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Chapter 1

Linear Equations and Matrices

Section 1.1, p. 8 2. x = 1, y = 2, z =−2.

4. No solution.

6. x = 13 + 10t, y =−8− 8t, t any real number.

8. Inconsistent; no solution.

10. x = 2, y =−1.

12. No solution.

14. x =−1, y = 2, z =−2.

16. (a) For example: s = 0, t = 0 is one answer.(b) For example: s = 3, t = 4 is one answer.

(c) s =t2.

18. Yes. The trivial solution is always a solution to a homogeneous system.

20. x = 1, y = 1, z = 4.

22. r =−3.

24. If x1 = s1, x2 = s2, ..., xn = sn satisfy each equation of (2) in the original order, then those same numbers satisfy each equation of (2) when the equations are listed with one of the original ones interchanged, and conversely.

25. If x1 = s1, x2 = s2, . . . , xn = sn is a solution to (2), then the pth and qth equations are satisfied. That is,

ap1s1 + ··· + apnsn = bp

aq1s1 + ··· + aqnsn = bq.

Thus, for any real number r,

(ap1 + raq1)s1 + · · · + (apn + raqn)sn = bp + rbq .

Then if the qth equation in (2) is replaced by the preceding equation, the values x1 = s1, x2 = s2, . . . , xn = sn are a solution to the new linear system since they satisfy each of the equations.

Full file at http://testbanksite.eu/Elementary-Linear-Algebra-with-Applications-9th-Edition--Solution2 Chapter 1

26. (a) A unique point.

(b) There are infinitely many points. (c) No points simultaneously lie in all three planes.

C2

28. No points of intersection: C1 C2 C1

One point of intersection: C1 C2

C1 C2Two points of intersection:

Infinitely many points of intersection: C1= C2

30. 20 tons of low-sulfur fuel, 20 tons of high-sulfur fuel.

32. 3.2 ounces of food A, 4.2 ounces of food B, and 2 ounces of food C.

34. (a) p(1) = a(1)2 + b(1) + c = a + b + c =−5 p(−1) = a(−1)2 + b(−1) + c = a − b + c = 1 p(2) = a(2)2 + b(2) + c = 4a + 2b + c = 7.

(b) a = 5, b =−3, c =−7.

Section 1.2, p. 19 ⎡ ⎤ ⎡ ⎤

0 1 0 0 1 0 1 1 1 1⎢⎢1 0 1 1 1 ⎥⎥ ⎢⎢1 0 1 0 0 ⎥⎥

2. (a) A =⎢0 ⎢⎣01

⎥ ⎢ ⎥1 0 0 0⎥ (b) A = ⎢1 1 0 1 0⎥.⎥ ⎢ ⎥1 0 0 0⎦ ⎣1 0 1 0 0⎦1 0 0 0 1 0 0 0 0

4. a = 3, b = 1, c = 8, d =−2.⎡5

6. (a) C + E = E + C =⎣ 45

⎡ ⎤

⎤−5 8

2 9⎦. (b) Impossible. (c)3 4

⎡ ⎤

[7 −7

0 1

]

−9 3 −9(d)⎣−12 −3 −15⎦.

−6 −3 −9

⎤

0 10 −9(e)⎣ 8 −1 −2⎦. (f) Impossible.

−5 −4 3

⎡ ⎤⎡1 28. (a) AT =⎣ 2 1⎦, (AT )T =

3 4

[1 2 32 1 4

] 5 4 5 ][−6 10(b)⎣−5 2 3⎦. (c)

11 178 9 4


Full file at http://testbanksite.eu/Elementary-Linear-Algebra-with-Applications-9th-Edition--SolutionSection 1.3

[(d)

04

]−4

0

]

⎡ ⎤3 4 [

(e)⎣ 6 3⎦. (f)9 10

] ]

3

]17 2

−16 6

10. Yes: 2[1 0

0 1

[1 0 [3 0+1 =

0 0 0 2⎤⎡λ−1 −2 −3

12.⎣ −6 λ+2 −3 ⎦.−5 −2 λ−4

14. Because the edges can be traversed in either direction.⎡

x1

⎤

⎢x2 ⎥16. Let x = ⎢⎢⎣

xn

⎥⎥e an n-vector. Then ⎦b

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤x1 0 x1 + 0 x1

⎢x2 ⎥ ⎢0⎥ ⎢x2 + 0⎥ ⎢ x2⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥x+0= ⎢ ⎥

∑ ∑

⎢ ⎥ ⎢ ⎥ ⎢ ⎥ x.⎣ ⎦+ ⎣ .⎦= ⎣ ⎦= ⎣ ⎦=

xn 0 xn + 0 xn

18.

19.

aij = (a11 + a12 + ··· + a1m) + (a21 + a22 + ··· + a2m) + ··· + (an1 + an2 + ··· + anm)i=1 j=1

= (a11 + a21 + · · · + an1) + (a12 + a22 + · · · + an2) + · · · + (a1m + a2m + · · · + anm) ∑ ∑

= aij.j=1 i=1

∑ ∑(a) True. (ai + 1) = ai + 1 = ai + n.

i=1 i=1 i=1 i=1⎛ ⎞∑ ∑ ∑

(b) True. ⎝ 1⎠ = m = mn.i=1 j=1 i=1⎡ ⎤⎡ ⎤∑ ∑

(c) True.⎣ ai⎦⎣ bj⎦ = a1 bj + a2

i=1 j=1 j=1

∑ ∑bj + ··· + an bj

j=1 j=1∑

= (a1 + a2 + · · · + an) bj

j=1( )

∑ ∑= ai bj = aibj

i=1 j=1 j=1 i=1

20. “new salaries” = u + .08u = 1.08u.

Section 1.3, p. 30 2. (a) 4. (b) 0. (c) 1. (d) 1.

4. x = 5.

n m

m n

n n∑ n ∑ n

n m n

n m m∑ m m

m

n∑ m ∑ m n


Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Full file at http://testbanksite.eu/Elementary-Linear-Algebra-with-Applications-9th-Edition--Solution4

√6. x = ± 2, y = ±3.

8. x = ±5.

10. x =65, y=12 5 .⎡

0 −1 1⎤ ⎡ 15 −7 14

Chapter 1

⎤ ⎡ ⎤8 8

12. (a) Impossible. (b)⎣ 1219

]

5 17⎦. (c)⎣ 23 −50 22 13 −1

29⎦. (d)⎣ 14 13⎦. (e) Impossible.17 13 9

]14. (a)

[ 58 12 66 13

(b) Same as (a). (c)

]

[ 28 8 38 34 4 41

](d) Same as (c). (e)

[ 28 32 16 18

[−16 −8 −26 ; same.

(f)−30 0 −31

⎤16. (a) 1.

⎡(f) ⎣

90

[ ](b)−6. (c) −3 0 1

⎤0 −3 0 0⎦. (g) Impossible.

⎡−1 4 2(d)⎣−2 8 4⎦. (e) 10.

3 −12 −6

−3 0 1

18. DI2 = I2D = D. ]

20.[0 0

0 0⎡

1⎤ ⎡ ⎤

0 ⎢14⎥ ⎢18⎥

22. (a) ⎢ ⎥ (b) ⎢ ⎥⎣0⎦.

13⎡1

⎣3⎦. 13

⎤ ⎤ ⎤⎡−2 ⎡−1 ⎤ ⎤ ⎤⎡1 ⎡−2 ⎡−1 24. col1(AB) = 1⎣ 2⎦+3⎣ 4⎦+2⎣ 3⎦; col2(AB) =−1⎣2⎦+2⎣ 4⎦+4⎣ 3⎦.

3 0 −2 3 0 −2

26. (a)−5. (b) BAT

[ ] [ ]28. Let A = aij be m × p and B = bij be p × n.

(a) Let the ith row of A consist entirely of zeros, so that aik = 0 for k = 1, 2, . . . , p. Then the (i, j) entry in AB is

∑aikbkj = 0 for j = 1, 2, . . . , n.

k=1

(b) Let the jth column of A consist entirely of zeros, so that akj = 0 for k = 1, 2, . . . , m. Then the (i, j) entry in BA is

∑

⎡ ⎤

bikakj = 0 for i = 1, 2, . . . , m.k=1 ⎡ ⎤

⎡ ⎤ x1 ⎡ ⎤2 3 −3 1

⎢3 0 2 030. (a) ⎢

1 2 3 −33⎥ ⎢3 0 2⎥ (b) ⎢

1 1 ⎢ ⎥ 70 3⎥⎢⎢x2 ⎥⎥ ⎢−2⎥

⎥⎢x3⎥= ⎢ ⎥⎣2 3 0 −4 0⎦. ⎣2 3 0 −4 0⎦⎢ ⎥ ⎣⎣x4⎦

3⎦.

0 0 1 1 1 0 0 1 1 1 x5 5

p

m




⎡2 3 −3 1⎢⎢3 0 2 0

5

⎤1 7 3 −2 ⎥⎥

(c) ⎣2 3 0 −4 0 3 ⎦0 0 1 1 1 5

][−2 3][x1[5]

32. =1 −5 x2 4

2x1 + x2 + 3x3 + 4x4 = 034. (a) 3x1− x2 + 2x3 =3

−2x1 + x2− 4x3 + 3x4 = 2(b) same as (a).

⎡ ⎤ ⎡ ⎤ ⎡ ⎤[[3] 2

36. (a) x1 +x21 −1

⎤

] [ ][1] 4+x3 = (b) x1⎣

4 −2⎡ ⎤

−1 1 32⎦+x2⎣−1⎦=⎣−2⎦. 3 1 1

⎤38. (a) [1 2

0 2 5 3

]⎡x1⎣x2⎦=

x3

1 2 1⎤⎡x1⎡0[1] (b)⎣ 1 1 2⎦⎣x2⎦=⎣0⎦.

12 0 2 x3 0

39. We have⎡ ⎤

v1

u·v=

⎤

∑ [ ]⎢v2 ⎥⎥uivi = u1 u2 ··· un ⎢ ⎥ uT v.⎣ ⎦=

i=1vn

⎡1 040. Possible answer:⎣ 2 0

3 0

42. (a) Can say nothing.

00⎦.0

(b) Can say nothing.

∑ n ∑43. (a) Tr(cA) =

i=1

(b) Tr(A + B) =

caii = c aii = cTr(A).i=1

∑(aii + bii) = aii +

i=1 i=1

∑

i=1

bii = Tr(A) + Tr(B).

[ ](c) Let AB = C = cij . Then

∑∑ n ∑ ∑Tr(AB) = Tr(C) = cii = aikbki = bkiaik = Tr(BA).

i=1 i=1 k=1 k=1 i=1

∑ n ∑(d) Since aTii = aii, Tr(AT ) =

[ ]

aTii = aii = Tr(A).i=1 i=1

(e) Let AT A = B = bij . Then

∑ n ∑ ∑ ∑bii = aTij aji = a2

ji =⇒ Tr(B) = Tr(AT A) = bii = a2ij≥ 0.

j=1 j=1 i=1 i=1 j=1

Hence, Tr(AT A)≥ 0.

n

n

n n∑ n

n∑ n n n

n

n n∑ n n




44. (a) 4. (b) 1. (c) 3.

45. We have Tr(AB − BA) = Tr(AB)− Tr(BA) = 0, while Tr([1

0

] )0

= 2.1 ⎡ b

1j

Chapter 1

⎤[ ] [ ] ⎢b2j ⎥

46. (a) Let A = aij

entry of Abj is

and B = bij be m × n and n × p, respectively. Then bj =

∑aikbkj, which is exactly the (i,j) entry of AB.

k=1

⎢⎢⎣bnj

⎥⎥nd the ith ⎦ a

[∑ ∑ ∑ ] [ ](b) The ith row of AB is

we havek aik bk1 k aik bk2 ···

[∑ ∑k aik bkn . Since ai = ai1 ai2 ··· ain ,

∑ ] aib = k aik bk1 k aik bk2 ··· k aik bkn

This is the same as the ith row of Ab.[ ] [ ]

47. Let A = aij and B = bij be m × n and n × p, respectively. Then the jth column of AB is⎡(AB)j =

⎣ a11b1j + ··· + a1nbnj

am1b1j + ··· + amnbnj ⎡ ⎤ ⎡⎤⎥⎦

⎤a11

=b1j⎣am1

a1n⎥ ⎢ ⎥⎦+ ··· + bnj ⎣ ⎦amn

= b1jCol1(A) + ··· + bnjColn(A).

Thus the jth column of AB is a linear combination of the columns of A with coefficients the entries in bj.

48. The value of the inventory of the four types of items.

50. (a) row1(A) · col1(B) = 80(20) + 120(10) = 2800 grams of protein consumed daily by the males.

(b) row2(A) · col2(B) = 100(20) + 200(20) = 6000 grams of fat consumed daily by the females.

51. (a) No. If x = (x1, x2, . . . , xn), then x · x = x21 +x2 + ···+x2n≥ 0.

(b) x = 0.

52. Let a = (a1, a2, . . . , an), b = (b1, b2, . . . , bn), and c = (c1, c2, . . . , cn). Then

∑ ∑(a) a · b = aibi and b·a = biai, so a·b = b·a.

i=1 i=1

∑ ∑ n ∑(b) (a + b) · c = (ai + bi)ci = aici + bici = a·c + b·c.

i=1 i=1 i=1

∑ ∑(c) (ka) · b = (kai)bi = k aibi = k(a·b).

i=1 i=1

nn

nn

nn

n



Full file at http://testbanksite.eu/Elementary-Linear-Algebra-with-Applications-9th-Edition--SolutionSection 1.4 7

53. The i, ith element of the matrix AAT is

∑ n ∑ ∑aikaTki = aikaik = (aik )2.

k=1 k=1 k=1

∑Thus if AAT = O, then each sum of squares (aik )2 equals zero, which implies aik = 0 for each i

k=1and k. Thus A = O.

54. AC =[ 17 2

22 18 3 23

] . CA cannot be computed.

55. BT B will be 6 × 6 while BBT is 1 × 1.

Section 1.4, p. 40[ ] [ ] [ ]

1. Let A = aij , B = bij , C = cij . Then the (i, j) entry of A + (B + C) is aij + (bij + cij ) andthat of (A + B) + C is (aij + bij ) + cij . By the associative law for addition of real numbers, these twoentries are equal.

[ ] [ ]2. For A = aij , let B = −aij

[ ] [ ] [ ] ∑4. Let A = aij ,B= bij ,C = cij . Then the (i, j) entry of (A + B)C is (aik + bik )ckj and that of

k=1∑ n ∑

AC + BC is aik ckj + bik ckj . By the distributive and additive associative laws for real numbers,k=1 k=1

these two expressions for the (i, j) entry are equal.[ ] [ ]

6. Let A = aij , where aii = k and aij = 0 if i = j, and let B = bij . Then, if i = j, the (i, j) entry of∑ ∑

AB is aisbsj = kbij , while if i = j, the (i, i) entry of AB is aisbsi = kbii. Therefore AB = kB.s=1 s=1

[ ] [ ] ∑7. Let A = aij and C = c1 c2 ··· cm⎡ ⎤ . Then CA is a 1 × n matrix whose ith entry is cjaij.

j=1

a1j

Since Aj = ⎢⎢⎢ a2j ⎥ ∑⎥⎥ the ith entry of cjAj is cjaij.

[8. (a)

⎣ . ⎦,

amj

cos 2θ sin 2θ− sin 2θ cos 2θ

j=1 j=1

] [ ] [ ]cos 3θ sin 3θ cos kθ sin kθ

(b) (c)− sin 3θ cos 3θ − sin kθ cos kθ

(d) The result is true for p = 2 and 3 as shown in parts (a) and (b). Assume that it is true for p = k.Then

[Ak+1 = AkA =

[

cos kθ sin kθ− sin kθ cos kθ

] [ ]cosθ sinθ

− sinθ cosθ]

=

[=

cos kθ cosθ− sin kθ sinθ cos kθ sinθ + sin kθ cosθ− sin kθ cosθ− cos kθ sinθ cos kθ cosθ− sin kθ sinθ

]cos(k + 1)θ sin(k + 1)θ

− sin(k + 1)θ cos(k + 1)θHence, it is true for all positive integers k.

∑ mn

n

nn

n

n

n

nn




][1 0 [0 110. Possible answers: A = ;A=

0 1 1 0[ ]

Chapter 1⎡ ⎤

] 1 1√ √

;A=⎣ 2 2 ⎦.1√ √

2 2] ]

12. Possible answers: A =

[ ]

1 1 −1 −1

[0 0 [0 1;A= ;A=

0 0 0 0

13. Let A = aij . The (i, j) entry of r(sA) is r(saij ), which equals (rs)aij and s(raij ).[ ]

14. Let A = aij . The (i, j) entry of (r + s)A is (r + s)aij , which equals raij + saij , the (i, j) entry ofrA + sA.

[ ] [ ]16. Let A = aij , and B =

[ ] [18. Let A = aij and B =

(i, j) entry of r(AB).

1

bij . Then r(aij + bij ) = raij + rbij .

] ∑bij . The (i, j) entry of A(rB) is aik(rbkj), which equals r

k=1

∑

k=1

aikbkj, the

20. 6A,k=16.22. 3.

24. If Ax = rx and y = sx, then Ay = A(sx) = s(Ax) = s(rx) = r(sx) = ry.

26. The (i, j) entry of (AT )T is the (j, i) entry of AT , which is the (i, j) entry of A.

27. (b) The (i, j) entry of (A + B)T is the (j, i) entry of [ ]

[ ]aij + bij , which is to say, aji + bji.

[ ] (d) Let A = aij and let bij = aji. Then the (i, j) entry of (cA)T is the (j, i) entry of caij , which

is to say, cbij .⎤ ⎡ ⎤⎡5 0 −4 −828. (A + B)T =⎣ 5 2⎦, (rA)T =⎣−12 −4⎦.

1 2 −8 12⎤ ⎤⎡−34 ⎡−3430. (a) ⎣ 17⎦. (b)⎣ 17⎦.

−51 −51]

(c) BT C is a real number (a 1 × 1 matrix).

] ]32. Possible answers: A =

] [

[1 −30 0]

[1;B=

23

]

2 [−1 2;C =

1 0 1

A= [2 0 3 0

0 0 [0 0;B= ;C =

1 0 0 1 33. The (i, j) entry of cA is caij , which is 0 for all i and j only if c = 0 or aij = 0 for all i and j.

34. Let A =[a bc d

]

be such that AB = BA for any 2 × 2 matrix B. Then in particular,

] [

so b = c = 0, A =[a

0

[a b][1 0c d 0 0

[a 0c 0

]0d .

=

] [=

1 0][a b] 0 0 c d

]a b0 0

−1

n n




Also[a

0

]0][1 1d 0 0

[a a]

[

=

[

9

]1 1][a 0 0 0 0 d

]a d

which implies that a = d. Thus A =[a 0

0 0

]

= ,0 0

0 a forsomenumbera.

35. We have

(A − B)T = (A + (−1)B)T

= AT + ((−1)B)T

= AT + (−1)BT = AT− BT by Theorem 1.4(d)).

36. (a) A(x1 + x2) = Ax1 + Ax2 = 0 + 0 = 0. (b) A(x1− x2) = Ax1− Ax2 = 0− 0 = 0. (c) A(rx1) = r(Ax1) = r0 = 0. (d) A(rx1 + sx2) = r(Ax1) + s(Ax2) = r0 + s0 = 0.

37. We verify that x3 is also a solution:

Ax3 = A(rx1 + sx2) = rAx1 + sAx2 = rb + sb = (r + s)b = b.

38. If Ax1 = b and Ax2 = b, then A(x1− x2) = Ax1− Ax2 = b− b = 0.

Section 1.5, p. 52 [ ]

1. (a) Let Im = dij so dij = 1 if i = j and 0 otherwise. Then the (i, j) entry of ImA is

∑dikakj = diiaij (since all other d’s = 0)

k=1

=aij (since dii = 1).

[ ]2. We prove that the product of two upper triangular matrices is upper triangular: Let A = aij with

[ ] [ ] ∑aij = 0 for i > j; let B = bij with bij = 0 for i > j. Then AB = cij where cij = aikbkj. For

k=1i > j, and each 1≤ k ≤ n, either i > k (and so aik = 0) or else k ≥ i > j (so bkj = 0). Thus every[ ]term in the sum for cij is 0 and so cij = 0. Hence cij is upper triangular.

[ ] [ ] [ ]3. Let A = aij and B = bij , where both aij = 0 and bij = 0 if i = j. Then if AB = C = cij , we

∑have cij = aikbkj = 0 if i = j.

k=1 ⎤ ⎤⎡9 −1 1 ⎡ 18 −5 114. A + B =⎣ 0 −2 7⎦ and AB =⎣ 0 −8 −7⎦.

0 0 3 0 0 0

5. All diagonal matrices.

m

n

n

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[6. (a) 7 −2

−3

10

] ][−9 −11(b) (c)

22 13

Chapter 1

][ 20 −204 76

q summands

8. ApAq = (A · A · · · A) (A · A · · · A) =Ap+q; (Ap)q = Ap A p Ap · · · Ap = Ap+p+···+p = Apq .p factors q factors q factors

p + q factors

9. We are given that AB = BA. For p = 2, (AB)2 = (AB)(AB) = A(BA)B = A(AB)B = A2B2. Assume that for p = k, (AB)k = Ak Bk . Then

(AB)k+1 = (AB)k (AB) = Ak Bk · A · B = Ak (Bk−1AB)B = Ak(Bk−2AB2)B = ··· = Ak+1Bk+1.

Thus the result is true for p = k + 1. Hence it is true for all positive integers p. For p = 0, (AB)0 = In = A0B0.

10. For p = 0, (cA)0 = In = 1 · In = c0 · A0. For p = 1, cA = cA. Assume the result is true for p = k: (cA)k = ck Ak , then for k + 1:

(cA)k+1 = (cA)k (cA) = ck Ak · cA = ck (Ak c)A = ck (cAk )A = (ck c)(Ak A) = ck+1Ak+1.

11. True for p = 0: (AT )0 = In = ITn = (A0)T . Assume true for p = n. Then

(AT )n+1 = (AT )nAT = (An)T AT = (AAn)T = (An+1)T .

12. True for p = 0: (A0)−1 = I−1n = In. Assume true for p = n. Then

(An+1)−1 = (AnA)−1 = A−1(An)−1 = A−1(A−1)n = (A−1)n+1.

(1

13. kA−1)(kA)=(k ·k)A−1A=In and(kA)(kA−1)=(k·k)AA−1 =In. Hence, (kA)−1 =1

kA−1 fork = 0.

14. (a) Let A = kIn. Then AT = (kIn)T = kITn = kIn = A.(b) If k = 0, then A = kIn = 0In = O, which is singular. If k = 0, then A−1 = (kA)−1 =1

kA−1, so A

is nonsingular.(c) No, the entries on the main diagonal do not have to be the same.

]16. Possible answers:

[a b0 a .

Infinitely many.

] [ ] ]17. The result is false. Let A =

[1 2 3 4

5 11 [ 10 14. Then AAT = and AT A =

11 25 14 20 18. (a) A is symmetric if and only if AT = A, or if and only if aij = aTij = aji.

(b) A is skew symmetric if and only if AT =−A, or if and only if aTij =aji =−aij.(c) aii =−aii, so aii = 0.

19. Since A is symmetric, AT = A and so (AT )T = AT .

20. The zero matrix.

21. (AAT )T = (AT )T AT = AAT .

Full file at http://testbanksite.eu/Elementary-Linear-Algebra-with-Applications-9th-Edition--Solution22. (a) (A + AT )T = AT + (AT )T = AT + A = A + AT .


Full file at http://testbanksite.eu/Elementary-Linear-Algebra-with-Applications-9th-Edition--SolutionSection 1.5 11

(b) (A − AT )T = AT− (AT )T = AT− A =−(A − AT ).

23. (Ak )T = (AT )k = Ak .

24. (a) (A + B)T = AT + BT = A + B.(b) If AB is symmetric, then (AB)T = AB, but (AB)T = BT AT = BA, so AB = BA. Conversely, if

AB = BA, then (AB)T = BT AT = BA = AB, so AB is symmetric.[ ] [ ]

25. (a) Let A = aij be upper triangular, so that aij = 0 for i > j. Since AT = aTij , where aTij =aji,we have aTij = 0 for j > i, or aTij = 0 for i < j. Hence AT is lower

triangular.

(b) Proof is similar to that for (a).

26. Skew symmetric. To show this, let A be a skew symmetric matrix. Then AT =−A. Therefore(AT )T = A =−AT . Hence AT is skew symmetric.

27. If A is skew symmetric, AT =−A. Thus aii =−aii, so aii = 0.

28. Suppose that A is skew symmetric, so AT =−A. Then (Ak )T = (AT )k = (−A)k =−Ak if k is apositive odd integer, so Ak is skew symmetric.

29. Let S = (1)2

(1)(A + AT ) and K =

2(A − AT ). Then S is symmetric and K is skew symmetric, by

Exercise 18. Thus S+K =

(1)2

(1)(A + AT + A − AT ) = (2A) = A.

2

Conversely, suppose A = S + K is any decomposition of A into the sum of a symmetric and skewsymmetric matrix. Then

AT = (S + K)T = ST + KT = S − KA + AT = (S + K) + (S − K) = 2S, S = (1)

2( )

(A + AT ),

1⎡ ⎤ A − AT = (S + K) −(S − K) = 2K, K =

⎤(A − AT )

2

30. S = 12

2 7 3⎣7 12 3⎦ and K =3 3 6

] ]

⎡0 −1 −7 1 ⎣1 0 1⎦. 2

7 −1 0

31. Form[2 3][w x

4 6 y z[1 0

= . Since the linear systems0 1

2w + 3y = 1 2x + 3z = 04w + 6y = 0 and

4x + 6z = 1

have no solutions, we conclude that the given matrix is singular.⎡1⎢ ⎤

0 0⎥32. D−1 =

⎡⎢4⎢⎣0

0

12

−12

0⎤12

⎥0⎥⎦.13

34. A =⎣− ⎦.2 −1

][1 2

Full file at http://testbanksite.eu/Elementary-Linear-Algebra-with-Applications-9th-Edition--Solution36. (a) ][41 3 6

[ 16] [ 38]

= (b)22 53



38.[−9−6

]

Chapter 1

]

40.[8

9

[1 0] ] ][0 0 [1 0

42. Possible answer:

43. Possible answer:

0 0

][1 2 3 4

+ =0 1 0 1

] ][−1 −2 [0 0+ =

3 4 6 8

44. The conclusion of the corollary is true for r = 2, by Theorem 1.6. Suppose r ≥ 3 and that theconclusion is true for a sequence of r − 1 matrices. Then

(A1A2 · · · Ar )−1 = [(A1A2 · · · Ar−1)Ar ]−1 = A−1r (A1A2 · · · Ar−1)−1 =

A−1r A −1 · ··A−12 A−11

45. We have A−1A = In = AA−1 and since inverses are unique, we conclude that (A−1)−1 = A.

46. Assume that A is nonsingular, so that there exists an n × n matrix B such that AB = In. Exercise 28in Section 1.3 implies that AB has a row consisting entirely of zeros. Hence, we cannot have AB = In.

47. Let

A=

⎡⎢⎢⎢a11 0 0 ··· 00 a22 0 ··· 0

⎤⎥⎥⎥,⎣0 0

where aii = 0 for i = 1, 2, . . . , n. Then ⎡1⎢a11⎢

⎦··· ··· ann

0 0 ··· 01

⎤⎥⎥0

A−1 =⎢⎢ a22 0 ··· 0 ⎥⎥⎥⎢ ⎥⎣0 0 ··· ···

as can be verified by computing AA−1 = A−1A = In. ⎤⎦

1ann

⎡ 16 0 048. A4 =⎣ 0 81 0⎦.

0 0 625⎡ ⎤ap

11⎢ 00 0 ··· 0p

a 0 ··· 0 ⎥49. Ap = ⎢ 22 ⎥⎢ ⎥⎣

0 0 ··· ··· ap ⎦.nn

50. Multiply both sides of the equation by A−1.

51. Multiply both sides by A−1.

r1




52. Form

[a b][w x

c d y z

13

] ][1 0= . This leads to the linear systems

0 1

aw + by = 1 ax + bz = 0cw + dy = 0 and

cx + dz = 1.

A solution to these systems exists only if ad − bc = 0. Conversely, if ad − bc = 0 then a solution tothese linear systems exists and we find A−1.

53. Ax = 0 implies that A−1(Ax) = A0 = 0, so x = 0.

54. We must show that (A−1)T = A−1. First, AA−1 = In implies that (AA−1)T = ITn= In. Now(AA−1)T = (A−1)T AT = (A−1)T A, which means that (A−1)T = A−1.⎡ ⎤

4 5 055. A + B =⎣ 0 4 1

6 −2 6⎡⎦ is one possible answer.

⎤ ⎡ ⎤2×2

2×256. A =⎣ 2 × 2 2×2

2×2 2×2[

2×1 2×2 2×3 2×1 ⎦ and B =⎣ 2 × 2 2×3 ⎦.2×1 1×2 1×3

] [ ]

A= 3×3 3×2 3×3 3×2 ⎡

21 48 41

⎢⎢18 26 34⎢

3×3 3×2 and B =

2×3 2×2 ⎤48 4033 5⎥⎥⎥

AB = ⎢24 26 42 47 16⎥⎢ ⎥⎢28⎢ 38 54 70 35⎥.⎥⎣33 33 56 74 42⎦34 37 58 79 54

57. A symmetric matrix. To show this, let A1, . . . , An be symmetric matrices and let x1, . . . , xn be scalars.Then AT

1 =A1,...,ATn =An.Therefore

(x1A1 + · · · + xnAn)T = (x1A1)T + · · · + (xnAn)T

=x1AT1 + ···+xnAT n

=x1A1 + ··· + xnAn.

Hence the linear combination x1A1 + · · · + xnAn is symmetric.

58. A scalar matrix. To show this, let A1, . . . , An be scalar matrices and let x1, . . . , xn be scalars. Then Ai = ciIn for scalars c1,...,cn. Therefore

x1A1 + ··· + xnAn = x1(c1I1) + ··· + xn(cnIn) = (x1c1 + ··· + xncn)In

which is the scalar matrix whose diagonal entries are all equal to x1c1 + · · · + xncn. ] [5 [ 19] [ 65] [ 214]

59. (a) w1 = ,w2 =1 5

(b) wn−1 = An−1w0. ]

,w3 = ,w4 =19 65

; u2 = 5, u3 = 19, u4 = 65, u5 = 214.

[4 [8] [ 16]

60. (a) w1 = ,w2 =2 (b) wn−1 = An−1w0.

Full file at http://testbanksite.eu/Elementary-Linear-Algebra-with-Applications-9th-Edition--Solution,w3 =4 8


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