Chapter 6 Take determinants on both sides of the equation

Chapter 6

We will evaluate the determinant of Bby expanding across the ith row (where iis neither pnor q).

6.2.61 We follow the hint: In0

−C A A B

C D =A B

−CA +AC −CB +AD 

=A B

0AD −CB .So, det  In0

−C A A B

C D = det(A) det(AD −CB).

6.2.62 a We compute In0

−CA−1InA B

C D =A B

0−CA−1B+D. Since the matrix In0

−CA−1Inis invert-

ible (its determinant is 1),the product A B

6.2.63 Let Mnbe the number of multiplications required to compute the determinant of an n×nmatrix by Laplace

expansion. We will use induction on nto prove that Mn> n!,for n≥3.

6.2.64 To compute det(A) for an n×nmatrix Aby Laplace expansion, det(A) = a11 det(A11)−a21 det(A21) + ···+

(−1)n+1an1det(An1),we first need to compute the nminors, which requires nLn−1operations; then we compute

the nproducts ai1det(Ai1); and finally we have to do n−1 additions. Altogether,

294

Section 6.2

6.2.65 a. Using Laplace expansion along the first row, we find dn= det (Mn) = det (Mn−1)−det −1∗

0Mn−2=

6.2.66 a. Using Laplace expansion along the first row, we find dn= det (Mn) = det (Mn−1)−det 1∗

0Mn−2=

det (Mn−1)−det (Mn−2) = dn−1−dn−2.

6.2.67 Let kbe the number of pattern entries to the left and above aij . Then the number of pattern entries to

the right and above aij is i−1−k, since there are i−1 rows above aij , each of which contains exactly one

6.2.68 Using Exercise 67 and the terminology introduced in the proof of Theorem 6.2.10, we have

sgnP = (−1)#(inversions in P)= (−1)#(inversions involving aij )(−1)#(inversions not involving aij )

= (−1)i+j(−1)#(inversions in Pij )= (−1)i+jsgn (Pij ).

6.2.69 a. The integers 0, 1, 2, 4, 5, 8, 9, and 10 are in G.

6.2.70 a.

f0f1f2f3f4f5f6f7f8

0 1 1 2 3 5 8 13 21

295

Chapter 6

Section 6.3

6.3.1By Theorem 2.4.10, the area equals |det 3 8

7 2 |=| − 50|= 50.

Figure 6.1: for Problem 6.3.3.

Figure 6.2: for Problem 6.3.4.

On the other hand, det 



a1b1c1

a2b2c2

1 1 1 

= det 



a1b1−a1c1−a1

a2b2−a2c2−a2

1 0 0 

,by subtracting the first column from the

second and third.

296

Section 6.3

6.3.5The volume of the tetrahedron T0defined by ~e1, ~e2, ~e3is 1

3(base)(height) = 1

6.

Here we are using the formula for the volume of a pyramid. (See Figure 6.3.)

6.3.6From Exercise 5 we know that volume of tetrahedron = 1

6

det 



a1b1c1

a2b2c2



, and Exercise 4 tells us that

Figure 6.4: for Problem 6.3.6.

6.3.8We need to show that both sides of the equation in Theorem 6.3.3 give zero.

297

Chapter 6

6.3.9Using linearity in the second column, we find that det ~v1~v2= det h~v1~vk

2+~v⊥

2 i= det h~v1~vk

2i

|{z }

noninvertible

+ det ~v1~v⊥

2= det ~v1~v⊥

2. Thus the two determinants are equal.

6.3.12 Denote the columns by ~v1,~v2,~v3,~v4. By Theorem 6.3.3 and Exercise 6.3.8 we know that |det(A)| ≤

k~v1kk~v2k k~v3kk~v4k; equality holds if the columns are orthogonal. Since the entries of the ~viare 0, 1, and −1, we

have k~vik ≤ √1 + 1 + 1 + 1 = 2. Therefore, |det A| ≤ 16.

6.3.13 By Theorem 6.3.6, the desired 2-volume is

298

Section 6.3

6.3.14 By Theorem 6.3.6, the desired 3-volume is

v

u

tdet 







1 0 0 0

1 1 1 1

1 2 3 4 







1 1 1

0 1 2

0 1 3

0 1 4









=v

u

tdet 



111

1 4 10

1 10 30 

=√6.

6.3.15 If ~v1, ~v2, . . . , ~vmare linearly dependent and if A= [~v1···~vm],then det(ATA) = 0 since ATAand Ahave

equal and nonzero kernels (by Theorem 5.4.2), hence ATAfails to be invertible.

6.3.16 False

6.3.17 a Let ~w =~v1×~v2×~v3.Note that ~w is orthogonal to ~v1, ~v2and ~v3,by Exercise 6.2.44c. Then V(~v1, ~v2, ~v3, ~w) =

V(~v1, ~v2, ~v3)k~w⊥k=V(~v1, ~v2, ~v3)k~wk.

↑

6.3.18 a (See Figure 6.6.)p0

0qcos(t)

sin(t)=p·cos(t)

q·sin(t), the ellipse with semi-axis ±p

0and ±0

q.

(area of the ellipse) = |det(A)|(area of the unit circle) = pqπ

299

Chapter 6

T(~x) = cos(t) 2√21

1

|{z }

+ sin(t)√21

−1

|{z }

semi-major axis semi-minor axis

.(See Figure 6.7.)

Figure 6.7: for Problem 6.3.18c.

6.3.19 det[~v1~v2~v3] = ~v1·(~v2×~v3) = k~v1kk~v2×~v3kcos θwhere θis the angle between ~v1and ~v2×~v3so det[~v1~v2~v3]>0

if and only if cos θ > 0,i.e., if and only if θis acute (0 ≤θ≤π

2).(See Figure 6.8.)

300

Section 6.3

v1

v2v3

θ

×

but T(~v1) = −~v1, T (~v2) = ~v2, T (~v3) = ~v3is negatively oriented.

b Preserves

Consider ~v2and ~v3orthogonal to the line (not parallel), and let ~v1=~v2×~v3; then ~v1, ~v2, ~v3is a positively oriented

basis, and T(~v1) = ~v1, T (~v2) = −~v2, T (~v3) = −~v3is positively oriented as well.

c Reverses

6.3.23 Here A=5−3

−6 7 ,det(A) = 17,~

b=1

0,so by Theorem 6.3.8

x1=

det 1−3

0 7 

17 =7

17, x2=

det 5 1

−6 0 

17 =6

17.

301

Chapter 6

6.3.25 By Theorem 6.3.9, the ijth entry of adj(A) is given by (−1)i+jdet(Aji), so since

6.3.26 By Theorem 6.3.9, A−1=1

det(A)adj(A), so if det(A) = 1, A−1= adj(A). If Ahas integer entries then

(−1)i+jdet(Aji) will be an integer for all 1 ≤i, j ≤n, hence adj(A) will have integer entries. Therefore, A−1

will also have integer entries.

6.3.27 By Theorem 6.3.8, using A=a−b

b a ,det(A) = a2+b2,~

b=1

0,we get

6.3.28 Here A=s a

m−h,det(A) = −sh −ma,~

b=I◦+G

Ms+M◦so, by Theorem 6.3.8

6.3.29 By Theorem 6.3.8,

dx1=

det





0R1−(1 −α)

0 1 −α−(1 −α)2

−R2de2−R2−(1−α)2

α







302

Section 6.3

6.3.30 Using the procedure outlined in Exercise 25, we find adj(A) = 



18 0 0

−12 6 0

−2−5 3 

.

6.3.33 Using the procedure outlined in Exercise 25, we find that adj(A) = 





24 0 0 0

0 12 0 0

0 0 8 0

0 0 0 6





.Note that the matrix

adj(A) is diagonal, and the ith diagonal entry of adj(A) is the product of all ajj where j6=i.

6.3.36 adj(adjA) = adj(det(A)A−1)

= det(det(A)A−1)(det(A)A−1)−1= (det A)ndet(A−1)(det(A)A−1)−1

= (det A)n−1(det(A)A−1)−1= (det A)n−11

det(A)(A−1)−1

= (det A)n−2A.

6.3.39 Yes, let Sbe an invertible matrix such that AS =SB, or SB−1=A−1S. Multiplying both sides by

det(A) = det(B),we find that S(det(B)B−1) = (det(A)A−1)S, or, S(adjB) = (adjA)S, as claimed.

303

Chapter 6

6.3.42 By Theorem 6.3.9, adj(A) = 0 if (and only if) all the minors Aji of Aare zero. By Exercise 41, this is the

case if (and only if) rank(A)≤n−2.

6.3.43 A direct computation shows that A(adjA) = (adjA)A= (det A)(In) for all square matrices. Thus we have

A(adjA) = (adjA)A= 0 for noninvertible matrices, as claimed.

Let’s write B= adj(A),and let’s verify the equation AB = (det A)(In) for the diagonal entries; the verification

for the oﬀ-diagonal entries is analogous. The ith diagonal entry of AB is

6.3.44 The equation A(adjA) = 0 from Exercise 43 means that im(adjA) is a subspace of ker(A). Thus rank(adjA) =

dim(im(adjA)) ≤dim(kerA) = n−rank(A) = n−(n−1) = 1,implying that rank(adjA)≤1. Since adj(A)6= 0,

by Exercise 42, we can conclude that rank(adjA) = 1.

6.3.45 Let A=a b

c d .We want AT= adj(A),or a c

b d =d−b

−c a .So, a=dand b=−c. Thus, the

equation AT= adj(A) holds for all matrices of the form a b

−b a .

304

Section 6.3

6.3.47 Note that 1

2det x1x2

y1y2is the area of the triangle OP1P2,where Odenotes the origin. This is likewise

true for one-half the second matrix. See Theorem 2.4.10. However, because of the reversal in orientation,

6.3.48 In what follows, we will freely use the fact that an invertible linear transformation Lfrom R2to R2maps an

ellipse into an ellipse (see Exercise 2.2.52).

Now consider a linear transformation Lthat transforms our 3-4-5 right triangle Rinto an equilateral triangle T.

R

2

0

3

[

√3

1

A =

L

0

1

21

3

1

√3

6.3.49 We will use the terminology introduced in the solution of Exercise 48 throughout. Note that the trans-

formation L−1,with matrix A−1=2−2/√3

0√3, maps the circle C(with radius 1/√3) into the ellipse E.

305

Chapter 6

18 ≈0.79.

True or False

Ch 6.TF.1T, by Theorem 6.2.3a, applied to the columns.

Ch 6.TF.2T, by Theorem 6.2.6.

Ch 6.TF.6F; We have det(4A) = 44det(A), by Theorem 6.2.3a.

Ch 6.TF.7F; Let A=B=I5, for example

Ch 6.TF.8T; We have det(−A) = (−1)6det(A) = det(A), by Theorem 6.2.3a.

Ch 6.TF.12 F. There is only one pattern with a nonzero product, containing all the 1’s. Since there are three

inversions in this pattern, det A=−1.

Ch 6.TF.13 T. Without computing its exact value, we will show that the determinant is positive. The pattern that

contains all the entries 100 has a product of 1004= 108, with two inversions. Each of the other 4! −1 = 23

Ch 6.TF.14 F; The correct formula is det(A−1) = 1

det(AT),by Theorems 6.2.1 and 6.2.8.

True or False

Ch 6.TF.17 T, by Theorem 6.2.7.

Ch 6.TF.21 F; Note that det 



1 1 0

0 1 1

1 0 1 

= 2.

Ch 6.TF.22 T, by Theorem 6.3.9.

Ch 6.TF.26 F; Let A= 2I2, for example

Ch 6.TF.27 T; Let A=





1 1 1 1

1 1 −1−1

1−1 1 −1

1−1−1 1





. The column vectors of Aare orthogonal and they all have length 2.

Ch 6.TF.31 F; Note that det(S−1AS) = det(A) but det(2A) = 23(det A) = 8(det A).

Ch 6.TF.32 F; Note that det(STAS) = (det S)2(det A) and det(−A) = −(det A) have opposite signs.

307

Chapter 6

Ch 6.TF.35 F; Let A=I2and B=−I2, for example.

Ch 6.TF.38 T; Note that det(A) and det(A−1) are both integers, and (det A)(det A−1) = 1. This leaves only the

possibilities det(A) = det(A−1) = 1 and det(A) = det(A−1) = −1.

Ch 6.TF.42 F; Let A=



2 1 1

1 2 1

1 5 2 

, for example

Ch 6.TF.43 T; Let A=a b

c d . If a6= 0, let B=0 0

0 1 ; if b6= 0, let B=0 0

1 0 ; if c6= 0, let B=0 1

0 0 ,

Ch 6.TF.45 T; A(adjA) = A(det(A)A−1) = det(A)In= det(A)A−1A= adj(A)A.

Ch 6.TF.46 T; Laplace expansion along the second row gives det(A) = −kdet 



1 2 4

8 9 7

0 0 5 

+C= 35k+C, for some

308

True or False

309