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1.8 • Solutions 61
25. Any point x on the line through p in the direction of v satisfies the parametric equation
x = p + tv for some value of t. By linearity, the image T(x) satisfies the parametric equation
26. a. From the figure following Exercise 22 in Section 1.5, the line through p and q is in the direction
of q – p, and so the equation of the line is x = p + t(q – p) = p + tq – tp = (1 – t)p + tq.
b. Consider x = (1 – t)p + tq for t such that 0 < t < 1. Then, by linearity of T,
27. Any point x on the plane P satisfies the parametric equation x = su + tv for some values of s and t.
By linearity, the image T(x) satisfies the parametric equation
28. Consider a point x in the parallelogram determined by u and v, say x = au + bv for 0 < a < 1,
0 < b < 1. By linearity of T, the image of x is
29.
30. Given any x in R
n
, there are constants c
1
, …, c
p
such that x = c
1
v
1
+ c
p
v
p
, because v
1
, …, v
p
span
R
n
. Then, from property (5) of a linear transformation,
31. (The Study Guide has a more detailed discussion of the proof.) Suppose that {v
1
, v
2
, v
3
} is linearly
dependent. Then there exist scalars c
1
, c
2
, c
3
, not all zero, such that
32. Take any vector (x
1
, x
2
) with x
2
≠
0, and use a negative scalar. For instance, T(0, 1) = (–2, –4), but
33. One possibility is to show that T does not map the zero vector into the zero vector, something that
34. Take u and v in R
3
and let c and d be scalars. Then
cu + dv = (cu
1
+ dv
1
, cu
2
+ dv
2
, cu
3
+ dv
3
). The transformation T is linear because
35. Take u and v in R
3
and let c and d be scalars. Then
cu + dv = (cu
1
+ dv
1
, cu
2
+ dv
2
, cu
3
+ dv
3
). The transformation T is linear because
36. Suppose that {u, v} is a linearly independent set in R
n
and yet T(u) and T(v) are linearly dependent.
Then there exist weights c
1
, c
2
, not both zero, such that c
1
T(u) + c
2
T(v) = 0 . Because T is linear,
1.8 • Solutions 63
23 5 5 0 1010 0
77 0 0 0 0110 0
−
⎡⎤⎡⎤
⎢⎥⎢⎥
−
⎢⎥⎢⎥
13
23
–
–
xx
xx
=
⎧
⎪=
⎪
1
1
−
⎡
⎤
⎢
⎥
−
⎢
⎥
34700 10010
58740 01010
−
⎡⎤⎡⎤
⎢⎥⎢⎥
−
⎢⎥⎢⎥
14
24
xx
xx
=−
⎧
⎪=−
⎪
1
1
−
⎡
⎤
⎢
⎥
−
⎢
⎥
23 5 5 8 1010 1
77 0 0 7 01102
−
⎡⎤⎡⎤
⎢⎥⎢⎥
−
⎢⎥⎢⎥
because the augmented matrix shows a consistent system. In fact,
13
1–
2–
xx
xx
=
⎧
⎪=
1
⎡
⎤
⎢
⎥
34704 10011
5 8 74 4 01012
−
⎡⎤⎡⎤
⎢⎥⎢⎥
−−
⎢⎥⎢⎥
because the augmented matrix shows a consistent system. In fact,
1
xx
=−
⎧
1
⎡
⎤
Notes:
At the end of Section 1.8, the Study Guide provides a list of equations, figures, examples, and
connections with concepts that will strengthen a student’s understanding of linear transformations. I
encourage my students to continue the construction of review sheets similar to those for “span” and
“linear independence,” but I refrain from collecting these sheets. At some point the students have to
assume the responsibility for mastering this material.
.
r
1.9 SOLUTIONS
Notes
: This section is optional if yo
u
instructors will want to cover at least
T
illustrate a fast way to solve Exercises
basis.
Exercises 25–28 and 31–36 offer
important links to earlier material.
35
12
−
⎡⎤
⎢⎥
3. T(e
1
) = e
1
– 3e
2
=
1
3
⎡⎤
⎢⎥
−
⎣⎦
, T(e
2
) = e
2
,
A
5. T(e
1
) = e
2
, T(e
2
) = –e
1
. A =
[
2
−e
e
7. Follow what happens to e
1
and e
2
.
S
circle in the plane, it rotates throug
h
point on the unit circle that lies in t
h
d
r
u plan to treat linear transformations only lightl
y
T
heorem 10 and a few geometric examples. Exercis
e
17–22 without explicitly computing the images of
fairly easy writing practice. Exercises 31, 32, and
A
=
10
31
⎡⎤
⎢⎥
−
⎣⎦
]
1
01
10
=−
⎡⎤
⎢⎥
⎣⎦
e
S
ince e
1
is on the unit
h
–3 /4
radians into a
h
e third quadrant and
d
y
, but many
e
s 15 and 16
the standard
35 provide
1.9 • Solutions 65
line
21
xx=− , namely,
(1 / 2 , –1 / 2 )
. Then this image reflects in the horizontal axis to
⎣
⎦
8. The horizontal shear maps e
1
into e
1
, and then the reflection in the line x
2
= –x
1
maps e
1
into –e
2
. (See
Table 1.) The horizontal shear maps e
2
into e
2
into e
2
+ 2e
1
. To find the image of e
2
+ 2e
1
when it is
reflected in the line x
2
= –x
1
, use the fact that such a reflection is a linear transformation. So, the
image of e
2
+ 2e
1
is the same linear combination of the images of e
2
and e
1
, namely,
–e
+ 2(–e
) = – e
– 2e
. To summarize,
9.
,
01
and so 10
A
→ → →− →−
−
⎡
⎤
−=
eee eee
11. The transformation T described maps
11 1
→→−
ee e
and maps
222
.
→− →−
eee
A rotation through
.
R
12. The transformation T in Exercise 10 maps
11 2
→→
eee
and maps
221
→− →−
eee
. A rotation about
the origin through
/
2
π
radians also maps e
1
into e
2
and maps e
2
into –e
1
. Since a linear
13. Since (2, 1)=2 e
1
+ e
2
, the image of (2, 1) under T is 2T(e
1
) + T(e
2
), by linearity of T. On the figure in
the exercise, locate 2T(e
1
) and use it with T(e
2
) to form the parallelogram shown below.
66 CHAPTER 1 • Linear Equations in Linear Algebra
112
240 24
xxx
−−
⎡⎤⎡ ⎤⎡⎤
12
1
32 32
xx
xxx
−−
⎡⎤ ⎡ ⎤
⎡⎤
⎢⎥ ⎢ ⎥
17. To express T(x) as Ax , write T(x) and x as column vectors, and then fill in the entries in A by
inspection, as done in Exercises 15 and 16. Note that since T(x) and x have four entries, A must be a
4×4 matrix.
12 1 1
2120 0
xx x x
+
⎡⎤ ⎡⎤ ⎡⎤
⎡⎤⎡ ⎤
18. As in Exercise 17, write T(x) and x as column vectors. Since x has 2 entries, A has 2 columns. Since
T(x) has 4 entries, A has 4 rows.
12
414
xx
+
⎡⎤
⎡⎤ ⎡ ⎤
19. Since T(x) has 2 entries, A has 2 rows. Since x has 3 entries, A has 3 columns.
11
54 154
xx
xx x Ax x
⎡⎤ ⎡⎤
−+ −
⎡⎤
⎡⎤⎡ ⎤
20. Since T(x) has 1 entry, A has 1 row. Since x has 4 entries, A has 4 columns.
11
xx
⎡⎤ ⎡⎤
1.9 • Solutions 67
21. T(x) =
12 1 1
11
45 45
xx x x
A
xx x x
+
⎡⎤⎡⎤ ⎡⎤⎡⎤ ⎡ ⎤
==
⎢⎥⎢⎥ ⎢⎥
⎢⎥ ⎢ ⎥
+⎣⎦ ⎣ ⎦
. To solve T(x) = 3
8
⎡
⎤
⎢
⎥
, row reduce the augmented
12
221
xx xx
−−
⎡⎤⎡⎤⎡⎤
0
⎡
⎤
augmented matrix:
210 2 10 210 210 202 101
−−−−
⎡⎤⎡ ⎤⎡⎤⎡⎤⎡⎤⎡⎤
23. a. True. See Theorem 10.
b. True. See Example 3.
24. a. False. See Theorem 12.
b. True. See Theorem 10.
25. A row interchange and a row replacement on the standard matrix A of the transformation T in
120 0
⎡
⎤
26. The standard matrix A of the transformation T in Exercise 2 is 2×3. Its columns are linearly
dependent because A has more columns than rows. So T is not one-to-one, by Theorem 12. Also, A is
27. The standard matrix A of the transformation T in Exercise 19 is 154
−
⎡
⎤
⎢
⎥
. The columns of A
28. The standard matrix A of the transformation T in Exercise 14 has linearly independent columns,
because the figure in that exercise shows that a
1
and a
2
are not multiples. So T is one-to-one, by
29. By Theorem 12, the columns of the standard matrix A must be linearly independent and hence the
equation Ax = 0 has no free variables. So each column of A must be a pivot column:
**
⎡⎤
30. By Theorem 12, the columns of the standard matrix A must span R
3
. By Theorem 4, the matrix must
have a pivot in each row. There are four possibilities for the echelon form:
*** *** *** 0 **
⎡⎤⎡⎤⎡⎤⎡⎤
31. “T is one-to-one if and only if A has n pivot columns.” By Theorem 12(b), T is one-to-one if and only
32. The transformation T maps R
n
onto R
m
if and only if the columns of A span R
m
, by Theorem 12. This
33. Define :
nm
T→
RR by T(x) = Bx for some m×n matrix B, and let A be the standard matrix for T.
34. Take u and v in R
p
and let c and d be scalars. Then
1.10 • Solutions 69
35. If :
nm
T→
RR maps
n
R
onto
m
R
, then its standard matrix A has a pivot in each row, by Theorem
36. The transformation T maps R
n
onto R
m
if and only if for each y in R
m
there exists an x in R
n
such that
y = T(x).
5656 1001
8338 0101
−−−
⎡⎤⎡⎤
⎢⎥⎢⎥
−−
⎢⎥⎢⎥
38. [M]
7599 1000
5644 0100
~~
−
⎡⎤⎡⎤
⎢⎥⎢⎥
−
⎢⎥⎢⎥
⋅⋅⋅
⎢⎥⎢⎥
. Yes. There is a pivot in every column of the
39. [M]
47375 10050
685128 01010
~~
710 8 914 0 0 1 2 0
−
⎡⎤⎡⎤
⎢⎥⎢⎥
−−
⎢⎥⎢⎥
⎢⎥⎢⎥
⋅⋅⋅
−−− −
. There is not a pivot in every row,
40. [M]
9 43 5 6 1 10000
14 15 7 5 4 0 1 0 0 0
~~
861259 00100
−
⎡⎤⎡⎤
⎢⎥⎢⎥
−−
⎢⎥⎢⎥
⎢⎥⎢⎥
⋅⋅⋅
−− −−
⎢⎥⎢⎥
. There is a pivot in every row, so the
1.10 SOLUTIONS
1. a. If x
1
is the number of servings of Cheerios and x
2
is the number of servings of 100% Natural
Cereal, then x
1
and x
2
should satisfy
70 CHAPTER 1 • Linear Equations in Linear Algebra
That is,
110 130 295
⎡⎤ ⎡⎤⎡⎤
⎢⎥ ⎢⎥⎢⎥
b. The equivalent matrix equation is
1
2
110 130 295
43 9
20 18 48
25 8
x
x
⎡
⎤⎡⎤
⎢
⎥⎢⎥
⎡⎤
⎢
⎥⎢⎥
=
⎢⎥
⎢
⎥⎢⎥
⎣⎦
⎢
⎥⎢⎥
⎢
⎥⎢⎥
⎣
⎦⎣⎦
. To solve this, row reduce the
augmented matrix for this equation.
110 130 295 2 5 8 1 2.5 4
⎡⎤⎡⎤⎡⎤
1 2.5 4 1 2.5 4 1 0 1.5
0 7 7 011 011
⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥
−−
⎢⎥⎢⎥⎢⎥
2. Set up nutrient vectors for one serving of Shredded Wheat (SW) and Kellogg's Crispix (Crp):
Nutrients: SW Crp
calories 160 110
⎡⎤⎡⎤
160 110
⎡⎤
⎢⎥
b. [M] Let u
1
and u
2
be the number of servings of Shredded Wheat and Crispix, respectively. Can
1
120
3.2
⎡
⎤
⎢
⎥
⎡⎤
1.10 • Solutions 71
160 110 130 1 .4 .64 1 .4 .64 1 .4 .64 1 0 .4
5 23.20 0 004627.601.601.6
~~~~
⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥
⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥
3. a. [M] Let x
1
, x
2
, and x
3
be the number of servings ofAnnies’s Mac and Cheese, broccoli, and
chicken, respectively, needed for the lunch. The values of x
1
, x
2
, and x
3
should satisfy
nutrients nutrients nutrients quantities
⎡ ⎤⎡⎤⎡⎤⎡ ⎤
From the given data,
270 51 70 400
⎡⎤ ⎡⎤ ⎡⎤⎡⎤
To solve, row reduce the corresponding augmented matrix:
270 51 70 400 1 0 0 .99
⎡⎤⎡⎤
.99 servings of Mac and Cheese
⎡⎤⎡ ⎤
b. [M] Changing from Annie’s Mac and Cheese to Annie’s Whole Wheat Shells and White Cheddar
changes the vector equation to
260 51 70 400
To solve, row reduce the corresponding augmented matrix:
260 51 70 400 1 0 0 1.09
⎡⎤⎡⎤
⎣⎦⎣⎦
1.09 servings of Shells
⎡⎤⎡ ⎤
4. Here are the data, assembled from Table 1 and Exercise 4:
Mg of Nutrients/Unit Nutrients
Required
soy soy
Nutrient (milligrams)
milk flour whey prot.
protein 36 51 13 80 33
a. Let x
1
, x
2
, x
3
, x
4
represent the number of units of nonfat milk, soy flour, whey, and isolated soy
protein, respectively. These amounts must satisfy the following matrix equation
1
2
36 51 13 80 33
52 34 74 0 45
x
x
⎡⎤
⎡⎤⎡⎤
⎢⎥
⎢⎥⎢⎥
36 51 13 80 33 0 0 0 .64
1
52 34 74 0 45 0 0 0 .54
1
⎡⎤⎡⎤
⎢⎥⎢⎥
⎢⎥⎢⎥
5. Loop 1: The resistance vector is
1
Total of RI voltage drops for current
11
I
⎡⎤
Loop 2: The resistance vector is
11
4
5Voltage drop for is negative; flows in opposite direction
II
⎡⎤
−
0
⎡⎤
0
⎡
⎤
500
11
⎡
⎤
−
1.10 • Solutions 73
Notice that each off-diagonal entry of R is negative (or zero). This happens because the loop current
directions are all chosen in the same direction on the figure. (For each loop j, this choice forces the
currents in other loops adjacent to loop j to flow in the direction opposite to current I
j
.)
50
⎡⎤
each loop is opposite to the direction chosen for positive current flow. Thus, the equation Ri = v
becomes
1
500 50
11
I
⎡⎤
⎡⎤⎡⎤
−⎢⎥
1
3.68
I
⎡⎤
⎡
⎤
6. Loop 1: The resistance vector is
1
6Total of RI voltage drops for current
I
⎡⎤
Loop 2: The resistance vector is
11
Voltage drop for is negative; flows in opposite direction
1
II
⎡⎤
−
0
4
⎡⎤
⎢⎥
−
⎢⎥
0
0
⎡
⎤
⎢
⎥
⎢
⎥
6100
1940
−
⎡
⎤
⎢
⎥
−−
⎢
⎥
30
20
⎡⎤
⎢⎥
⎢⎥
v becomes
1
6100 30
1940 20
I
I
−⎡⎤
⎡⎤⎡⎤
⎢⎥
⎢⎥⎢⎥
−−
1
6.36
8.14
I
I
⎡⎤
⎡
⎤
⎢⎥
⎢
⎥
