .
3 5 4 21 35 28
TbA
So, we see that (bA)T = bAT.
(d)
0 2 3 2 1 3 6 5 2 6 6 12
= 1 7 4 0 4 5 = 6 31 54 = 5 31 26 .
3 5 9 2 1 4 12 26 70 2 54 70
T T
TCA
Chapter 1: Systems of Linear Equations and Matrices 27
2 0 2 0 1 3 6 6 12
= 1 4 1 2 7 5 = 5 31 26
3 5 4 3 4 9 2 54 70
T TA C
So, we see that (CA)T = ATCT.
90. Since = 3 ∙ 3 − 4 ∙ 2 = 1, the inverse is
91. Using the formula for the inverse of a 2 2 matrix, we get ad bc = 6( 2) 3 ( 5) = 3, so
1 13
2 3=
5 6B .
92. Since = 4 3 9 1 = 3, the inverse is
93. 1 121
2 3=
7 0D
95.
Then ad – bc = 4∙3 – 1∙9 = 3, so that
,
Using the value of A−1 from Exercise 92, we have
,
and the two are equal.
96. We have
Chapter 1: Systems of Linear Equations and Matrices 28
so that
97.
ABC
1 23 151 2 2
49 323 3
64 45= = ,
98 69
23 1511 1 1 2 22 1
3 49 323 3
2 5 2 12= =
3 6 5 31 3C B A
98. Since , we must find the inverse of the given matrix. Since , the inverse is
99. The inverse of
4 2
1 3 is 5A, and by Theorem 1.4.5 we find this to be 1
10
3 25 =
1 4A
,
so 150
3 2=
1 4A
.
100. Invert both sides. On the right-hand side we get so that
Then
102. 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.)
Chapter 1: Systems of Linear Equations and Matrices 29
103. Let 1 1
=2 3
A
(a) 3 11 15=
30 41A
(b)
3 41 15=
30 11A (Note: 41(11) ( 15)( 30) = 1)
(c)
2 3 4 2 2 1 0 2 22 + = + =
8 11 4 6 0 1 4 6A A I
(d)
1 1 1 0 1 1= 2, = 2 =
2 3 0 1 2 1p x x p A
(e)
2 6 8 1 1 1 0 6 7=2 +1, = + =
16 22 2 3 0 1 14 20p x x x p A
(f)
3 11 15 1 1 4 0 13 13= 2 +4, = 2 + =
30 41 2 3 0 4 26 39p x x x p A
104. (a)
(b)
(c)
(d)
(e)
(f)
110. If A2 + 5A 2I = 0, then A2 5A = 2I, so .12 5 =A A I I Thus the inverse of A is .1
2 5A I
112. Multiplying both sides on the left by and on the right by
yields
.
114. Yes, it is true. From part (e) of Theorem 1.4.8, it follows that .
This statement can be extended to n factors so that
115. C
1 111 1 1 1 1 1 1 1=D CBA BA C D D CBA A B C C D
11 1 1 1 1
11 1 1 1
11 1 1
11 1 1 2
=
=
=
= = .
D CB AA B C C D
D C BB C C D
D CC C D
D D C D C
Chapter 1: Systems of Linear Equations and Matrices 30
116. Let
Then we wish to find values for the such that
AX =
is the identity matrix
This gives nine equations in the nine unknowns
Consider the equations resulting from the first column:
The second equation gives substituting that into the first equation gives the pair of
equations and , which is obviously inconsistent. Since the system has
no solutions, we conclude that A is not invertible.
117. If X = A 1 exists, then AX = I, so we attempt to solve the linear system that arises from setting
the coordinates equal to each other.
.
11 12 13 11 31 12 32 13 33
21 22 23 21 22 23
31 32 33 11 31 12 32 13 33
1 0 1 + + + 1 0 0
= 0 1 0 = = 0 1 0
1 0 1 0 0 1
x x x x x x x x x
AX x x x x x x
x x x x x x x x x
Notice that we immediately see that x21 = 0, x22 = 1, x23 = 0, so we can just work with the linear
system given by the other 6 variables.
11 31
12 32
13 33
11 31
12 32
13 33
x x 1
x x 0
x x 0
x x 0
x x 0
x x 1
This augmented matrix
1 0 0 1 0 0 1
0 1 0 0 1 0 0
0 0 1 0 0 1 0
1 0 0 1 0 0 0
0 1 0 0 1 0 0
0 0 1 0 0 1 1
reduces to
12
12
12
12
1 0 0 0 0 0
0 1 0 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0 0
0 0 0 0 0 1
.
Thus 111 2
=x , x12 = 0, etc.
Chapter 1: Systems of Linear Equations and Matrices 31
Since there is a solution, the matrix A is invertible, and the inverse is
1 12 2
1 12 2
0
0 1 0
0
.
119.
,
.
120. This linear system is
1
2
7 2 3=
3 1 0
x
x. To solve, multiply both sides of the equation by
1 1 2=
3 7A to get .
1
2
1 2 3 3= =
3 7 0 9
x
x
121.
,
Elementary Matrices and a Method for Finding A−1
129. (a) Elementary matrix (corresponds to multiplying the first row by ).
(b) Elementary matrix (corresponds to interchanging the second and third rows).
(c) Elementary matrix (corresponding to subtracting twice the first row from the second row).
(d) Not an elementary matrix (interchanges the first and second rows and multiplies the third
row by −1).
130. (a) Multiply R2 by 14
,
14
1 0=
0E . Then
214
1 0 1 0=
0 0 4I .
(b) Add 9 R2 to R1.
1 9 0
= 0 1 0
0 0 1
E .
(c) Interchage R2 and R3.
1 0 0
= 0 0 1
0 1 0
E
(d) Add R4 to R2.
1 0 0 0
0 1 0 1=
0 0 1 0
0 0 0 1
E
131. (a) Add 2 times the first row to the second row:
(b) Multiply the first row by
:
Chapter 1: Systems of Linear Equations and Matrices 32
(c) Interchange the second and third rows:
(d) Add twice the first row to the second row:
133. (a) Multiply the first row by : EA =
(b) Add four times the first row to the second row:
(c) Multiply the second row by :
134. (a)
1 0 0
= 0 1 0
2 0 1
E
(b)
1 0 0
= 0 1 0
2 0 1
E
(c)
E
12
0 0
= 0 1 0
0 0 1
(d)
2 0 0
= 0 1 0
0 0 1
E
135. (a)
1 0 0
= 0 1 1
0 0 1
E (b)
1 0 0
= 0 1 1
0 0 1
E
(c)
1 0 0
= 3 1 0
0 0 1
E (d)
1 0 0
= 3 1 0
0 0 1
E
137.
3 6 1 0
4 5 0 1
The identity matrix was adjoined to the given matrix.
13
1 2 0
4 5 0 1
The first row was multiplied by . 1
3
13
43
01 2
10 13
times the first row was added to the second row.
13
4 139 13
01 2
0 1
The second row was multiplied by
.
5 239 13
4 139 13
1 0
0 1
2 times the second row was added to the first row.
Chapter 1: Systems of Linear Equations and Matrices 33
The inverse is
.
138. No inverse exists because the discriminant is 4(3) 6(2) = 0.
139.
6 4 1 0
3 2 0 1
The identity matrix was adjoined to the given matrix.
0 0 1 2
3 2 0 1
2 times the second row was added to the first row.
A row of zeros was obtained on the left side, therefore the matrix is not invertible.
140.
2 1 1 1 0 0
0 6 4 0 1 0
0 2 2 0 0 1
Divide R1 by 2, and divide R2 by 6.
1 1 12 2 2
2 13 6
1 0 0
0 1 0 0
0 2 2 0 0 1
Add 2 R2 to R3.
1 1 12 2 2
2 13 6
10 13 3
1 0 0
0 1 0 0
0 0 0 1
Multiply R3 by 310
.
1 1 12 2 2
2 13 6
3110 10
1 0 0
0 1 0 0
0 0 1 0
Add 23
R3 to R2, and add 12
R3 to R1.
31 1 12 2 20 20
1 110 5
3110 10
1 0
0 1 0 0
0 0 1 0
Finally; add 12
R2 to R1.
1 12 4
1 110 5
3110 10
1 0 0 0
0 1 0 0
0 0 1 0
Thus,
1 12 4
1 1 110 5
3110 10
0
= 0
0
A .
141.
1 2 0 1 0 0
2 1 2 0 1 0
0 2 1 0 0 1
The identity matrix was adjoined to the given matrix.
Chapter 1: Systems of Linear Equations and Matrices 34
1 2 0 1 0 0
0 3 2 2 1 0
0 2 1 0 0 1
times the first row was added to the second row.
2 2 13 3 3
1 2 0 1 0 0
0 1 0
0 2 1 0 0 1
The second row was multiplied by
.
2 2 13 3 3
7 4 23 3 3
1 2 0 1 0 0
0 1 0
0 0 1
times the second row was added to the third row.
2 2 13 3 3
34 27 7 7
1 2 0 1 0 0
0 1 0
0 0 1
The third row was multiplied by
.
2 1 27 7 7
34 27 7 7
1 2 0 1 0 0
0 1 0
0 0 1
times the third row was added to the
second row.
3 2 47 7 7
2 1 27 7 7
34 27 7 7
1 0 0
0 1 0
0 0 1
times the second row was added to the first row.
The inverse is
.
143.
1 1 25 5 5
1 1 15 5 10
1 4 15 5 10
1 0 0
0 1 0
0 0 1
The identity matrix was adjoined to the given matrix.
12
12
1 1 2 5 0 0
1 1 0 5 0
1 4 0 0 5
Each row was multiplied by 5.
52
52
1 1 2 5 0 0
0 0 5 5 0
0 5 5 0 5
times the first row was added to the second row and times the first row was added to the third row.
52
52
1 1 2 5 0 0
0 5 5 0 5
0 0 5 5 0
The second and third rows were interchanged.
Chapter 1: Systems of Linear Equations and Matrices 35
12
1 1 2 5 0 0
0 1 1 0 1
0 0 1 2 2 0
The second row was multiplied by
and the
third row was multiplied by
.
1 1 0 1 4 0
0 1 0 0 1 1
0 0 1 2 2 0
times the third row was added to the
second row and 2 times the third row was added to the first row.
1 1 0 1 3 1
0 1 0 0 1 1
0 0 1 2 2 0
times the second row was added to the first row.
The inverse is
.
145. 3
4
0 0 0 01 1
2 20 0 01
2 20 0 0 1
Adjoin the identity matrix to the given matrix
1
0
1
2
1
2
1 0 0 0 0
3 1 0 0
0 1 4 0 0
Multiplied the second and third rows each
by
4
1
2
1
2
1 0 0 1 0 0
0 1 0 0
0 3 1 0 0
The second and third rows were exchanged
4
1
2
31
2 2
1 0 0 1 0 0
0 1 0 0
0 0 13 0
−3 times the second row was added to the third row
2
13 2 13 2
0
4
1
31
1 0 0 1 0
0 1 0 0
0 0 1 0
The third row was multiplied by
4 1
13 13
31
13 13
0
2 2
2 2
1 0 0 1 0 0
0 1 0
0 0 1 0
4 times the third row was added to the second row
Chapter 1: Systems of Linear Equations and Matrices 36
The inverse is
.
146.
1 4 4 1 0 0
1 2 4 0 1 0
1 3 2 0 0 1
Subtract R1 from R2 and also from R3.
1 4 4 1 0 0
0 2 0 1 1 0
0 1 2 1 0 1
Interchange R3 and R2.
1 4 4 1 0 0
0 1 2 1 0 1
0 2 0 1 1 0
Multiply R2 by 1.
1 4 4 1 0 0
0 1 2 1 0 1
0 2 0 1 1 0
Add 2 R2 to R3.
1 4 4 1 0 0
0 1 2 1 0 1
0 0 4 1 1 2
Divide R3 by 4.
1 1 14 4 2
1 4 4 1 0 0
0 1 2 1 0 1
0 0 1
Add 2 R3 to R2 and 4 R3 to R1.
1 12 2
1 1 14 4 2
1 4 0 0 1 2
0 1 0 0
0 0 1
Finally, add 4 R2 to R1. .1 12 2
1 1 14 4 2
1 0 0 2 1 2
0 1 0 0
0 0 1
Thus, .1 1 12 2
1 1 14 4 2
2 1 2
= 0A
147.
1 0 0 1 0 0
1 4 0 0 1 0
1 4 7 0 0 1
Adjoined the identity matrix to the given matrix
Chapter 1: Systems of Linear Equations and Matrices 37
1 0 0 1 0 0
0 4 0 1 1 0
0 4 7 1 0 1
Added −1 times the first row to each of the other rows
1 0 0 1 0 0
0 4 0 1 1 0
0 0 7 0 1 1
Added −1 times the second row to the third row.
1 0 0 1 0 0
0 1 0 0
0 0 1 0
1 14 4
1 17 7
Multiplied the second row by
and
the third by
The inverse is
.
148. This matrix is singular because its third row contains only zeros. (Refer to Example 6 in Section
1.4.)
150.
0 0 1 0 1 0 0 0
1 0 0 1 0 1 0 0
0 1 3 0 0 0 1 0
2 1 4 3 0 0 0 1
Adjoined the identity matrix to the given matrix
1 0 0 1 0 1 0 0
0 0 1 0 1 0 0 0
0 1 3 0 0 0 1 0
2 1 4 3 0 0 0 1
Interchanged the first and second rows
1 0 0 1 0 1 0 0
0 0 1 0 1 0 0 0
0 1 3 0 0 0 1 0
0 1 4 5 0 2 0 1
Added −2 times the first row to the fourth row
0
0 0
1 0 0 1 0 1 0 0
0 1 4 5 0 2 0 1
0 1 3 0 0 1 0
0 0 1 1 0 0
Interchanged the second and fourth rows
4 0
0 0
1 0 0 1 0 1 0 0
0 1 5 0 2 1
0 0 7 5 0 2 1 1
0 0 1 1 0 0
Added the second row to the third row
Chapter 1: Systems of Linear Equations and Matrices 38
4 0
1 0 0 1 0 1 0 0
0 1 5 0 2 1
0 0 1 0 1 0 0 0
0 0 7 5 0 2 1 1
Interchanged the third and fourth rows
4 0
1 0 0 1 0 1 0 0
0 1 5 0 2 1
0 0 1 0 1 0 0 0
0 0 0 5 7 2 1 1
Added −7 times the third row to the fourth row
04
7 2 1 15 5 5 5
0 1 0 01 0 0 1
0 2 10 1 5
1 0 0 00 0 1 0
0 0 0 1
Multiplied the fourth row by
0
37 1 15 5 5 5
7 2 1 15 5 5 5
1 0 0 0
0 2 10 1 4 5
1 0 0 00 0 1 0
0 0 0 1
Added −1 times the fourth row to the first row
0
37 1 15 5 5 5
7 2 1 15 5 5 5
1 0 0 0
7 1 00 1 4 0
1 0 0 00 0 1 0
0 0 0 1
Added 5 times the fourth row to the second row
0
37 1 15 5 5 5
7 2 1 15 5 5 5
1 0 0 0
3 1 00 1 0 0
1 0 0 00 0 1 0
0 0 0 1
Added −4 times the third row to the second row
The inverse is
.
151.
(a)
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
1
2
3
k
k
k
Adjoined the identity matrix to the given matrix
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
3
2
1
k
k
k
Interchanged the first and third rows
Chapter 1: Systems of Linear Equations and Matrices 39
0 01 0 0
0 1 0 0 0
0 0 1 0 0
3
2
1
1
1
1
k
k
k
Multiplied the first row by
, the second by
, the third by
The inverse
.
(b)
1 0 1 0 0
0 1 0 1 0
0 0 0 0 1
k
k
k
Adjoined the identity matrix to the given matrix
1 0 0 0
0 1 0 0
0 0 1 0 0
1 1
1 1
1
k k
k k
k
Multiplied each row by
1 0 0
0 1 0 0
0 0 1 0 0
2 2
1 1 1
1 1
1
kk k
k k
k
Added −
times the second row to the
first row
31 0 0
0 1 0 0
0 0 1 0 0
2
1 1 1
1 1
1
k k k
k k
k
Added
times the third row to the
first row
31 0 0
0 1 0
0 0 1 0 0
2
2
1 1 1
1 1 1
1
k k k
k k k
k
Added
times the third row to the
socond row
The inverse is
.
152. Form the augmented matrix with I3, and reduce to echelon form.
1 0 0
1 1 0 1 0
0 0 0 0 1
c c c
c
c
reduces to
1 1 11 1
1 1(1 ) 1
1
1 0 0
0 1 0 0
0 0 1 0 0
c c c
c c c
c
Chapter 1: Systems of Linear Equations and Matrices 40
unless c = 0 or c = 1. Thus as long as c 0 or c 1, the matrix is invertible.
153. 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 the identity matrix.
2 0
1 2
0 1
c
c
c
1 2
2 0
0 1
c
c
c
The first and second rows were interchanged
1 2
0 1
2 0
c
c
c
The second and third rows were interchanged
2
2
1 2
0 1
0 2
c
c
c c
times the first row was added to the third row
3
1 2
0 1
0 0 4
c
c
c c
times the second row was added to the third row
This matrix can be reduced to the identity matrix if and only if c3 − 4c ≠ 0, i.e., if c ≠ 0 and c ≠ ±2. For any other value of c, the matrix is invertible.
154. Keeping track of the elementary matrices used to row reduce
2 3=
1 0A to I, we see that
13
0 1 1 2 2 30=
1 0 0 1 1 00 1I, so taking inverses (and reversing the order of multiplication)
we get
1 2 3 0 0 1 2 3=
0 1 0 1 1 0 1 0
155. 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:
3 times the first row was added to the
second row
Multiplied the second row by
Chapter 1: Systems of Linear Equations and Matrices 41
Since , then
157. From Exercise 154 above, we can quickly see that
1 13
2 3 0 1 1 20=
1 0 1 0 0 10 1
158. We use the elementary matrices already obtained in the solution of Exercise 155. Since
, then
160. Keeping track of the elementary matrices used to row reduce A to B, we see that
1 0 0 1 2 0 3 11 18 7 1 2
0 0 1 0 1 0 5 6 8 = 1 3 4
0 1 0 0 0 1 1 3 4 5 6 8
, and so
1 2 0 1 0 0 7 1 2 3 11 18
0 1 0 0 0 1 1 3 4 = 5 6 8
0 0 1 0 1 0 5 6 8 1 3 4
161. Let us begin by finding a sequence of elementary row operations that will reduce to :
6 9 4
= 5 1 0
1 2 1
B
1 2 1
5 1 0
6 9 4
(a)
The first and third rows were interchanged.
1 2 1
5 1 0
6 9 4
(b)
The first row was multiplied by .
1 2 1
0 9 5
0 3 2
(c)
5 times the first row was added to the second row and times the first row was added to the third row.
59
1 2 1
0 1
0 3 2
(d)
The second row was multiplied by
.
Chapter 1: Systems of Linear Equations and Matrices 42
59
13
1 2 1
0 1
0 0
(e)
times the second row was added to the third row.
59
1 2 1
0 1
0 0 1
(f)
The third row was multiplied by .
1 2 0
0 1 0
0 0 1
(g)
times the third row was added to the
second row and times the third row was added to the first row.
1 0 0
0 1 0
0 0 1
(h)
times the second row was added to the first row.
Now we reduce to , then continue applying the inverse operations of the ones listed above,
in reverse order, to obtain .
2 1 0
= 1 1 0
3 0 1
A
1 1 0
2 1 0
3 0 1
The first and second rows were interchanged.
1 1 0
2 1 0
3 0 1
The first row was multiplied by .
1 1 0
0 3 0
0 3 1
times the first row was added to the second row and times the first row was added to the third row.
1 1 0
0 1 0
0 3 1
The second row was multiplied by
.
Chapter 1: Systems of Linear Equations and Matrices 43
1 1 0
0 1 0
0 0 1
times the second row was added to the third row.
1 1 0
0 1 0
0 0 1
The third row was multiplied by .
1 0 0
0 1 0
0 0 1
The second row was added to the first row.
1 2 0
0 1 0
0 0 1
inverted operation (h): 2 times the second row was added to the first row
59
1 2 1
0 1
0 0 1
inverted operations (g): The third row was added to the first row and
times the third row was added to the second
row.
59
13
1 2 1
0 1
0 0
inverted operation (f):
The third row was multiplied by
.
59
1 2 1
0 1
0 3 2
inverted operation (e): times the second row was added to the third row.
1 2 1
0 9 5
0 3 2
inverted operation (d): The second row was multiplied by 9.
1 2 1
5 1 0
6 9 4
inverted operations (c): 6 times the first row was added to the third row and times the first row was added to the second row.
1 2 1
5 1 0
6 9 4
inverted operation (b): The first row was multiplied by .
Chapter 1: Systems of Linear Equations and Matrices 44
6 9 4
= 5 1 0
1 2 1
B
inverted operation (a): The first and third rows were interchanged.
More on Linear Systems and Invertible Matrices
165. If ,
3 5=
1 2A then |A| = 1 and ,
1 2 5=
1 3A so .
2 5 2 19= =
1 3 3 11x
166. We begin by inverting the coefficient matrix
4 3 1 0
6 5 0 1
Adjoined the identity matrix to the coefficient matrix
0
1
3 1441
06 5
Multiplied the first row by
0
10
3 144
3122
1
Added 6 times the first row to the second row
2
3 144 01
30 1
Multiplied the second row by 2
2
5 32 2
1 0
0 1 3
Added
times the second row to the first row
Since the inverse of the coefficient matrix is ,2
5 32 2
3 Theorem 1.6.2 states that the system has
exactly one solution:
1
2
7,
23
x
x
295 322 2
2 17
i.e.,
.
167.
1 1 13 3 3
1 1 1 23 3 3
2 1 13 3 3
6 3
= = 3 = 9
12 9
x bA
168.
1 1 21
2 1 2
b 6b b6 1= = =
b 5b +b5 1x bA
169. We begin by inverting the coefficient matrix:
Chapter 1: Systems of Linear Equations and Matrices 45
6
1 4 1 1 0 0
1 3 2 0 1 0
2 6 0 0 1
Adjoined the identity matrix to the coefficient matrix
1 4 1 1 0 0
0 1 1 1 1 0
2 6 6 0 0 1
Added the first row to the second row
1 4 1 1 0 0
0 1 1 1 1 0
0 2 4 2 0 1
Added times the first row to the third row
0 2 1
1 4 1 1 0 0
0 1 1 1 1 0
0 0 2
Added 2 times the second row to the third row
0 1
12
1 4 1 1 0 0
0 1 1 1 1 0
0 0 1
Multiplied the third row by
12
1 0 5 3 4 0
0 1 1 1 1 0
0 0 1 1 1
Added times the second row to the first row
52
12
1 0 0 3 9
0 1 1 1 1 0
0 0 1 0 1
Added times the third row to the first row
52
12
12
1 0 0 3 9
0 1 0 1 2
0 0 1 0 1
Added the third row to the second row
Thus the inverse of the coefficient matrix is .
52
12
12
3 9
1 2
0 1
(a)
51 2
12 2
13 2
3 9 1 12
1 2 1 3 ,
0 1 0 1
x
x
x
so that the solution is x1 = 12, x2 = 3, and x3 = 1.
Chapter 1: Systems of Linear Equations and Matrices 46
(b)
51 2
12 2
13 2
3 9 1 57
1 2 5 12 ,
0 1 6 8
x
x
x
so that the solution is x1 = 57, x2 = 12, and x3 = 8.
170.
1 4 5= 1
5 6A , so we can find solutions for parts (a)-(d) by multiplying A 1 by a matrix
whose columns are the b vectors from parts (a)-(d).
.
1 0 4 1 5 4 5 0 4 1 5 5 46 19 25= = =
1 6 3 1 5 6 1 6 3 1 6 56 23 31A
Thus (a) ,
5=
6x (b) ,
46=
56x (c) ,
19=
23x (d) .
25=
31x
171.
1 3 5 1 0 1
1 2 0 0 1 1
2 5 4 1 1 0
We augmented the coefficient matrix with three columns of constants on the right hand sides of the systems (i), (ii) and (iii) – refer to Example 2 on p. 50-51.
1 3 5 1 0 1
0 1 5 1 1 2
0 1 6 3 1 2
The first row was added to the second row and times the first row was added to the third row.
1 3 5 1 0 1
0 1 5 1 1 2
0 0 1 2 2 0
The second row was added to the third row.
1 3 5 1 0 1
0 1 5 1 1 2
0 0 1 2 2 0
The third row was multiplied by .
1 3 0 9 10 1
0 1 0 9 11 2
0 0 1 2 2 0
times the third row was added to the first row and to the second row.
1 0 0 18 23 5
0 1 0 9 11 2
0 0 1 2 2 0
times the second row was added to the first row.
We conclude that the solutions of the three systems are:
(i) ,
Chapter 1: Systems of Linear Equations and Matrices 47
(ii) , ,
(iii) ,
172. Row reduce the augmented matrix 1
2
.b
b
1 3
4 12
Add 4 R1 to R2. 1
1 24
1 3
0 0
b
b b. We can see that this system will be consistent only if
4b1 + b2 = 0, or b2 = 4b1.
173. Row reduce the augmented matrix 1
2
.b
b
2 5
3 6
Multiply R1 by 12
. 1
2
b
b
1522
1
3 2
Add 3R1 to R2. 1
1 2
.b
b b
5 122
32722
1
0 Multiply R2 by 2
27.
1
1 2
.b
b b
1522
1 29 27
1
0 1
Add 52
R2 to R1. 1 2
1 2
529 27
1 29 27
1 0
0 1
b b
b b This system is consistent for all values of b1 and b2.
174. Find the row-reduced form of the augmented matrix:
1
2
3
1 2 2
4 5 4
4 7 8
b
b
b
The augmented matrix of the system
1
1 2
1 3
1 2 2
0 3 4 4 +
0 1 0 4 +
b
b b
b b
Added 4 times the first row to the second and third rows
1
1 3
1 2
1 2 2
0 1 0 4 +
0 3 4 4 +
b
b b
b b
Interchanged the second and third rows
1
1 3
1 2
1 2 2
0 1 0 4
0 3 4 4 +
b
b b
b b
Multiplied the second row by
2
4
1
1 3
1 2 3
1 2
0 1 0 4
0 0 8 + 3
b
b b
b b b
Added times the second row to the third row
Chapter 1: Systems of Linear Equations and Matrices 48
1 2 38 3
4
b b b
1
1 3
1 2 2
0 1 0 4
0 0 1
b
b b
Multiplied the third row by
The row-reduced form of the coefficient matrix is the identity, so the system is consistent for all
values of , , and .
175. The augmented matrix
1
2
3
4
1 3 1 2
2 1 5 1
3 2 2 1
5 7 3 0
b
b
b
b
row reduces to
1
1 2
1 3
1 3 4
1 3 1 2
2 +0 7 3 5
3 +0 11 1 5
b 2 +0 0 0 0
b
b b
b b
b b
. There is no need to reduce this to row echelon form. Thus
the system is consistent only if
b1 2b3 + b4 = 0, or b1 = 2b3 b4.
176. (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.
1 1 2 0
2 1 2 0
3 1 0 0
The augmented matrix for the homogeneous system .
1 1 2 0
0 1 6 0
0 2 6 0
times the first row was added to the second row and times the first row was added to the third row.
1 1 2 0
0 1 6 0
0 2 6 0
The second row was multiplied by .
1 1 2 0
0 1 6 0
0 0 6 0
2 times the second row was added to the third row.
Chapter 1: Systems of Linear Equations and Matrices 49
1 1 2 0
0 1 6 0
0 0 1 0
The third row was multiplied by
.
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
2 1 2 0
2 2 2 0
3 1 3 0
The augmented matrix for the homogeneous system .
2 2 2 0
2 1 2 0
3 1 3 0
The first and second rows were interchanged.
1 1 1 0
2 1 2 0
3 1 3 0
The first row was multiplied by
.
1 1 1 0
0 1 0 0
0 4 0 0
2 times the first row was added to the second row and times the first row was added to the third row.
1 1 1 0
0 1 0 0
0 4 0 0
The second row was multiplied by .
1 0 1 0
0 1 0 0
0 0 0 0
times the second row was added to the third row and the second row was added to the first row.
If we assign an arbitrary value , the general solution is given by the formulas
, , and .
Chapter 1: Systems of Linear Equations and Matrices 50
177. The inverse of
1 2 3
3 7 6
1 0 8
is
56 16 9
18 5 3
7 2 1
, so multiplying the equation on the left by this
matrix gives
56 16 9 1 4 2 0 3 83 186 264 113 56
= 18 5 3 0 1 5 2 7 = 27 59 85 37 19
7 2 1 3 6 8 9 0 10 24 32 13 7
X
178. If the matrix on the left is invertible, then
.
Find the inverse of that matrix:
0
2 0 1 1 0 0
0 1 1 0 1 0
1 1 0 0 1
Adjoined the identity matrix to the coefficient matrix
01 1 0 0 1
0 1 1 0 1 0
2 0 1 1 0 0
Interchanged the first and third rows
0
1
1 1 0 0 1
0 1 1 0 1 0
0 2 1 0 2
Added twice the first row to the third row
0
1
1 1 0 0 1
0 1 1 0 1 0
0 2 1 0 2
Multiplied the second row by
01 1 0 0 1
0 1 1 0 1 0
0 0 1 1 2 2
Added times the second row to the third row
01 1 0 0 1
0 1 1 0 1 0
0 0 1 1 2 2
Multiplied the third row by
0 0
1
1 1 0 1
0 1 0 1 2
0 0 1 1 2 2
Added times the third row to the second row
Chapter 1: Systems of Linear Equations and Matrices 51
1 2
1 0 0 1 1 1
0 1 0 1
0 0 1 1 2 2
Added times the second row to the first row
Thus the inverse is 1 2 ,
1 1 1
1
1 2 2
so that
1 2
1 1 1
1
1 2 2
1 2 3 4
9 8 7 6
2 1 0 1
12 11 10 9
14 12 10 8 .
23 20 17 14
Diagonal, Triangular, and Symmetric Matrices
183. Since all of the diagonal entries are nonzero, the matrix is invertible.
184. The matrix is not invertible because one diagonal entry is 0.
186. A diagonal matrix is invertible if and only if its diagonal entries are all nonzero. All four diagonal
entries of this diagonal matrix are nonzero, therefore the matrix is invertible.
188. The first column of the first matrix is multiplied by 7, the 2nd column of the first matrix is
multiplied by 1 and the third column of the first matrix is multiplied by .12
Thus,
.
12
7 0 03 2 8 21 2 4
0 1 0 =4 1 6 28 1 3
0 0
190.
191. If
2 0=
0 1A ,
( )
22
2
2 0 4 0= =
0 10 1A , ,
12 4
0=
0 1A .
( 1)
1
20
=0
kk
kA
192.
,
,
Chapter 1: Systems of Linear Equations and Matrices 52
193. ,
12
15
3 0 0
= 0 0
0 0
A
2 14
125
9 0 0
= 0 0
0 0
A ,
19
2
0 0
= 0 4 0
0 0 25
A , .
1
30 0
= 0 2 0
0 0 25
k
k k
k
A
194.
,
,
195. Symmetric
196. Not symmetric since a12 a21
198. Symmetric
199. Symmetric
200. Symmetric
202. Not invertible since this triangular matrix has a zero on the diagonal.
203. Invertible.
204. From part (c) of Theorem 1.7.1, since the matrix is lower diagonal and has a zero entry on the
diagonal, it is not invertible.
205. For
23
4 0
a to be symmetric, a2 must equal 4, so a = 2, a = 2.
206. In order for A to be symmetric, we must have a + b c = 4, a b = 1, and
2a b c = 3. We solve this system by Gauss-Jordan elimination:
1 4
1
1 1
1 1
1 0 1
2 3
The augmented matrix of the system
41 1 1
0 2 1 3
2 1 1 3
times the first row was added to the second row
1 4
1
1 1
0 2 3
0 3 1 5
times the first row was added to the third row
32 2
4
1
1
1 1 1
0 1
0 3 5
The second row was multiplied by
Chapter 1: Systems of Linear Equations and Matrices 53
32 2
2 2
4
1
1 1
1 1 1
0 1
0 0
times the second row was added to the third row
3
2 2
1
1
1
1 1 4
0 1
0 0 1
The third row was multiplied by 2
1
0 2
1
1 1 4
0 1
0 0 1
times the third row was added to the
second row
1
0 2
1
1 0 2
0 1
0 0 1
times the second row was added to the first row
0
0 2
1
1 0 3
0 1
0 0 1
The third row was added to the first row
The only solution to the system is a = 3, b = 2, and c = 1, so these are the only possible values of a, b, and c.
207. For
22 5
= 0 +3 1
0 0
x x
A x x
x
to be invertible, all of the diagonal entries need to be different
from 0, so as long as x 2, x 3, and x 0, the matrix will be invertible.
208. For example
There are seven other possible answers, found by changing the
signs on the nonzero entries of the matrix. For example,
is one of them.
209.
2 0 0 1 0 0 2 0 0
= 1 1 0 6 2 0 = 5 2 0
3 4 5 1 1 5 32 13 25
AB . Thus for these two lower triangle matrices, it
is true that the product is again a lower triangular matrix.
211. Such a diagonal matrix A =
. Then
Chapter 1: Systems of Linear Equations and Matrices 54
2
2
0 0 0 1 02
0 0 0 0 1
0 2 00
0 0 20
0
0
0.
0
2
2
2
3 2 = 3
3=
3
3 2=
3 2
( 1)( 2)=
( 1)( 2)
A A I
a
a a
b b
a a
b b
a a a
b b b
a
bb
This is a zero matrix whenever (a 1)(a 2) = 0 and (b 1)(b 2) = 0, which is when both a and b are
either 1 or 2. Thus the four such matrices are
.
212. (a) From Theorem 1.4.1, the properties (page 41) and the assumption ,
we have
This shows that .
(b) From Theorem 1.4.1, the properties (page 41) and the assumption ,
we have
215.
217. The condition is equivalent to the linear system
The augmented matrix
has the reduced row echelon form
.
Chapter 1: Systems of Linear Equations and Matrices 55
If we assign the arbitrary value , the general solution is given by the formulas
, , , .
219. (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 second equation
yields .
The first equation
yields
.
(b) Step 1. Solve
The first equation yields .
The second equation yields .
The third equation yields
Step 2. Solve
using back-substitution:
The third equation yields .
The second equation yields
.
The first equation
yields
.
220. If ,11
21 22
0=
dA
d d
then the diagonal entries of 3 8 0=
9 1A
will be 3
11d and d 322 . Thus
311 = 8d and 3
22 = 1d , so d11 = 2 and d22 = 1. Now, we find d21:
3
3
21 21
2 0 2 0= =
1 3 1A
d d
. Since 3d21 = 9, d21 = 3, and we have found that
2 0=
3 1A
.
Applications of Linear Systems
222. (a) There are five nodes – each of them corresponds to an equation.
Chapter 1: Systems of Linear Equations and Matrices 56
This system can be rearranged 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
, , , , ,
(c) When and , the remaining flow rates become , ,
, and . The directions of the flow agree with the arrow orientations in the
diagram.
224. 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
If we assign an arbitrary value , the general solution is given by the formulas
However, all three unknowns must be nonnegative integers.
The nonnegativity of requires the inequality
, i.e.,
.
Likewise for ,
yields
.
Chapter 1: Systems of Linear Equations and Matrices 57
When
, all three variables are nonnegative. Of the four integer values inside this
interval (4, 5, 6, and 7), only yields integer values for , , and .
We conclude that the box has to contain 3 pennies, 4 nickels, and 6 dimes.
226. From Kirchhoff's current law, we have Kirchhoff's voltage law yields
op nside Loop o om nside Loop utside Loop
The corresponding linear system can be rewritten as
Its augmented matrix has the reduced row echelon form
.
The solution is
,
, and
.
227. We are looking for positive integers , and such that
The number of atoms of each element on the two sides must be equal:
ygen
This gives the linear system
The row-reduced form of the corresponding augmented matrix is
,
which has the general solution
,
,
, and . Choosing t = 6 gives the
smallest solution in positive integers, which is , , , and . The balanced
equation is
Chapter 1: Systems of Linear Equations and Matrices 58
228. We are looking for positive integers , and such that
The number of atoms of each element on the two sides must be equal:
Carbon
This gives the linear system
The row-reduced form of the corresponding augmented matrix is
,
which has the general solution
,
,
, and Choosing gives
the smallest solution in positive integers, which is , , , and . The
balanced equation is
229. The three points must satisfy the quadratic equation y = ax2 + bx + c, giving us 3 equations in
three unknowns a, b, c.
c
a b c
a b c
= 1
+ + = 2
+ = 0
Solving the linear system ,
0 0 1 1
1 1 1 = 2
1 1 1 0
a
b
c
yields
1 12 2
1 1 12 2
1 1 1 2
= 2 = 0 2 = 1
0 1 0 0 0 1
A
a
b
c
.
Thus the quadratic equation is y = 2x2 + x 1.
230. We are looking for a polynomial of the form
such that , and . Substituting these values into p(x) gives the
Chapter 1: Systems of Linear Equations and Matrices 59
linear system
The augmented matrix of this system has the reduced row echelon form
.
There is a unique solution , , . The polynomial is
232. The graph goes through the points ( 1, 1), (0, 4), (1, 3), and (2, 4). Writing the cubic
equations (y = ax3 + bx2 + cx + d) satisfied by these three points and forming the corresponding
linear system we get
1 1 1 1 1
0 0 0 1 4=
1 1 1 1 3
8 4 2 1 4
a
b
c
d
. Solving using A 1, we get
1 1 1 16 2 2 6
1 12 2
1 1 13 2 6
1 1
4 21 0
3 01
4 40 1 0 0
a
b
c
d
.
The cubic equation is y = x3 2x2 + 4.
Leontief Input-Output Models
233. (a) The columns of the consumption matrix list the inputs required by C and P, respectively, to
produce $1 worth of output, so
C0.1 0.2
=0.3 0.1
.
(b) Solving the system x C x = d, where d is the demand, means solving (I C )x = d.
Since 0
,0
0.9 .2=
.3 0.9I C
use ,
1 43
0.9 0.2=
0.3 0.9I C
to get x = (I C ) 1
,9000 12400
=6000 10800
so the copy store needs to produce $12,400 and the paper company
needs to produce $10,800.
234. (a)
(b) The Leontief matrix is
, and the outside
demand vector is
. The Leontief equation leads to the linear
Chapter 1: Systems of Linear Equations and Matrices 60
system with the augmented matrix
. Its reduced row echelon form is
. To meet the consumer demand, the economy must produce $225,000
worth of food and $225,000 worth of housing.
235. (a)
C
0.2 0.6 0.5
= 0.4 0.2 0.2
0.2 0.1 0.2
(b) The augmented matrix for the system (I C ) x = d is
0 0 0
0 0 0
0 0 0
1 .2 .6 .5 6000
.4 1 .2 .2 3000
.2 .1 1 .2 2000
and
it row reduces to .
1 0 0 35277.8
0 1 0 25000
0 0 1 14444.4
The economy should produce $35,277.80 of housing,
$25,000 of food, and $14,444.40 of utilities.
236. (a)
(b) The Leontief matrix is
;
the outside demand vector is
.
The Leontief equation leads to the linear system with the augmented matrix
.
Its reduced row echelon form is
.
To meet the consumer demand, the company must produce approximately $19,578.29 worth of
Web design, $16,346.56 worth of software, and $6839.25 worth of networking.
238.
;
. Thus
.
Chapter 1: Systems of Linear Equations and Matrices 61
239. Since the row sums are all less than 1, the economy is productive.
240.
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.