Solution manual for linear algebra for engineers and scientists using matlab by hardy

Apply forward and backward reduction on the augmented matrix M = [A| b] of S.. Apply forward and backward reduction on the augmented matrix M = [A| b] of S... Equations E1, E2, E3 in S a

Chapter 1

Linear Systems

1 (x + 2)2+ (y − 1)2= x2+ 4x + 4 + y2− 2y + 1 = 3 + x2+ y2 ⇒ 4x − 2y = −2 ⇒ 2x − y = −1.Linear

2 (x − y)2= x2− 2xy + y2= x2+ y2 ⇒ −2xy = 0 ⇒ xy = 0 Nonlinear

3 (x − y)(x + y) = x2+ xy − xy − y2= x2+ y2 ⇒ 2y2= 0 ⇒ y = 0 Linear

4 x + (x − 1)(y − 1) = x + xy − x − y + 1 = xy + y − 1 ⇒ −2y2= −2 ⇒ y = 1 Linear

5 Pivot variables: x1, x2, x3 Free variable: x4 No variables uniquely determined Let x4 = t,where t is a real parameter Back-substituting, third equation: x3= 3 − 2t, second equation: x2=

−1+x3−2x4= −1+(3−2t)−2t = 2−4t, first equation: x1= 2x2+10x4= 2(2−4t)+10t = 4+2t.Solution: (x1, x2, x3, x4) = (4 + 2t, 2 − 4t, 3 − 2t, t)

6 Pivot variables: x1, x2, x4 Free variable: x3 Uniquely determined: x4 Let x3 = t, where t is areal parameter Back-substituting, third equation: x4= 2.4, second equation: x2= 1 + x3− 7x4=1+t−7(2.4) = −15.8+t, first equation: x1= 2−2x2−x3+x4= 2−2(−15.8+t)−t+2.4 = 36−3t.Solution: (x1, x2, x3, x4) = (36 − 3t, −15.8 + t, t, 2.4)

7 Pivot variables: x1, x3 Free variables: x2, x4 No variables uniquely determined Let x2= s and

x4= t, where s, t are real parameters Back-substituting, second equation: x3= −1.5 + 0.5t, firstequation: x1= 4 − 0.5s Solution: (x1, x2, x3, x4) = (4 − 0.5s, s, −1.5 + 0.5t, t)

8 Pivot variables: x1, x3, x4 Free variables: x2, x5 Uniquely determined: x3 Let x2 = s and

x5 = t, where s, t are real parameters Back-substituting, third equation: x4 = 2.0 − t, second

equation: x3= 2.1, first equation: x1= 7.0+ x2−2x3−x4= 7.0+s−2(2.1)−(2.0−t) = 0.8+s+t.Solution: (x1, x2, x3, x4, x5) = (0.8 + s + t, s, 2.1, 2.0 − t, t).

9

3x1 = 6.0 E1

x1− x2= 5.5 E2Solve E1: x1= 2.0 Solve E2: −x2= 5.5 − x1=3.5 ⇒ x2= −3.5 Solution: (x1, x2) = (2, −3.5)

Refer to Figure 1.1 [9] Equations E1, E2are resented, respectively, by lines L1, L2in R2 0 2 4 6

rep-−6

−4

−2 0

r3−2r1→ r3, r3−r2→ r3, to obtain an echelon form U for M and a linear system (S)′equivalent

x1= 1 − x2+ x3= 4 Solution: (x1, x2, x3) = (4, −2, 1) (S) is represented by three planes in R3with a single point in common

12 Let (S) denote the linear system M = [A| b] =

 Perform the following

EROs on M: r1↔ r2, r2− 3r1→ r2, r3+ 2r1→ r3, r3+ r2→ r3, to obtain an echelon form U

for M and a linear system (S)′ equivalent to (S).

,

the zero equation occuring in (S)′ because some equation in (S) is a linear combination of otherequations in (S) Pivot variables: x1, x3 Free: x2 Uniquely determined: x3 Let x2= t be a realparameter Back-substituting, second equation: x3 = 3.0, first equation: x1 = 0.5 − x2− x3 =

−2.5 − t Solution: (x1, x2, x3) = (−2.5 − t, t, 3.0) (S) is represented by three planes in R3 thatintersect in a line

13 Let (S) denote the linear system M = [A| b] =

14 Let (S) denote the linear system M = [A| b] =

1 3 0 1 Perform the followingEROs on M: r1↔ r2, to obtain an echelon form U for M and a linear system (S)′ equivalent to(S)

15 Let (S) denote the linear system M = [A| b] =

integer coefficients, perform the following EROs on M: r1 ↔ r3, r2− 3r1 → r2, r3− 5r1 → r3,

r4− 9r1→ r4, r3− 7r2 → r3, r4− 8r2 → r4, r4− r3 → r4, to obtain an echelon form U for M

and a linear system (S)′ equivalent to (S).

in R3with no point in common

16 Let (S) denote the linear system M = [A| b] =

 Perform the following

EROs on M: r2− 2r1→ r2, r3+ r1→ r3, r3−35r2→ r3, to obtain an echelon form U for M and

a linear system (S)′ equivalent to (S)

5x3= 215Pivot variables: x1, x2, x3 Free variables: none All variables uniquely determined Back-substituting, third equation: x3 = 1, second equation: x2= 1

Pivot variables: x1, x2, x3 Free variables: none All variables uniquely determined Reading thesolution off from M*, we have x1= x2= 0, x3= 2 Solution: (x1, x2, x3) = (0, 0, 2).

19 Let (S) denote the linear system Then (S) is 2 × 4 and will be either inconsistent or consistentwith at least one free variable, indicating infinitely many solutions Finding the reduced form M*

on M = [A| b] using the sequence of EROs shown below:

Clear below in column 1

r2− 2r1→ r2, r4− r1→ r4

Clear below in column 3

r2↔ r3, r3− 2r2→ r3, r4+ r2→ r4Clear below in column 5

−1

4r3→ r3, r4+ 2r3→ r4and we have

and the last row in U indicates that (S) is inconsistent

21 Finding the reduced form M* of M = [A| b], we have

22 Free variables: x2, x3 Let x2 = s and x3 = t, where s, t are real parameters Then x1 =

4 − 2s − 3t General solution: (x1, x2, x3) = (4 − 2s − 3t, s, t) Setting s = 2 and t = −1gives x1 = 4 − 4 + 3 = 3, showing that (3, 2, −1) is a particular solution Setting x1 = p and

x3 = q gives x2 = 2 − p − 32q, giving the general solution (x1, x2, x3) = (p, 2 − 12p − 32q, q) Forany particular solution corresponding to the ordered pair (s, t), for fixed s, t, the unique orderedpair (p, q) = (4 − 2s − 3t, t) corresponds to the same particular solution Conversely, for anyparticular solution corresponding to the ordered pair (p, q), for fixed p, q, the unique ordered pair(s, t) = (2 −12p − 32q, q) corresponds to the same particular solution.

, where A is 3 × 3, M is 3 × 4 and b is 3 × 1 Solving the

system by Gauss–Jordan elimination corresponds to reducing M to its reduced form M*

where t is a real parameter

Particular solutions in nonnegative integers must satisfy x1 = 12 − 5t ≥ 0, x2 = 4 − 3t ≥ 0 and

t ≥ 0 Hence 125 ≥ t ≥ 0 and 43 ≥ t ≥ 0 so that 1 ≥ t ≥ 0, where t is an integer Particularsolutions: (x1, x2, x3) = (12, 4, 0), (7, 1, 1)

Particular solutions in nonnegative integers must satisfy x1= 3 − 2s − t ≥ 0, x3= 3 − t ≥ 0, s ≥ 0,and t ≥ 0 Hence 3 ≥ t ≥ 0 ⇒ t = 0, 1, 2, 3 Particular solutions (x1, x2, x3, x4) are:

If α = 1, then the right side of (1.1) shows that (S) has infinitely many solutions Now assume

α = 1 and reduce further to obtain (1.2), noting that 1 − α2+ 1 − α = 2 − α − α2= (1 − α)(2 + α)

If α = −2, then (S) has a unique solution because A s I3 and if α = −2, the (S) is inconsistent

If α = 0, then the reduced form of M is

(a) α = 1, no solution, (b) α = 1, unique solution, (c) No value α gives infinitely many solutions

27 Apply forward and backward reduction on the augmented matrix M = [A| b] of (S)

28 Substituting x1= −1, x2= 1, x3= 0 into E1gives −3(−1) + 5(1) − 7(0) = 8 = −2 The particularsolution fails to satisfy E1 and so is not a particular solution to (S)

Apply forward and backward reduction on the augmented matrix M = [A| b] of (S)

−x1+ 6x2+ 8x3= −154x1+ 2x2+ 7x3= 21Perform forward reduction on M, maintaining integer coefficients.

32 M is 2 × 5, A is 2 × 4 (S)

x1− x2 + x4+ x5= 2

−x1+ 2x2− x3+ x4+ x5= 2Performing forward reduction using r2+ r1→ r2, we have

an inconsistent linear system may (or may not) result in a consistent subsystem See Exercise 60

33 Let (S) denote the linear system Performing forward and backward elimination on (S) usingaugmented matrices, we have

Unique solution: (x, y) = (−8, 2) Refer to Figure 1.1 [33]

Equations E1, E2, E3 in (S) are represented, respectively,

by lines L1, L2, L3 in R2 that intersect in a single point

−1 0 1 2 3

34 Rearranging the system into standard form, we have(S)

2x + 4y = − 6 E17x + 5y = 6 E2

Perform forward and ward elimination using augmented matrices The EROsare chosen to preserve integer coefficients

Unique solution: (x, y) = (3, −3) Refer to Figure 1.1 [34]

Equations E1, E2 in (S) are represented, respectively, bylines L1, L2in R2that intersect in a single point

2r1→ r1 1

2r2→ r2

1 0 1.5

Unique solution: (x, y) = (1.5, 2) Refer to Figure 1.1 [35]

Equations E1, E2 in (S) are represented, respectively, bylines L1, L2in R2that intersect in a single point

0 1 2

Row 2 in the last matrix shows that (S) is inconsistent

Refer to Figure 1.1 [36] Equations E1, E2 in (S) are resented, respectively, by parallel lines L1, L2 in R2 with

rep-no point in common

0 1 2 3

37 Perform forward elimination using augmented matrices.

(S) is consistent if and only if c = 4.5 and for this value of c the unique solution (x1, x2) = (0.5, 0.5)

is obtained by performing backward elimination on the linear system with augmented matrix U.Refer to Figure 1.1 [37] Equations E1, E2, E3 in (S) are represented, respectively, by lines L1,

L2, L3in R2 intersecting in a single point

38 Perform forward elimination using augmented matrices

1.6667 2

(S) is consistent if and only if c = ±√3 For these values of c the unique solution (x1, x2) = (1, 1)

is obtained by performing backward elimination on the linear system with augmented matrix U.Refer to Figure 1.1 [38] Equations E1, E2, E3 in (S) are represented, respectively, by lines L1,

L2, L3in R2 that intersect in a single point

39 Substitute x = 1, y = 2 and x = −1, y = 0.5 into mx+b =

y to obtain a 2 × 2 linear system

(S)

m + b = 2

−m + b = 0.5Perform forward and backward elimination on (S) usingaugmented matrices

40 Substitute x = −1, y = 1, z = 0 and x = 0, y = 2, z = 1 and x = 4, y = 5, z = 2 into ax+by+c = z

to obtain a 3 × 3 linear system

We will solve by Gauss-elimination beginning with the augmented matrix for (S) Forward nation:

b + c = 3 − 5 = −2 The equation of the plane P is

z = −2x+3y −5, alternatively, 2x−3y +z −5 = 0

Refer to Figure 1.1 [40] The plane intercepts thez-axis in the point (0, 0, −5) The point (3, 0, −11)lies in P and in the xz-plane The point (0, 3, 4)lies in P and in the yz-plane

0 1 2 3

0 1 2 3

−11

−5 0 4

x-valuesy-values

Figure 1.1 [40]

41 The values x1 = 0.7, x2 = 0 do not satisfy equation E1and so (0.7, 0) is not a particular solution to (S) Substi-tuting the values x1 = 1.2, x2 = −1 into each equation

of (S) in turn, we have E1: 1.2 + 1.5(−1) = −0.3, E2:2(1.2) + (−1) = 1.4, E3: 3(1.2) + 2.5(−1) = 1.1 All equa-tions in (S) are satisfied and so (1.2, −1) is a particularsolution Refer to Figure 1.1 [41] Equations E1, E2, E3are represented, respectively, by lines L1, L2, L3 that in-tersect in the single point (1.2, −1) in R2 It follows that(1.2, −1) is the unique solution to (S) and there is only oneparticular solution in this case

−2

−1 0

(c) Refer to Figure 1.1 [42] The equation −3x−

y +z = 0 is represented by a plane P in R3passingthrough the origin (0, 0, 0) The coordinates of anypoint (a, b, c) lying in P make (S) consistent, andconversely, any values of (a, b, c) which make (S)consistent correspond to a point on P

−1 0

1

−1

0 1

−2 0 2

x-valuesy-values

per-a unique point C forming isosceles triper-angles △DEC per-and

△EF C with base angles α and β, respectively, and mon side EC Hence r = |DC| = |EF | = |F C| is theradius of the circle passing through the three points

43 Substitute (x, y) = (0, −1), (1, 0), (2, 2) into the equation ax + by + c = −(x2+ y2) to obtain a 3 × 3linear system

2

− 72

2+



y −72

2

− 72

2+



y −72

2

−652 = 0Hence p = −3.5, q = −3.5 and r2=65

2 so that r =

65/2 ≃ 5.7

44 Substitute (x, y) = (1, 1), (−1, 1), (−1, 2) into the equation ax + by + c = −(x2+ y2) to obtain a

showing that (S) has a unique solution (a, b, c) = (0, −3, 1) giving x2+ y2− 3y + 1 = 0 as theequation of the circle Completing the square, we have

2+ 1

4 so that r =

5/4 ≃ 1.1180

45 Denote the system by (S) Setting X = xy, Y = y2, Z = yz in (S) gives a linear system (S)′,where

0

−1

0 0.5 1

π π

tan γ

tan γ sin α cos β

Figure 1.1 [46]

For the specified ranges, sin α = 1 ⇒ α = π/2 ≃ 1.5708, cos β = 0.5 ⇒ β = π3 ≃ 1.0472,5π3 ≃5.2360, tan γ = −1 ⇒ γ = 3π

4 ≃ 2.3562 There are two solutions: (α, β, γ) =  π2,π

48 (a) Assume m = 0, for otherwise no action is taken and (S2) is identical to (S1) Perform

E1− mE2→ E1 on (S1) to obtain

(S2)

(p1− mq1)x1+ (p2− mq2)x2+ (p3− mq3)x3= b1− mb2

q1x1+ q2x3+ q3x3= b2The second equation in (S2) is satisfied by the solution (x1, x2, x3) = (a1, a2, a3) and so is the firstbecause

(p1− mq1)a1+ (p2− mq2)a2+ (p3− mq3)a3 = p1a1+ p2a2+ p3a3− m(q1a1+ q2a2+ q3a3)

= b1− mb2(b) The solution set to the linear system (S2) obtained by performing E1 ↔ E2 is clearly thesame as (S1) More generally, rearranging (permuting) the equations in a linear system does notchange the solution set

(c) Performing cE1→ E1on (S1) we obtain a linear system (S2) whose first equation is satisfied

by the solution (x1, x2, x3) = (a1, a2, a3) because cp1a1+cp2a2+cp3a3= c(p1a1+p2a2+p3a3) = cb1.Note that scaling by c = 0 turns E1into a zero which is satisfied for any x1, x2, x3

49 The parametric form of the equation of L, given in (1.30) on page 16, is (x, y, z) = (1 − 0.6t, 1 −0.4t, t), where t is a real parameter (a) Setting x = 0 gives t = 5

3 and so L passes through theplane x = 0 (yz-plane) in the point 0,1

50 The problem stated in the Nine Chapters (200–100 B C ) is roughly this: There are three bundles

of corn, labeled B1, B2, B3, respectively Each bundle contains a certain (unknown) number of

measures of corn Three bundles of B1, two of B2 and one of B3contain 39 measures, two of B1,three of B2 and one of B3 contain 34 measures, and one of B1, two of B2 and three of B3 contain

26 measures How many measures of corn are contained in one bundle of each type?

Let x1, x2, x3 denote the measures of corn contained in one bundle of each type B1, B2, B3,respectively The problem as stated translates into the linear system (S) Performing forwardelimination on (S) using augmented matrices, and maintaining integer coefficients, we have

26 − 2(4.25) − 3(2.75) = 9.25 Unique solution: (x1, x2, x3) = (9.25, 4.25, 2.75) (measures)

51 Apply Gauss elimination using augmented matrices

Pivot variable: x1 Free variable: x2 Let x2= t, where t is a real parameter Back-substituting,

we have x2= 5 − 3t, x1=12−32x2−12t = 12−32(5 − 3t) −12t = −7 + 4t The equation of the line

53 True Each variable is a pivot variable

54 False For example, the equation 0x1+ 0x2= 0 has no pivot variables because there are no nonzerocoefficients

55 False For example, the linear system

x1+ x2= 12x1+ 2x2= 2 is equivalent to

x1+ x2= 10x1+ 0x2= 0,which has only one pivot variable However,

x1+ x2= 1

x1+ 2x2= 2 is equivalent to

x1+ x2= 10x1+ x2= 1,which has two pivot variables A consistent 2 × 2 linear system has two pivot variables

56 False For example, the linear system







x1+ x2+ x3 = 12x1+ 2x2+ 2x3 = 2

−x1− x2− x3= −1

has two free variables

57 False For example, the linear system







x1+ x2= 12x1+ 3x2= 33x1+ 3x2= 3

is consistent, having solution (x1, x2) =(0, 1)

58 False See Illustration 1.2

59 True Although scaling an equation can provide useful simplification during forward elimination,scaling only becomes necessary during back-substitution or backward elimination

60 False See Exercise 32

61 False According to the Computational Note on page 14, Gauss–Jordan elimination is about 30%less efficient than Gauss–elimination for solving large-scale linear systems

62 True The matrix contains m rows

63 True Just as a linear equation in three variables has two degrees of freedom and is represented by

a plane in R3, so a linear equation in eight variables has seven degrees of freedom and is represented

67 Apply forward and backward elimination on (S) Using gauss.m with the format rat option, the

4 × 6 augmented matrix M of (S) changes into the augmented matrix M* of an equivalent system

vari-3 + 2t, t,1

3, 0) The variables x4 and x5 are uniquely determined.The system (1.24) on page 14 is identical to (S) except that the variable x5(and its coefficients inall equations) is missing As x5= 0, the solution to (1.24) is (x1, x2, x3, x4) = (6−6t, −2

3+2t, t,1

3)

68 Apply forward and backward elimination on (S) Using gauss.m with the format rat option, the

5 × 6 augmented matrix M of (S) changes into the augmented matrix M* of an equivalent system(S)′

and we solve (S)′ which has the same solution set as (S) M* shows that (S)′ has pivot variables

x1, x3, x4and free variables x2, x5 Let x3= s and x5= t, where s, t is are real parameters Then(x1, x2, x3, x4, x5) = (−497 − s, s, −75,249

4 −3

2t, t) The variable x3 is uniquely determined

69 Solving the system

x + 2y = 2 E1

x + y = 3 E2

by Gauss elimination, we have (x1, x2) = (4, −1) In order to plot the lines and their intersectionpoint we need to define a (row) vector x of x-values that includes the value x = 4 (contrast withtext, page 20) Then y1 and y2 are vectors whose components are y-values corresponding to eachvalue x in the vector x

The commandtitle(′Two lines in 2-space′, ′FontSize′,10)adds a title to the plot and the commandsgtext(′ L 1′)

gtext(′ L 2′)manually position labels on the lines

−2

−1 0 1

L 1

L 2

Figure 1.1 [69]

Two lines in 2-space

72 We are given that a + b + c = 14 and a − b + c = 0 (a) We have n = 100a + 10b + c and

m = 100a + 10c + b and so n − m = 27 ⇒ b − c = 3 and we obtain a 3 × 3 linear system

(b) If n − m = k, then 9b − 9c = k implies that b − c = k9 and this equation replaces equation E3

in (1.3) to form a new system (S)′ The same sequence of EROs are used to reduce the augmentedmatrix of (S)′ We have

We must have 0 ≤ 7 −k9 and so 0 ≤ k ≤ 63 and k must be a multiple of 9 The solutions in terms

of k and n are given below

(k, n) = (9, 176), (18, 275), (27, 374), (36, 473), (45, 572), (54, 671), (63, 770)For each value k = 1, 2, , test if the triples of nonnegative integers (a, b, c) satisfy the constraints

a + b + c = 14, b = a + c and that 9b − 9c = k is an integer The range for b and c is 0, 1, , 9 and

so k = 9b − 9c ≤ 9(9) − 9(0) = 81 gives an upper bound for k

1.2 E chelon For ms, Rank

Exercises 1–8 Pivot positions in the matrix A are located using any echelon form for A The pivots inthese positions are nonzero numbers that are necessarily equal to 1 (circled) in the reduced form A*

We find r = rank A by counting the number of nonzero rows in an echelon form for A and this integercoincides with the number of pivots and number of pivot columns in A

Pivot positions: (1, 1) and (2, 2), pivot columns: 1 and 2, rank r = 2

3 A is not in echelon form— forward reduction is required Performing forward reduction, thenbackward reduction, we have

Pivot positions: (1, 1) and (2, 2), pivot columns: 1 and 2, rank r = 2

4 A is not in echelon form— forward reduction is required Performing forward reduction (nobackward reduction required), we have

2r1→ r1

1 −32

Pivot positions: (1, 1), pivot column: 1 and rank r = 1

5 A is in echelon form but not in reduced form— the pivot should have value 1 Computing thereduced form, we have

6 A is not in echelon form— forward reduction is required Performing forward reduction, thenbackward reduction, we have

Pivot positions: (1, 1) and (2, 2), pivot columns: 1 and 2, rank r = 2

7 A is not in echelon form— forward reduction is required Performing forward reduction, thenbackward reduction, we have

Pivot positions: (1, 1) and (2, 4), pivot columns: 1 and 4, rank r = 2

8 A is not in echelon form— forward reduction is required Computing an echelon form and thereduced form, we have

, which shows that columns 1 and 3 in A are pivot

columns and rank r = 2

3r2→ r2

4 8

2 8 3

s3

8r1→ r1 3

8r2→ r2

4 3

2 1 = Band so A s B It can be easily shown that A s I2s B and so rank A = rank B = 2

12 One approach is to show that the reduced forms of A and B are identical

14 We have A s A* and B s B*, where A* and B* are the reduced forms of A and B, respectively.

If A* = B*, then A s A* = B* s B, which shows that A s B We have

4r2→ r2

r3− 2r2→ r3 1

4r2→ r2

r3− 2r2→ r3 1

s1

(b) For α = 11 and any β, we have 3 = rank A = rank M = n the number of unknowns and sothere is a unique solution for every β.

(c) If α = 11 and β = 36, then 2 = rank A = rank M and so there are two pivots and one freevariable, giving infinitely many solutions

(a) If α = 4 and β =29, then 2 = rank A < rank M = 3 and so (S) is inconsistent

(b) If α= 4, then 3 = rank A = rank M = n the number of unknowns and so there is a uniquesolution for any β

(a) If p = −4, then sys is inconsistent If p = −4, then p + 4 = 0 and we can continue forwardreduction.

3r3→ r3

r4+ r3→ r4 1

3r2→ r2

r3− r2→ r3

r4− r3→ r4 1

37 Perform forward elimination using augmented matrices.

(S) is consistent if and only if c = 4.5 and for this value of c the unique solution (x1,... z = and x = 0, y = 2, z = and x = 4, y = 5, z = into ax +by+ c = z

to obtain a × linear system

We will solve by Gauss-elimination beginning with the augmented matrix for (S) Forward... not in echelon form— forward reduction is required Computing an echelon form and thereduced form, we have

, which shows that columns and in A are pivot

columns and rank r =

3r2→