Solution manual for applied linear algebra by olver

On the other hand, the i, j entry of ckrk equalsthe product of the ithentry of ck, namely aik, with the jthentry of rk, namely bkj.Summing these entries, aikbkj, over k yields the usual

Solutions — Chapter 1

1.1.1

(a) Reduce the system to x − y = 7, 3y = −4; then use Back Substitution to solve for

x = 173 , y = −43.(b) Reduce the system to 6 u + v = 5, −52v = 52; then use Back Substitution to solve for

(a) With Forward Substitution, we just start with the top equation and work down Thus

2 x = −6 so x = −3 Plugging this into the second equation gives 12 + 3y = 3, and so

y = −3 Plugging the values of x and y in the third equation yields −3 + 4(−3) − z = 7,and so z = −22

(b) We will get a diagonal system with the same solution

(c) Start with the last equation and, assuming the coefficient of the last variable is 6= 0, usethe operation to eliminate the last variable in all the preceding equations Then, againassuming the coefficient of the next-to-last variable is non-zero, eliminate it from all butthe last two equations, and so on

(d) For the systems in Exercise 1.1.1, the method works in all cases except (c) and (f ).Solving the reduced system by Forward Substitution reproduces the same solution (as

it must):

(a) The system reduces to 32x = 172 , x + 2 y = 3

(b) The reduced system is 152 u = 152 , 3 u − 2v = 5

(c) The method doesn’t work since r doesn’t appear in the last equation

(d) Reduce the system to 32u = 12, 72u − v = 52, 3 u − 2w = −1

(e) Reduce the system to 23x1= 83, 4 x1+ 3 x2 = 4, x1+ x2+ x3 = −1

(f ) Doesn’t work since, after the first reduction, z doesn’t occur in the next to lastequation

(g) Reduce the system to 5521x1= 57, x2+ 218 x3= 34, x3+ 83x4 = 23, x3+ 3 x4= 1

1.2.1 (a) 3 × 4, (b) 7, (c) 6, (d) ( −2 0 1 2 ), (e)

0

B 02

−6

1

C

C, b =

0

B036

C, x =

0 B B

xyzw

1 C

C, b =

0 B B

−3321

1 C

C;

(g) A =

0 B B

C, x =

0 B B

C, b =

0 B B

1111

1 C

C

1.2.5

(a) x − y = −1, 2x + 3y = −3 The solution is x = −65, y = −15.(b) u + w = −1, u + v = −1, v + w = 2 The solution is u = −2, v = 1, w = 1.(c) 3 x1− x3 = 1, −2x1− x2 = 0, x1+ x2− 3x3 = 1

The solution is x1 = 15, x2 = −25, x3 = −25.(d) x + y − z − w = 0, −x + z + 2w = 4, x − y + z = 1, 2y − z + w = 5

The solution is x = 2, y = 1, z = 0, w = 3

1.2.6

(a) I =

0 B B

C, O =

0 B B

C

(b) I + O = I , I O = O I = O No, it does not

1.2.7 (a) undefined, (b) undefined, (c) 3 6 0

1.2.8 Only the third pair commute.

D = diag (d1, , dn) The (i, j) entry of A D is aijdj The (i, j) entry of D A is diaij If

di6= dj, this requires aij = 0, and hence, if all the di’s are different, then A is diagonal.1.2.13 We need A of size m × n and B of size n × m for both products to be defined Further,

A B has size m × m while B A has size n × n, so the sizes agree if and only if m = n

1.2.14 B = x y

0 x

!

where x, y are arbitrary

1.2.15 (a) (A + B)2 = (A + B)(A + B) = AA + AB + BA + BB = A2 + 2AB + B2, since

1.2.16 If A B is defined and A is m × n matrix, then B is n × p matrix and AB is m × p matrix;

on the other hand if B A is defined we must have p = m and B A is n × n matrix Now,since A B = B A, we must have p = m = n

1.2.20 False — unless they commute: A B = B A

1.2.21 Let v be the column vector with 1 in its jthposition and all other entries 0 Then A v

is the same as the jthcolumn of A Thus, the hypothesis implies all columns of A are 0and hence A = O

1.2.22 (a) A must be a square matrix (b) By associativity, A A2 = A A A = A2A = A3.(c) The na¨ıve answer is n − 1 A more sophisticated answer is to note that you can com-pute A2 = A A, A4 = A2A2, A8 = A4A4, and, by induction, A2r with only r matrixmultiplications More generally, if the binary expansion of n has r + 1 digits, with s nonzerodigits, then we need r + s − 1 multiplications For example, A13 = A8A4A since 13 is 1101

in binary, for a total of 5 multiplications: 3 to compute A2, A4 and A8, and 2 more to tiply them together to obtain A13

mul-1.2.23 A = 0 1

0 0 .

♦ 1.2.24 (a) If the ithrow of A has all zero entries, then the (i, j) entry of A B is ai1b1j + · · · +

ainbnj = 0 b1j + · · · + 0bnj = 0, which holds for all j, so the ithrow of A B will have all 0’s.(b) If A = 1 1

They are not the same

1.2.27 (a) X = O (b) Yes, for instance, A = 1 2

(c) z =

0

B111

0

B111

1

C=

0

B260

1

C, while B = A W =

0 B

0 B B

C=

0 B

A, and so B z =

0

B000

l= 1

bklclj, so the (i, j) entry of A (B C) is

n X

k= 1

p X

l= 1

aikbklclj

On the other hand, the (i, l) entry of A B is

k X

i = 1

aikbkl, so the (i, j) entry of (A B) C is

p X

l = 1

0

@ n X

k= 1

p X

= c(A B − B A) + d(AC − C A) = c [ A, B ] + d [ A, C ]

(ii) [ A, B ] = A B − B A = − (B A − AB) = − [ B, A ]

(iii) h[ A, B ], Ci= (A B − B A)C − C (AB − B A) = AB C − B AC − C AB + C B A,

h

[ C, A ], Bi= (C A − AC)B − B (C A − AB) = C AB − AC B − B C A + B AC,

h

[ B, C ], Ai= (B C − C B)A − A(B C − C B) = B C A − C B A − AB C + AC B.Summing the three expressions produces O

♦ 1.2.32 (a) (i) 4, (ii) 0, (b) tr(A + B) =

n X

i = 1

(aii+ bii) =

n X

i = 1

aii+

n X

i = 1

bii= tr A + tr B

(c) The diagonal entries of A B are

n X

j = 1

aijbji, so tr(A B) =

n X

i = 1

n X

j = 1

aijbji; the diagonal

entries of B A are

n X

i = 1

bjiaij, so tr(B A) =

n X

i = 1

n X

j = 1

bjiaij These double summations are

clearly equal (d) tr C = tr(A B − B A) = tr AB − tr B A = 0 by part (a)

(e) Yes, by the same proof

♦ 1.2.33 If b = A x, then bi = ai1x1 + ai2x2 + · · · + ainxn for each i On the other hand,

cj = (a1j, a2j, , anj)T, and so the ithentry of the right hand side of (1.13) is

x1ai1+ x2ai2+ · · · + xnain, which agrees with the expression for bi

1

C( 2 3 0 ) +

0

B−121

1

C( 3 −1 4 ) +

0

B 11

(c) If we set B = x, where x is an n × 1 matrix, then we obtain (1.14)

(d) The (i, j) entry of A B is

n X

k = 1

aikbkj On the other hand, the (i, j) entry of ckrk equalsthe product of the ithentry of ck, namely aik, with the jthentry of rk, namely bkj.Summing these entries, aikbkj, over k yields the usual matrix product formula

p(x)q(x) = 2 x5− 5x3+ 4 x2− 3x + 2

(d) True, since powers of A mutually commute For the particular matrix from (b),

p(A) q(A) = q(A) p(A) = 2 8

−4 −6

!

(b) S2 is only defined if S is square.

(c) Any of the matrices ±1 0

C (c) Since matrix addition is

done entry-wise, adding the entries of each block is the same as adding the blocks (d) Xhas size k × m, Y has size k × n, Z has size l × m, and W has size l × n Then AX + B Zwill have size i × m Its (p, q) entry is obtained by multiplying the pthrow of M times the

qthcolumn of P , which is ap1x1q + · · · + apixiq + bp1z1q + · · · + bplzlq and equals thesum of the (p, q) entries of A X and B Z A similar argument works for the remaining threeblocks (e) For example, if X = (1), Y = ( 2 0 ), Z = 0

C The individual block products are

04

!

= 13

−9

1 C

0 B B

−5

1 C

Creduces to

0 B B

1 C

C

Solution: x4= −3, x3 = −32, x2 = −1, x1 = −4

(f )

0 B B

1 C

Creduces to

0 B B

−48

1 C

C

Solution: w = 2, z = 0, y = −1, x = 1

1.3.2

(a) 3 x + 2 y = 2, −4x − 3y = −1; solution: x = 4, y = −5,(b) x + 2 y = −3, −x + 2y + z = −6, −2x − 3z = 1; solution: x = 1, y = −2, z = −1,(c) 3 x − y + 2z = −3, −2y − 5z = −1, 6x − 2y + z = −3;

solution: x = 23, y = 3, z = −1,(d) 2 x − y = 0, −x + 2y − z = 1, −y + 2z − w = 1, −z + 2w = 0;

solution: x = 1, y = 2, z = 2, w = 1

1.3.3 (a) x = 173 , y = −43; (b) u = 1, v = −1; (c) u = 32, v = −13, w = 16; (d) x1 =

11

3 , x2 = −103 , x3 = −23; (e) p = −23, q = 196, r = 52; (f ) a = 13, b = 0, c = 43, d = −23;(g) x = 13, y = 76, z = −83, w = 92

C−→

0 B B

C−→

0 B B

C−→

0 B B

1 − 2 i

1

C

solution: z = i , y = −12 − 32i , x = 1 + i (c) 1 − i 2

solution: z = −12 − 12i , y = −52 + 12i , x = 52 + 2 i 1.3.7 (a) 2 x = 3, −y = 4, 3z = 1, u = 6, 8v = −24 (b) x = 32, y = −4, z = 13,

u = 6, v = −3 (c) You only have to divide by each coefficient to find the solution

♦ 1.3.8 0 is the (unique) solution since A0 = 0

♠ 1.3.9

Back Substitution

startset xn = cn/unnfor i = n − 1 to 1 with increment −1set xi = 1

nextjend

a11b12+ a12b22= a22b12+ a12b11, or (a11− a22)b12= a12(b11− b22).1.3.11 Clearly, any diagonal matrix is both lower and upper triangular Conversely, A beinglower triangular requires that aij = 0 for i < j; A upper triangular requires that aij = 0 for

i > j If A is both lower and upper triangular, aij = 0 for all i 6= j, which implies A is adiagonal matrix

by the jthcolumn of Ak, whose first j − k − 1 entries are non-zero, and all the rest arezero, according to the induction hypothesis; therefore, if i > j − k − 1, every term in thesum producing this entry is 0, and the induction is complete In particular, for k = n,every entry of Ak is zero, and so An= O.

(a) Add −2 times the second row to the first row of a 2 × n matrix

(b) Add 7 times the first row to the second row of a 2 × n matrix

(c) Add −5 times the third row to the second row of a 3 × n matrix

(d) Add 12 times the first row to the third row of a 3 × n matrix

(e) Add −3 times the fourth row to the second row of a 4 × n matrix

1.3.15 (a)

0 B B

C, (b)

0 B B

C, (c)

0 B B

C, (d)

0 B B

C The second is easier to predict

since its entries are the same as the corresponding entries of the Ei.1.3.18

(a) Suppose that E adds c 6= 0 times row i to row j 6= i, while E adds d 6= 0 times row k toerow l 6= k If r1, , rn are the rows, then the effect of E E is to replacee

(i) rj by rl+ c ri+ d rk for j = l;

(ii) rj by rj + c ri and rl by rl+ (c d) ri+ d rj for j = k;

(iii) rj by rj + c ri and rl by rl+ d rk otherwise

On the other hand, the effect of EE is to replacee

(i) rj by rl+ c ri+ d rk for j = l;

(ii) rj by rj + c ri+ (c d) rk and rl by rl+ d rk for i = l;

(iii) rj by rj + c ri and rl by rl+ d rk otherwise

Comparing results, we see that EE =e E E whenever i 6= l and j 6= k.e(b) E1E2 = E2E1, E1E3 6= E3E1, and E3E2= E2E3

(c) See the answer to part (a)

1.3.19 (a) Upper triangular; (b) both special upper and special lower triangular; (c) lowertriangular; (d) special lower triangular; (e) none of the above

1.3.20 (a) aij = 0 for all i 6= j; (b) aij = 0 for all i > j; (c) aij = 0 for all i > j and aii = 1for all i; (d) aij = 0 for all i < j; (e) aij = 0 for all i < j and aii = 1 for all i

♦ 1.3.21

(a) Consider the product L M of two lower triangular n × n matrices The last n − i entries

in the ithrow of L are zero, while the first j − 1 entries in the jthcolumn of M are zero

So if i < j each summand in the product of the ithrow times the jth column is zero,

and so all entries above the diagonal in L M are zero.

(b) The ithdiagonal entry of L M is the product of the ithdiagonal entry of L times the ith

diagonal entry of M (c) Special matrices have all 1’s on the diagonal, and so, by part (b), does their product

A, U =

0 B B

A, (h) L =

0 B B

C,

U =

0 B B

C, (i) L =

0 B B B

C, U =

0 B B B

(b) (1) Add −2 times first row to second row (2) Add −3 times first row to third row.(3) Add −5 times first row to fourth row (4) Add −4 times second row to third row.(5) Add −6 times second row to fourth row (6) Add −7 times third row to fourth row.(c) Use the order given in part (b)

♦ 1.3.25 See equation (4.51) for the general case

C=

0 B B B

0 B B B

C

1.3.26 False For instance 1 1

1 0 is regular Only if the zero appear in the (1, 1) positiondoes it automatically preclude regularity of the matrix

♦ 1.3.30

(a) Let u11, , unn be the pivots of A, i.e., the diagonal entries of U Let D be the nal matrix whose diagonal entries are dii = sign uii Then B = A D is the matrix ob-tained by multiplying each column of A by the sign of its pivot Moreover, B = L U D =

diago-LU , wheree U = U D, is the L U factorization of B Each column ofe U is obtained byemultiplying it by the sign of its pivot In particular, the diagonal entries of U , which areethe pivots of B, are uiisign uii = | uii| > 0

(b) Using the same notation as in part (a), we note that C = D A is the matrix obtained

by multiplying each row of A by the sign of its pivot Moreover, C = D L U ever, D L is not special lower triangular, since its diagonal entries are the pivot signs.But L = D L D is special lower triangular, and so C = D L D D U =b LbU , whereb

How-b

U = D U , is the L U factorization of B Each row ofU is obtained by multiplying itb

by the sign of its pivot In particular, the diagonal entries of U , which are the pivots ofb

!

, (b) x =

0

@14 1 4

1

A, (c) x =

0

B010

1

C, (d) x =

0 B B

− 47

2 7 5 7

1 C

−1

1

C, (g) x =

0 B B

2110

1 C

C, (h) x =

0 B B B

@

−3712

−1712

1 4

2

1 C C C A

, (i) x =

0 B B B

@

3 35 6 35 1 7 8 35

1 C C C A

C, x2 =

0 B B

− 16

− 32

5 3

1 C

C;

(c) L =

0 B

A, U =

0 B

A; x1=

0

B123

1

C, x3=

0

B−9.305668.6111

−4.9306

1 C

(e) L =

0 B B

C, U =

0 B B

A; x1=

0 B B B

5 4

−14

1 4 1 4

1 C C

C, x2=

0 B B B

1 14

−145

1 14 1 2

1 C C

C;

(f ) L =

0 B B

C, U =

0 B B

C;

x1 =

0 B B

1040

1 C

C, x2 =

0 B B

1132

1 C

C, x3 =

0 B B

108414

1 C

C

1.4.1 The nonsingular matrices are (a), (c), (d), (h)

1.4.2 (a) Regular and nonsingular, (b) singular, (c) nonsingular, (d) regular and nonsingular.1.4.3 (a) x1 = −53, x2= −103 , x3 = 5; (b) x1 = 0, x2 = −1, x3 = 2;

(c) x1 = −6, x2= 2, x3 = −2; (d) x = −132 , y = −92, z = −1, w = −3;

(e) x1 = −11, x2 = −103 , x3 = −5, x4= −7

1.4.4 Solve the equations −1 = 2b + c, 3 = −2a + 4b + c, −3 = 2a − b + c, for a = −4, b = −2,

c = 3, giving the plane z = −4x − 2y + 3

1.4.5

(a) Suppose A is nonsingular If a 6= 0 and c 6= 0, then we subtract c/a times the first rowfrom the second, producing the (2, 2) pivot entry (a d − bc)/a 6= 0 If c = 0, then thepivot entry is d and so a d − bc = ad 6= 0 If a = 0, then c 6= 0 as otherwise the firstcolumn would not contain a pivot Interchanging the two rows gives the pivots c and b,and so a d − bc = bc 6= 0

(b) Regularity requires a 6= 0 Proceeding as in part (a), we conclude that ad − bc 6= 0 also.1.4.6 True All regular matrices are nonsingular

♦ 1.4.7 Since A is nonsingular, we can reduce it to the upper triangular form with nonzero nal entries (by applying the operations # 1 and # 2) The rest of argument is the same as

diago-in Exercise 1.3.8

1.4.8 By applying the operations # 1 and # 2 to the system Ax = b we obtain an equivalentupper triangular system U x = c Since A is nonsingular, uii 6= 0 for all i, so by Back Sub-stitution each solution component, namely xn= cn

k = i+1

uikxk

1

A,for i = n − 1, n − 2, , 1, is uniquely defined

1.4.9 (a) P1=

0 B B

C, (b) P2 =

0 B B

C,

(c) No, they do not commute (d) P1P2 arranges the rows in the order 4, 1, 3, 2, while

P2P1 arranges them in the order 2, 4, 3, 1

C, (c)

0 B B

C, (d)

0 B B B

C,

0 B B

C,

0 B B

C,

0 B B

C,

0 B B

C,

0 B B

C; (b)

0 B B

C,

0 B B

C,

0 B B

C,

0 B B

C,

0 B B

C,

0 B B

C; (c)

0 B B

C,

0 B B

C

1.4.13 (a) True, since interchanging the same pair of rows twice brings you back to where

you started (b) False; an example is the non-elementary permuation matrix

C (b) True (c) False — A P permutes the columns of A according to

the inverse (or transpose) permutation matrix P−1 = PT

C, (iii)

0 B B

C, (iv )

0 B B B

1.4.18 Let ri, rj denote the rows of the matrix in question After the first elementary row eration, the rows are ri and rj + ri After the second, they are ri − (rj + ri) = −rj and

op-rj+ ri After the third operation, we are left with −rj and rj+ ri+ (−rj) = ri

C;

(d)

0 B B

0 B B

C=

0 B B

0 B B

C, x =

0 B B

C;

(e)

0 B B

0 B B

C=

0 B B

0 B B

C, x =

0 B B

−1

−113

1 C

C;

(f )

0 B B B

0 B B B

C=

0 B B B

0 B B B

C, x =

0 B B B

100

−10

1 C C

0 B

A;

solution: x1 = 54, x2= 74, x3 = 32

(b)

0 B B

0 B B

C=

0 B B

0 B B

C;

solution: x = 4, y = 0, z = 1, w = 1

(c)

0 B B

0 B B

C=

0 B B

0 B B

C;

solution: x = 193 , y = −53, z = −3, w = −2

♦ 1.4.21

(a) They are all of the form P A = L U , where P is a permutation matrix In the first case,

we interchange rows 1 and 2, in the second case, we interchange rows 1 and 3, in thethird case, we interchange rows 1 and 3 first and then interchange rows 2 and 3

(b) Same solution x = 1, y = 1, z = −2 in all cases Each is done by a sequence of tary row operations, which do not change the solution

elemen-1.4.22 There are four in all:0

B0 1 0

1 0 0

0 0 1

1 C

The other two permutation matrices are not regular.

1.4.23 The maximum is 6 since there are 6 different 3 × 3 permutation matrices For example,

ifukj = 0 for all k ≥ j, stop; print “A is singular”

ifujj = 0 but ukj 6= 0 for some k > j theninterchange rowsj and k of U

interchange rowsj and k of Pfor m = 1 to j − 1 interchange ljm and lkm nextmfor i = j + 1 to n

set lij = uij/ujjadd − uij times row j to row i of Anexti

nextjend

0 B

C, (f )

0 B B

xy

(c) Yes: P =

0 B B

Cinterchanges two pairs of rows

1.5.9 (a)

0 B B

C, (b)

0 B B

C, (c)

0 B B

C, (d)

0 B B B

C

1.5.10

(a) If i and j = π(i) are the entries in the ithcolumn of the 2 × n matrix corresponding tothe permutation, then the entries in the jthcolumn of the 2 × n matrix corresponding tothe permutation are j and i = π−1(j) Equivalently, permute the columns so that thesecond row is in order 1, 2, , n and then switch the two rows

(b) The permutations correspond to(i) 14 23 32 41

1.5.11 If a = 0 the first row is all zeros, and so A is singular Otherwise, we make d → 0 by

an elementary row operation If e = 0 then the resulting matrix has a row of all zeros.Otherwise, we make h → 0 by another elementary row operation, and the result is a matrixwith a row of all zeros

1.5.12 This is true if and only if A2 = I , and so, according to Exercise 1.2.36, A is either ofthe form ±1 0

row of U and the ithcolumn of U−1 is the product of their diagonal entries, which mustequal 1 since U U−1 = I

♦ 1.5.18 (a) A = I−1A I (b) If B = S−1AS, then A = S B S−1 = T−1B T , where T = S−1.(c) If B = S−1A S and C = T−1B T , then C = T−1(S−1AS)T = (S T )−1A(S T )

0 B B

C,

0 B B

(b) A X = I does not have a solution Indeed, the first column of this matrix equation is

the linear system

!

=

0

B100

, where z, w are arbitrary

1.5.21 The general solution to A X = I is X =

C, where y, v are arbitrary

Any of these matrices serves as a right inverse On the other hand, the linear system

Y A = I is incompatible and there is no solution

(b) Yes A simple example is A = −1 1

The general solution to the

2 × 2 matrix equation has the form A = B M, where M = xz wy

!

is any matrix with

tr M = x + w = −1, and det M = xw − y z = 1 To see this, if we set A = B M,then ( I + M )−1 = I + M−1, which is equivalent to I + M + M−1 = O Writing thisout using the formula (1.38) for the inverse, we find that if det M = x w − y z = 1 then

tr M = x+w = −1, while if det M 6= 1, then y = z = 0 and x+x−1+1 = 0 = w+w−1+1,

in which case, as in part (a), there are no real solutions

1.5.23 E =

0 B B

C, E−1=

0 B B

C, (g)

0 B

A,

0 B B

C, (i)

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

C=

0 B B

C,

(i)

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

0 B B

C=

0 B B

C