2.8 • Solutions 147
5. The vector w is in the subspace generated by v
1
and v
2
if and only if the vector equation x
1
v
1
+ x
2
v
2
6. The vector u is in the subspace generated by {v
1
, v
2
, v
3
} if and only if the vector equation x
1
v
1
+ x
2
v
2
+ x
3
v
3
= u is consistent. The row operations below show that u is not in the subspace generated by
{v
1
, v
2
, v
3
}.
⎢
⎢
⎣
145114511451
3437 081210 012 1
[]~ ~ ~
−−−
⎡
⎤⎡ ⎤⎡ ⎤
⎢
⎥⎢ ⎥⎢ ⎥
−−−− − −
⎢
⎥⎢ ⎥⎢ ⎥
vv vu
7. a. There are three vectors: v
1
, v
2
, and v
3
in the set {v
1
, v
2
, v
3
}.
b. There are infinitely many vectors in Span{v
1
, v
2
, v
3
} = Col A.
c. Deciding whether p is in Col A requires calculation:
234 6 23 46 23 46
−− −− −−
⎡⎤⎡⎤⎡⎤
8.
2206 2206 2206
[]0351~0351~0351
A
−− − −− − −− −
⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥
=− − −
⎢⎥⎢⎥⎢⎥
p
9. To determine whether p is in Nul A, simply compute Ap. Using A and p as in Exercise 7,
234 6 2
−− −
⎡⎤⎡⎤⎡⎤
10. To determine whether u is in Nul A, simply compute Au. Using A as in Exercise 7 and u = (–5, 5, 3),
2205 0
−− −
⎡⎤⎡⎤⎡⎤
4
12. p = 3 and q = 5. Nul A is a subspace of R
3
because solutions of Ax = 0 must have 3 entries, to match
13. To produce a vector in Col A, select any column of A. For Nul A, solve the equation Ax = 0. (Include
an augmented column of zeros, to avoid errors.)
3 2 1 50 3 2 1 50 321 50
94170~02480~02480
92510 048160 00000
−−−
⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥
−− − −
⎢⎥⎢⎥⎢⎥
⎢⎥⎢⎥⎢⎥
−−−
⎣⎦⎣⎦⎣⎦
The general solution is x
1
= x
3
– x
4
, and x
2
= –2x
3
+ 4x
4
, with x
3
and x
4
free. The general solution in
parametric vector form is not needed. All that is required here is one nonzero vector. So choose any
values for x
3
and x
4
(not both zero). For instance, set x
3
= 1 and x
4
= 0 to obtain the vector (1, –2, 1,
0) in Nul A.
14. To produce a vector in Col A, select any column of A. For Nul A, solve the equation Ax = 0:
⎢
⎢
⎢
⎢
⎣
1230 10 1/30
1230 12 30
015/30 015/30
4570 0350
⎡⎤
−
⎡
⎤
⎡⎤⎡⎤
⎢⎥
⎢
⎥
⎢⎥⎢⎥
−− ⎢⎥
⎢
⎥
⎢⎥⎢⎥
15. Yes. Let A be the matrix whose columns are the vectors given. Then A is invertible because its
16. No. One vector is a multiple of the other, so they are linearly dependent and hence cannot be a basis
for any subspace.
2.8 • Solutions 149
17. Yes. Place the three vectors into a 3×3 matrix A and determine whether A is invertible:
05 6 24 2
−
⎡⎤⎡⎤
18. No. Place the three vectors into a 3×3 matrix A and determine whether A is invertible:
135 135 135
⎡⎤⎡⎤⎡⎤
19. No. The vectors cannot be a basis for R
3
because they only span a plan in R
3
. Or, point out that the
⎢
⎢
⎢
⎣
36
⎡
⎤
21. a. False. See the definition at the beginning of the section. The critical phrases “for each” are
missing.
b. True. See the paragraph before Example 4.
22. a. False. See the definition at the beginning of the section. The condition about the zero vector is
only one of the conditions for a subspace.
b. False. See the warning that follows Theorem 13.
150 CHAPTER 2 • Matrix Algebra
23. (Solution in Study Guide)
459 2 126 5
65112~015 6
348 3 000 0
A
−−
⎡⎤⎡⎤
⎢⎥⎢⎥
=−
⎢⎥⎢⎥
⎢⎥⎢⎥
−
⎣⎦⎣⎦
. The echelon form identifies
45
⎡⎤⎡⎤
01 5 60
00 0 00
⎢⎥
−
⎢⎥
⎢⎥
⎣⎦
. This corresponds to:
134
234
560
00
xxx
+−=
=
.
Solve for the basic variables and write the solution of Ax = 0 in parametric vector form:
47 47
xxx
−−
47
−
Notes
: (1) A basis is a set of vectors. For simplicity, the answers here and in the text list the vectors
without enclosing the list inside set brackets. This style is also easier for students. I am careful,
however, to distinguish between a matrix and the set or list whose elements are the columns of the
matrix.
(2) Recall from Chapter 1 that students are encouraged to use the augmented matrix when solving Ax
(3) Because the concept of a basis is just being introduced, I insist that my students write the
parametric vector form of the solution of Ax = 0. They see how the basis vectors span the solution space
24.
3690 1254
2472~0036
3666 0000
A
−−
⎡⎤⎡⎤
⎢⎥⎢⎥
=−
⎢⎥⎢⎥
⎢⎥⎢⎥
−−
⎣⎦⎣⎦
. Basis for Col A:
39
2,7
36
⎡
⎤⎡⎤
⎢
⎥⎢⎥
⎢
⎥⎢⎥
⎢
⎥⎢⎥
⎣
⎦⎣⎦
.
For Nul A, obtain the reduced (and augmented) echelon form for Ax = 0:
2.8 • Solutions 151
⎢
⎢
⎢
⎢
⎣
26 26
xxx
−−
⎡⎤⎡ ⎤ ⎡⎤ ⎡ ⎤
26
−
⎡
⎤⎡ ⎤
25.
148 3 7 1480 5
127 3 4 0250 1
~
229 5 5 0001 4
369 5 2 0000 0
A
−−
⎡⎤⎡⎤
⎢⎥⎢⎥
−−
⎢⎥⎢⎥
=⎢⎥⎢⎥
−
⎢⎥⎢⎥
−−
⎢⎥⎢⎥
⎣⎦⎣⎦
. Basis for Col A:
14 3
12 3
,,
22 5
36 5
−
⎡
⎤⎡ ⎤⎡ ⎤
⎢
⎥⎢ ⎥⎢ ⎥
−
⎢
⎥⎢ ⎥⎢ ⎥
⎢
⎥⎢ ⎥⎢ ⎥
−
⎢
⎥⎢ ⎥⎢ ⎥
−
⎢
⎥⎢ ⎥⎢ ⎥
⎣
⎦⎣ ⎦⎣ ⎦
.
For Nul A, obtain the reduced (and augmented) echelon form for Ax = 0:
10 20 70
−
⎡⎤
135
270
xxx
−+=
⎢
⎢
⎢
⎢
⎢
⎣
135
235
27 27
2.5 .5 2.5 .5
xxx
xxx
−−
⎡⎤⎡ ⎤
⎡
⎤⎡⎤
⎢⎥⎢ ⎥
⎢
⎥⎢⎥
−+ −
uv
Basis for Nul A: {u, v}.
Note
: The solution above illustrates how students could write a solution on an exam, when time is
precious, namely, describe the basis by giving names to appropriate vectors found in the calculations.
313 18 31 306
313 02 02 604
−− − − −
⎡⎤⎡⎤
⎢⎥⎢⎥
−
311
310
−−
⎡⎤⎡ ⎤⎡ ⎤
⎢⎥⎢ ⎥⎢ ⎥
For Nul A,
10004/30
0130 20
⎡⎤
⎢⎥
−
15
4/3 0
320
xx
xx x
+=
+−=
152 CHAPTER 2 • Matrix Algebra
u v
27. Construct a nonzero 3×3 matrix A and construct b to be almost any convenient linear combination of
the columns of A.
28. The easiest construction is to write a 3×3 matrix in echelon form that has only 2 pivots, and let b be
29. (Solution in Study Guide) A simple construction is to write any nonzero 3×3 matrix whose columns
30. Since Col A is the set of all linear combinations of a
1
, … , a
p
, the set {a
1
, … , a
p
} spans Col A.
31. If Col F ≠ R
5
, then the columns of F do not span R
5
. Since F is square, the IMT shows that F is not
32. If Col B = R
7
, then the columns of B span R
7
. Since B is square, the IMT shows that B is invertible
33. If Nul C = {0}, then the equation Cx = 0 has only the trivial solution. Since C is square, the IMT
34. If the columns of A form a basis, they are linearly independent. This means that A cannot have more
35. If Nul B contains nonzero vectors, then the equation Bx = 0 has nontrivial solutions. Since B is
36. If the columns of C are linearly independent, then the equation Cx = 0 has only the trivial (zero)
solution. That is, Nul C = {0}.
37. [M] Use the command that produces the reduced echelon form in one step (ref or rref depending
2.8 • Solutions 153
3501 3 102.54.53.5
−− −
⎡⎤⎡⎤
35
−
⎡⎤⎡⎤
For Nul A, obtain the solution of Ax = 0 in parametric vector form:
1345
2345
2.54.53.50
1.52.51.50
xxxx
xxxx
+−+=
+−+=
⎢
⎢
⎢
⎢
⎢
⎣
1345
2345
2.5 4.5 3.5 2.5 4.5 3.5
1.5 2.5 1.5 1.5 2.5 1.5
x xxx
xxxx
−+ − −−
⎡⎤⎡ ⎤
⎡
⎤⎡⎤⎡⎤
⎢⎥⎢ ⎥
⎢
⎥⎢⎥⎢⎥
−+ − −−
By the argument in Example 6, a basis for Nul A is {u, v, w}.
53268 10100
41387 01100
−−
⎡⎤⎡⎤
⎢⎥⎢⎥
−− −
⎢
⎢
⎢
⎣
53 68
41 87
−−
⎡
⎤⎡ ⎤⎡ ⎤⎡ ⎤
⎢
⎥⎢ ⎥⎢ ⎥⎢ ⎥
−−
13
0
xx
+=
13
23
xx
xx
=−
⎧
⎪=
⎪
⎪
⎩
154 CHAPTER 2 • Matrix Algebra
1
2
1
1
x
x
−
⎡⎤ ⎡⎤
⎢⎥ ⎢⎥
Note
: The Study Guide for Section 2.8 gives directions for students to construct a review sheet for the
concept of a subspace and the two main types of subspaces, Col A and Nul A, and a review sheet for the
concept of a basis. I encourage you to consider making this an assignment for your class.
2.9 SOLUTIONS
Notes
: This section contains the ideas from Sections 4.4–4.6 that are needed for later work in Chapters
5–7. If you have time, you can enrich the geometric content of “coordinate systems” by discussing crystal
lattices (Example 3 and Exercises 35 and 36 in Section 4.4.) Some students might profit from reading
Exercises 1–16 may be assigned after students have read as far as Example 2. Exercises 19 and 20 use
the Rank Theorem, but they can also be assigned before the Rank Theorem is discussed.
Exercises 9–16 include important review of techniques taught in Section 2.8 (and in Sections 1.2 and
2.5). They make good test questions because they require little arithmetic. My students need the practice
here. Nearly every time I teach the course and start Chapter 5, I find that at least one or two students
cannot find a basis for a two-dimensional eigenspace!
2. If [x]
B
= 1
2
−
⎡⎤
⎢⎥
⎣⎦
, then x is formed from b
1
and b
2
using weights –1 and 2:
r
3. To find c
1
and c
2
that satisfy x = c
1
b
u
4. As in Exercise 3,
12
1
[]
5
⎡
=⎢−
⎣
bb x
5.
12
122 1
[]479~0
357 0
−
⎡⎤⎡
⎢⎥⎢
=−
⎢⎥⎢
⎢⎥⎢
−−
⎣⎦⎣
bb x
7. Fig. 1 suggests that w = 2b
1
– b
2
an
[w]
B
= 2
1
⎡⎤
⎢⎥
−
⎣⎦
and [x]
B
= 1.5
.5
⎡⎤
⎢⎥
⎣⎦
. To
c
Note
: Figures 1 and 2 display what Sec
t
8. Fig. 2 suggests that x = b
1
+ b
2
, y =
r
b
1
+ c
2
b
2
, row reduce the augmented matrix:
21 1 2 1 10 3
~~
39 0 714 01 2
−− −
⎤⎡ ⎤⎡ ⎤
⎥⎢ ⎥⎢ ⎥
−−
⎦⎣ ⎦⎣ ⎦
,
22 104
11~011
11 000
−⎤⎡ ⎤
⎥⎢ ⎥
⎥⎢ ⎥
⎥⎢ ⎥
−−
⎦⎣ ⎦
. [x]
B
=
1
2
4.
1
c
c
⎡⎤⎡⎤
=
⎢⎥⎢⎥
⎣⎦
⎣⎦
n
d x = 1.5b
1
+ .5b
2
, in which case,
c
onfirm [x]
B
, compute
t
ion 4.4 calls B-graph paper.
1/3b
1
– b
2
, and z = –4/3 b
1
+2b
2
. If so, then
156 CHAPTER 2 • Matrix Algebra
12
02 2
(1 / 3) (1 / 3) 32 1
−
⎡⎤ ⎡⎤ ⎡ ⎤
−= − = =
⎢⎥ ⎢⎥ ⎢ ⎥
−
⎣⎦ ⎣⎦ ⎣ ⎦
bb y
and
12
024
(4/3) 2 4/3 2
320
⎡⎤ ⎡⎤ ⎡⎤
−+=− +==
⎢⎥ ⎢⎥ ⎢⎥
⎣⎦ ⎣⎦ ⎣⎦
bb z
.
⎡
⎢
⎢
⎢
⎢
⎣
12 6
−
⎡⎤⎡ ⎤⎡ ⎤
⎢⎥⎢ ⎥⎢ ⎥
Columns 1, 2 and 4, of the echelon form certainly cannot span Col A since those vectors all have zero
in their fourth entries. For Nul A, use the reduced echelon form, augmented with a zero column to
insure that the equation Ax = 0 is kept in mind:
⎢
⎢
⎢
⎢
⎣
13 0 0 0
⎡⎤
12
30
xx
+=
12
33
xx
−−
⎡⎤
⎡
⎤⎡⎤
3
−
⎡⎤
a basis for Nul A. From this information, dim Col A = 3 (because A has three pivot columns) and dim
Nul A = 1 (because the equation Ax = 0 has only one free variable).
⎢
⎢
⎢
⎣
12 154 12120
21156 01103
−− −−
⎡
⎤⎡ ⎤
⎢
⎥⎢ ⎥
−
10 1 000
⎡⎤
⎢⎥
13
23
0
0
xx
xx
+=
+=
⎢
⎢
⎢
⎢
13
1
xx
−−
⎡⎤⎡⎤ ⎡⎤
⎢⎥⎢⎥ ⎢⎥
1
−
⎡
⎤
⎢
⎥
2.9 • Solutions 157
From this, dim Col A = 4 and dim Nul A = 1.
11. The information
⎡
⎢
⎢
~
0090 9 00000
367310 00000
A
=
⎢
⎥⎢ ⎥
⎢
⎥⎢ ⎥
−−−−−
⎣
⎦⎣ ⎦
shows that columns 1and
25
−
⎡⎤⎡⎤
⎣⎦⎣⎦
120110
⎡⎤
⎣⎦
⎢
⎢
⎢
⎢
⎢
⎣
1245
2211
100
x xxx
xx
−−− −−−
⎡⎤⎡ ⎤ ⎡⎤ ⎡⎤ ⎡⎤
⎢⎥⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥
211
100
−−−
⎡
⎤⎡ ⎤⎡ ⎤
⎢
⎥⎢ ⎥⎢ ⎥
⎢
⎢
⎢
⎣
12 446 1284 6
51 9 210 0 234 1
−−
⎡
⎤⎡ ⎤
⎢
⎥⎢ ⎥
−−
12 4
−
⎡⎤⎡⎤⎡ ⎤
⎢⎥⎢⎥⎢ ⎥
1000 00
⎡⎤
1
0
x
=
⎢
⎢
⎢
⎢
⎢
⎣
1
000
221
x
xxx
⎡⎤⎡ ⎤ ⎡⎤ ⎡ ⎤
⎢⎥⎢ ⎥ ⎢⎥ ⎢ ⎥
−− −−
00
21
⎡
⎤⎡ ⎤
⎢
⎥⎢ ⎥
−−
158 CHAPTER 2 • Matrix Algebra
From this, dim Col A = 3 and dim Nul A = 2.
13. The four vectors span the column space H of a matrix that can be reduced to echelon form:
1324 1324 1324 1324
3915 0057 0057 0057
− − − − −− −−
⎡ ⎤⎡⎤⎡⎤⎡⎤
⎢ ⎥⎢⎥⎢⎥⎢⎥
−− − − −
14. The five vectors span the column space H of a matrix that can be reduced to echelon form:
12013 1201 3 1201 3
−− −
⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥
15. Col A = R
4
, because A has a pivot in each row and so the columns of A span R
4
. Nul A cannot equal
16. Col A cannot be R
3
because the columns of A have four entries. (In fact, Col A is a 3-dimensional
17. a. True. This is the definition of a B-coordinate vector.
b. False. Dimension is defined only for a subspace. A line must be through the origin in R
n
to be a
subspace of R
n
.
18. a. True. This fact is justified in the second paragraph of this section.
b. False. The dimension of Nul A is the number of free variables in the equation Ax = 0.
2.9 • Solutions 159
19. The fact that the solution space of Ax = 0 has a basis of three vectors means that dim Nul A = 3.
Since a 5×7 matrix A has 7 columns, the Rank Theorem shows that rank A = 7 – dim Nul A = 4.
20. A 6×8 matrix A has 8 columns. By the Rank Theorem, rank A = 8 – dim Nul A. Since the null space
is three-dimensional, rank A = 5.
21. A 9×8 matrix has 8 columns. By the Rank Theorem, dim Nul A = 8 – rank A. Since the rank is seven,
dim Nul A = 1. That is, the dimension of the solution space of Ax = 0 is one.
22. Suppose that the subspace H = Span{v
1
, …, v
5
} is four-dimensional. If {v
1
, …, v
5
} were linearly
23. A 3×5 matrix A with a two-dimensional column space has two pivot columns. The remaining three
columns will correspond to free variables in the equation Ax = 0. So the desired construction is
possible. There are ten possible locations for the two pivot columns, one of which is
24. A rank 1 matrix has a one-dimensional column space. Every column is a multiple of some fixed
25. The p columns of A span Col A by definition. If dim Col A = p, then the spanning set of p columns is
26. If columns a
1
, a
3
, a
4,
a
5
, and a
7
of A are linearly independent and if dim Col A = 5, then {a
1
, a
3
, a
4
,a
5
,
27. a. Start with B = [b
1
⋅ ⋅ ⋅ b
p
] and A = [a
1
⋅ ⋅ ⋅ a
q
], where q > p. For j = 1, …, q, the vector a
j
is
in W. Since the columns of B span W, the vector a
j
is in the column space of B. That is, a
j
= Bc
j
for some vector c
j
of weights. Note that c
j
is in R
p
because B has p columns.
28. If A contained more vectors than B, then A would be linearly dependent, by Exercise 27, because B
29. [M] Apply the matrix command ref or rref to the matrix [v
1
v
2
x]:
15 14 16 1 0 2
⎡⎤⎡⎤
30. [M] Apply the matrix command ref or rref to the matrix [v
1
v
2
v
3
x]:
68 911 100 2
304 2 0101
~
97 8 17 001 1
4 3 3 8 000 0
−− −
⎡⎤⎡⎤
⎢⎥⎢⎥
−
⎢⎥⎢⎥
⎢⎥⎢⎥
−−−
⎢⎥⎢⎥
−−
⎣⎦⎣⎦
Chapter 2 • Supplementary Exercises 161
Chapter 2 SUPPLEMENTARY EXERCISES
1. a. True. If A and B are m×n matrices, then B
T
has as many rows as A has columns, so AB
T
is
defined. Also, A
T
B is defined because A
T
has m columns and B has m rows.
b. False. B must have 2 columns. A has as many columns as B has rows.
c. True. The ith row of A has the form (0, …, d
i
, …, 0). So the ith row of AB is (0, …, d
i
, …, 0)B,
which is d
i
times the ith row of B.
i. True. An n×n elementary matrix is obtained by a row operation on I
n
.
j. False. Elementary matrices are invertible, so a product of such matrices is invertible. But not
every square matrix is invertible.
k. True. If A is 3×3 with three pivot positions, then A is row equivalent to I
3
.
l. False. A must be square in order to conclude from the equation AB = I that A is invertible.
m. False. AB is invertible, but (AB)
–1
= B
–1
A
–1
, and this product is not always equal to A
–1
B
–1
.
p. True. If the equation Ax =
1
0
0
⎡⎤
⎢⎥
⎢⎥
⎢⎥
⎣⎦
has a unique solution, then there are no free variables in this
2. C = (C
–1
)
–1
= 75 7/25/2
1
64 3 2
2
−−
⎡⎤⎡ ⎤
=
⎢⎥⎢ ⎥
−−
−⎣⎦⎣ ⎦
162 CHAPTER 2 • Matrix Algebra
4. From Exercise 3, the inverse of I – A is probably I + A + A
2
+ ⋅ ⋅ ⋅ + A
n–1
. To verify this, compute
11 11
()() ()
nn nnn
IAIAAIAAAIAAIAAIA
−− −−
− +++ =+++ − +++ =− =− 
5. A
2
= 2A – I. Multiply by A: A
3
= 2A
2
– A. Substitute A
2
= 2A – I: A
3
= 2(2A – I) – A = 3A – 2I.
10 01
⎡⎤ ⎡⎤
2
2
01
⎡⎤
7. (Partial answer in Study Guide) Since A
–1
B is the solution of AX = B, row reduction of [A B] to
[I X] will produce X = A
–1
B. See Exercise 15 in Section 2.2.
[]
13835 13835 13835
2 4 11 1 5~0 2 5 7 5~0 1 3 6 1
12 5 34 0 1 3 6 1 0 2 5 7 5
AB
−− −
⎡⎤⎡ ⎤⎡ ⎤
⎢⎥⎢ ⎥⎢ ⎥
=−−−−
⎢⎥⎢ ⎥⎢ ⎥
⎢⎥⎢ ⎥⎢ ⎥
−− − −− −
⎣⎦⎣ ⎦⎣ ⎦
8. By definition of matrix multiplication, the matrix A satisfies
12 13
37 11
A
⎡⎤⎡⎤
=
⎢⎥⎢⎥
⎣⎦⎣⎦
9. Given
54 7 3
and
23 2 1
AB B
⎡⎤ ⎡⎤
==
⎢⎥ ⎢⎥
−
⎣⎦ ⎣⎦
, notice that ABB
–1
= A. Since det B = 7 – 6 =1,
2.9 • Solutions 163
Note:
Variants of this question make simple exam questions.
10. Since A is invertible, so is A
T
, by the Invertible Matrix Theorem. Then A
T
A is the product of
invertible matrices and so is invertible. Thus, the formula (A
T
A)
–1
A
T
makes sense. By Theorem 6 in
Section 2.2,
11. a. For i = 1,…, n, p(x
i
) = c
0
+ c
1
x
i
+ ⋅ ⋅ ⋅ +
1
1n
in
cx
−
−
=
0
1
row ( ) row ( )
n
ii
c
VV
c
−
⎡⎤
⎢⎥
⋅=
⎢⎥
⎢⎥
⎣⎦
c
.
By a property of matrix multiplication, shown after Example 6 in Section 2.1, and the fact that c
b. Suppose x
1
, …, x
n
are distinct, and suppose Vc = 0 for some vector c. Then the entries in c are the
coefficients of a polynomial whose value is zero at the distinct points x
1
, …, x
n
. However, a
c. (Solution in Study Guide) When x
, …, x
are distinct, the columns of V are linearly independent,
12. If A = LU, then col
1
(A) = L⋅col
1
(U). Since col
1
(U) has a zero in every entry except possibly the first,
L⋅col
1
(U) is a linear combination of the columns of L in which all weights except possibly the first
13. a. P
2
= (uu
T
)(uu
T
) = u(u
T
u)u
T
= u(1)u
T
= P, because u satisfies u
T
u = 1.
b. P
T
= (uu
T
)
T
= u
TT
u
T
= uu
T
= P
14. Given
0
0
1
⎡⎤
⎢⎥
=
⎢⎥
⎢⎥
⎣⎦
u
, define P and Q as in Exercise 13 by
164 CHAPTER 2 • Matrix Algebra
⎡
⎢
⎢
⎢
⎣
15. Left-multiplication by an elementary matrix produces an elementary row operation:
i
16. Since A is not invertible, there is a nonzero vector v in R
n
such that Av = 0. Place n copies of v into
an n×n matrix B. Then AB = A[v ⋅ ⋅ ⋅ v] = [Av ⋅ ⋅ ⋅ Av] = 0.
matrix AB is not invertible, by the IMT. (Basically the same argument was used to solve Exercise 22
in Section 2.1.)
Note
: (In the Study Guide) It is possible that BA is invertible. For example, let C be an invertible 4×4
18. By hypothesis, A is 5×3, C is 3×5, and CA = I
3
. Suppose x satisfies Ax = b. Then CAx = Cb. Since
⎡
⎢
19. [M] Let
.3 .6 .3
.3 .2 .4
A
=
⎢⎥
⎢⎥
⎣⎦
. Then
.39 .48 .39
.30 .26 .31
A
=
⎢
⎥
⎢
⎥
⎣
⎦
. Instead of computing A
next, speed up
the calculations by computing
422 844
.2875 .2834 .2874 .2857 .2857 .2857
.4251 .4332 .4251 , .4285 .4286 .4285
.2874 .2834 .2875 .2857 .2857 .2857
AAA AAA
⎡⎤⎡⎤
⎢⎥⎢⎥
== ==
⎢⎥⎢⎥
⎢⎥⎢⎥
⎣⎦⎣⎦
⎡
⎢
⎢
⎢
⎣
2.9 • Solutions 165
⎡
⎢
⎢
⎢
48
.2119 .1998 .1998 .2024 .2022 .2022
⎡⎤⎡⎤
⎢⎥⎢⎥
⎣
⎦
20. [M] The 4×4 matrix A
4
is the 4×4 matrix of ones, minus the 4×4 identity matrix. The MATLAB
command is A4 = ones(4) – eye(4). For the inverse, use inv(A4).
0111 2/31/31/31/3
−
⎡⎤⎡ ⎤
1
1 0 1 1 1 1/ 4 3/4 1/ 4 1/ 4 1/ 4
,
1 1 0 1 1 1/ 4 1/4 3/ 4 1/ 4 1/4
AA
−
−
⎢⎥⎢ ⎥
⎢⎥⎢ ⎥
==
−
1
66
110111 1/51/54/51/51/51/5
,
AA
−
⎢⎥⎢ ⎥
−
==
⎢⎥⎢ ⎥
66
AI
conjecture is:
A
= J – I
and
1
1
AJI
−
=⋅−
n
A
n
((n – 1)
–1
J – I) = (n – 1)
–1
A
n
J – A
n
= J – (J – I) = I
n
n