Solution manual for an introduction to linear algebra for science and engineering 2nd edition by norman

i ii √2 √2 Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman... This, along with one of the points, may be used to obtain an equatio

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CHAPTER 1 Euclidean Vector Spaces

Practice Problems

A1 (a)

14

+

23



x1

x2

14



23



14

+

23



(b)

32



−

41



x1

x2

32



41



−

41



32



−

41



(c) 3



−14



3



−14



(d) 2

21



− 2

3

−1



=

42



−

6



21



3

−1



−2

3

−1

2

21



− 2

3

−1



A2 (a)

4

−2

+



−13



26



3

43



=

13

+

4/31



=

7/34



31



− 2

1/41/3



=

22/3



−

1/22/3



=

3/20



2

 √2

√3

+ 3

1

√6



=

2

√6

+

3

6



Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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√2

√2

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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+ t



−51



23

A8 Note that alternative correct answers are possible

2



=

3

−5

 This, along with one of the points, may be used to obtain an equation for the line

x =



−12



+ t

3

−5



Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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x =

41

x2= 3x1+ 2

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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11



+ t

3



−

12



=

3



−

12



=



−62

⎥⎥⎥⎦ Therefore, the points S, T, and U are not

Homework ProblemsB1 (a)

x1

x2

1

−2





−13



1

−2

+



−13



=

01

x1

x2

4

−3



23



−

23



4

−3



−

23



=

2

−6



  A10 (a) LetP, Q,andRbethreepointsinRn,withcorrespondingvectorsp, q,andr. IfP, Q,andRarecollinear,

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(c)

x1

x2

1

−2



3

1



52



33



−

33



−3

52



−

33

(b)

00

(c)



+ t

4



+ t



−21

B9 Other correct answers are possible.

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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2t+3

3/20



+ t



−1/21



+ t

1

−12



11



1

−1



11



13



1

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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2(√

2(√

D2 (a) PQ + QR+ RP can be described informally as “start at P and move to Q, then move from Q to R, then

from R to P; the net result is a zero change in position.”



00

−11

−22

To prove it is not a subspace we just need to find one example where the set is not closed under addition

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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(b) Since the condition of the set only contains linear variables, we suspect that this is a subspace To prove it

is a subspace we need to show that it satisfies the definition of a subspace

⎥⎥⎥⎦ in S Then they must satisfy the condition of

S , so x1 = x3and y1 = y3 We now need to show that x + y =

⎥⎥⎥⎦ satisfies the conditions of the set

⎥⎥⎥⎦ satisfies the conditions of the

(c) Since the condition of the set only contains linear variables, we suspect that this is a subspace Call



tx1

tx2



(d) The condition of the set involves multiplication of entries, so we suspect that it is not a subspace Observe

⎥⎥⎥⎦ is not in the set since

(e) At first glance this might not seem like a subspace since we are adding the vector

⎡

⎢⎢⎢⎢⎢

⎢⎢⎢⎣

234

⎤

⎥⎥⎥⎥⎥

subspace to prove this, but the point of proving theorems is to make problems easier Therefore, we instead

stentryof

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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⎪⎪⎪⎭ and hence is a subspace by Theorem 1.2.2

S By definition S is a subset ofR4and is non-empty since the zero vector satisfies the conditions of the

(b) The set clearly does not contain the zero vector and hence cannot be a subspace

(d) The conditions of the set involve a multiplication of variables, so we suspect that it is not a subspace Using

⎤

⎥⎥⎥⎥⎥

⎥⎥⎥⎥⎥

(e) Since the conditions of the set only contains linear variables, we suspect that this is a subspace Call the

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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A4 Alternative correct answers are possible



− 2

12

+ 1

13



=

00



A5 Alternative correct answers are possible

(a) We observe that neither vector is a scalar multiple of the other Hence, this is a linearly independent set of

−10

−20

−10

−10

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(d) Observe that the third vector is the sum of the first two vectors hence, by Theorem 1.2.3 we can write

−25

B4 Alternative correct answers are possible.

.,



Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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− 7

12

+ 5



−12



=

00



B5 Alternative correct answers are possible.

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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y



∈ V, but x + y is not in U and not in V, so it is not in U ∪ V Thus, U ∪ V is not a subspace.

⎤

⎥⎥⎥⎥⎥

⎥⎥⎥⎥⎥

⎥⎥⎥⎦.

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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D5 By definition of spanning, every x∈ Span B can be written as a linear combination of the vectors in B Now,

D6 Assume that{v1, ,v k , x} is linearly dependent Then, there exist coefficients t1, , t k , t k+1not all zero such that



non-zero Hence, by definition, the set is linearly dependent

(f) TRUE By Theorem 1.2.2

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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1.3 Length and Dot Products Practice Problems

√

3

−4



=

3/5

−4/5



11



2

11

⎤

⎥⎥⎥⎥⎥

⎥⎥⎥⎦

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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−21

⎤

⎥⎥⎥⎥⎥

⎥⎥⎥⎦ = (−3)(2) + 1(−1) + 7(1) = 0 Hence these vectors are orthogonal

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−3/21

3

−1



·

2

k



= 3(2) + (−1)k = 6 − k.

3

Therefore, the vectors are always orthogonal

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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ii

(b) The scalar equation of the plane is

ii

(c) The scalar equation of the hyperplane is

−11

−12

⎥⎥⎥⎦ Therefore, an equation for the

⎥⎥⎥⎦ Hence, an equation for the plane passing through

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⎥⎥⎥⎦ Hence, an equation for the plane passing through

P(1, 2, 1) with normal vector n is



·

22



= 2 =

10



·

2

−97



100

Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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https://TestbankDirect.eu/

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·



−22



= 1(−2) + 2(2) = 2, so they are not orthogonal

(b)

46



·



−32

−1

−21

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B9 (a)

23

−21

There is one solution for t (and thus, one point of intersection of the line and the plane) exactly when

n · d  0 If n · d = 0, there is no solution for t unless we also have n · p = k In this case the equation is satisfied for all t and the line lies in the plane Thus, to have no point of intersection, it is necessary and

D2 Since x = x − y + y,

x = x − y + y = (x − y) + y ≤ x − y + y

the previous one, we can conclude that

Conceptual Problems

D1 (a) Intuitively,ifthereis nopointofintersection, thelineis paralleltothe plane,sothedirectionvectorof



Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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D4 Let x be any point that is equidistant from P and Q Then x satisfies x − p = x − q , or equivalently,

D6 Let S denote the set of all vectors orthogonal to u By definition, a vector orthogonal to u must be inRn , so S is



Solution Manual for An Introduction to Linear Algebra for Science and Engineering 2nd Edition by Norman

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⎡

⎢⎢⎢⎢⎢

⎢⎢⎢⎣

cos αcos βcos γ

⎤

⎥⎥⎥⎥⎥

⎥⎥⎥⎦

cos αcos β



1.4 Projections and Minimum Distance Practice Problems



=

0

−5



3

−5



−

0

−5



=

30



Section1.4ProjectionsandMinimumDistance 25

D7 Bydefinition,avectororthogonaltoanyvectorin S mustbeinRn,soS⊥ isasubsetofRn Moreover,since

The fact that v i  0implies that c i = 0 Sincethis is valid for 1 ≤ i ≤ k we get {v1, ,v k} is linearlyindependent