13 may - aplicações de matrizes no ensino médio

8/17/2019 13 May - Aplicações de Matrizes No Ensino Médio

1/70

Aplicações de matrizes no ensino médio

Silvia da Rocha Izidoro Ferreira


2/70

Aplicações de matrizes no ensino médio

Silvia da Rocha Izidoro Ferreira

Orientador: Prof. Dr. Sérgio Henrique Monari Soares

Dissertação apresentada ao Instituto de CiênciasMatemáticas e de Computação - ICMC-USP, como

parte dos requisitos para obtenção do título de Mestreem Ciências – Programa de Mestrado Profissional emMatemática. VERSÃO REVISADA

USP – São Carlos

Maio de 2013

SERVIÇO DE PÓS-GRADUAÇÃO DO ICMC-USP

Data de Depósito:

Assinatura:________________________


3/70

Ficha catalográfica preparada pela Seção de Tratamentoda Informação da Biblioteca Prof. Achille Bassi – ICMC/USP

Ferreira, Silvia da Rocha Izidoro

F383a Aplicações de Matrizes no ensino médio / Silvia da

Rocha Izidoro Ferreira; orientador Sérgio Henrique

Monari Soares. –- São Carlos, 2013.

57 p.

Dissertação (Mestrado – Programa de Pós-Graduação em

Mestrado Profissional em Matemática em Rede Nacional

(PROFMAT)) -- Instituto de Ciências Matemáticas e de

Computação, Universidade de São Paulo, 2013.

1. Matrizes. 2. Produto de matrizes. 3. Aplicações

de matrizes. I. Soares, Sérgio Henrique Monari,orient. II. Título.


4/70


5/70


6/70


7/70


8/70


9/70


10/70


11/70


12/70


13/70


14/70

•

•

•

•

•

•


15/70

•

•

•


16/70

A = (aik) B = (bkj) n × p p × q

AB

C = (cij)

cij = ai1b1 j + · · · + aipb pj,

i = 1, · · · , n

j = 1, · · · , q

{P 1,...,P n}

(P i, P j)


17/70

{P 1,...,P n} P i → P j P i

P j (P i, P j)

P i → P j P i

P j P i P j

P i → P j P j → P i P i P j

P 1

P 2

P 3

P 4

n

M = (mij)

n

M

mij =

1,

P i → P j,

0,

M =

0 1 1 0

0 0 1 1

1 0 0 1

1 0 0 0

P 2 P 1

P 2 P 4 P 1 P 2 → P 4 → P 1

P 2 P 1 P 2 → P 4

P 2 → P 4 → P 1 → P 3

k

k ∈ N

P i P j


18/70

M

m

(k)ij

(i, j)

M k

m(k)ij k P i

P j

k = 1

P i P j mij

P i P j mij = m(1)ij k = 1

k

(i, j) M k

m

(k)ij

k

P i P j

k + 1

k + 1

M

m(k+1)ij (i, j) M

k+1

M k+1 = M kM

m(k+1)ij = m

(k)i1 m1 j + m

(k)i2 m2 j + · · · + m

(k)in mnj.

m

(k)i1 = 0 m

(k)i1 = 1 k

P i P 1 P 1 P j

k + 1

P i P j m

(k)i1 m1 j P i

P j k + 1 P 1 P i P j

P 1 k + 1 m

(k)i1 m1 j = 0

l = 1, 2, · · · , n

P i P j P l k + 1

m(k)il mlj

k + 1

k + 1

P i P j

k ∈ N

A

A =

0 1 1 1

1 0 0 0

0 1 0 10 1 1 0

.

P 1

P 4

A3 =

2 3 2 2

1 2 1 1

1 2 1 2

1 2 2 1

.

m(3)14 = 2 P 1 P 4


19/70

P i → P j

P i P j

P i P j P i → P j P j → P i

{P 1, P 2, P 3, P 4} {P 3, P 4, P 6}.

S = (sij)

sij =

1,

P i ↔ P j,

0,

S

S


20/70

S

s3ij (i, j) S

3

P i

s3ii = 0

s3ii = 0 P i

S

P i → P j → P k → P i

P i ↔ P j ↔ P k ↔ P i {P i, P j, P k}

P i

P i s

3ii = 0

M =

0 1 1 1

1 0 1 0

0 1 0 1

1 0 0 0

P 1

P 2

P 3

P 4

M

S =

0 1 0 1

1 0 1 0

0 1 0 01 0 0 0

S 3 =

0 3 0 2

3 0 2 0

0 2 0 12 0 1 0

.


21/70

s311, s

322, s

333, s

344

M =

0 1 0 1 1

1 0 0 1 0

1 1 0 1 0

1 1 0 0 0

1 0 0 1 0

.

S =

0 1 0 1 1

1 0 0 1 00 0 0 0 0

1 1 0 0 0

1 0 0 0 0

S 3 =

2 4 0 4 3

4 2 0 3 10 0 0 0 0

4 3 0 2 1

3 1 0 1 0

.

s311, s322, s

344 P 1, P 2, P 4

{P 1, P 2, P 4}

P i P j P i → P j P j → P i

n

P i → P j P i P j

P 2

P 1 P 3

P 3

P 2

P 1 P 4

P 2

P 1 P 3

P 5 P 4

P 1

P 1, P 2, P 3 P 1, P 3, P 5


22/70

P 1

P i

P 1 P i P 1 → P i

P i → P 1

P k P 1 → P k P k → P i

P 1 → P k → P i P 1 P i

P i → P k P i

P 1 P i

P 1

P i → P 1 P i

P 1

P 1 P i P 1

M

M 2

i

M

P i i M 2

P i

i M + M 2 P i

M + M 2

P i i A = M + M

2

M

• A

B

C

D


23/70

• B C E

• C

D

E

• D

B

• E

A D

A

B C

D E

M

M =

0 1 1 1 00 0 1 0 1

0 0 0 1 1

0 1 0 0 0

1 0 0 1 0

A = M + M 2 =

0 1 1 1 0

0 0 1 0 1

0 0 0 1 1

0 1 0 0 01 0 0 1 0

+

0 1 1 1 2

0 0 0 2 1

1 1 0 1 0

0 0 1 0 10 2 1 1 0

=

0 2 2 2 2

0 0 1 2 2

1 1 0 2 1

0 1 1 0 11 2 1 2 0

,

,

,

,

A − (

)

E − (

)

B

C − (

)


24/70

D − ( )

A

E

B

C

D

B

C

E

E

A

B

C

B

C

E

C

B

B

D

C

a, b

m

m > 0

a

b

m (a ≡ b m) m a − b a − b m

m

a

m

0, 1, 2, · · · , m−1

a

m

Z

m

[0] = {x ∈ Z; x ≡ 0 m}

[1] = {x ∈ Z; x ≡ 1 m}

[m − 1] = {x ∈ Z; x ≡ m − 1

m}.

[a] = {x ∈ Z; x ≡ a m} m

a ∈ Z

m

Zm

Zm = {[0], [1], · · · , [m − 1]}.

a

m

a m a


25/70

m

a

R

|a| m r a

r =

R,

a ≥ 0

m − R, a


26/70

Z4

[2] = [0] [2] · [2] = [0] Z3

Zm Z4 [2]

Zm

[a] ∈ Zm a m

(a, m) = 1.

[a]

[b] ∈ Zm [1] = [a] · [b] = [a · b]

ab ≡ 1

m

(a, m) = 1

(a, m) = 1

b

t

ab − mt = 1

[1] = [1] + [mt] = [1 + mt] = [a · b] = [a] · [b].

[a]

3

26

(3, 26) = 1

4

26

4 26

2

[a]

[a]−1

m

A

Zm m B

Zm

AB = BA = I ,

I

B

B

A

A−1

2 × 2 A Zm m

det(A)

m

m

A =

a b

c d

∈ M 2(Zm) det(A) = D = ad − bc ∈ Zm

A

m

A−1

A

AA−1 = A−1A = I .

det(A)det(A−1) = det(AA−1) = det(I ) = 1

m.


27/70

det(A−1)

m

det(A)

(m, D) = 1

D−1 ∈ Zm

DD−1 = 1

m.

A−1 =

D−1d −D−1b

−D−1c D−1a

,

A

A

Zm m

m

det(A)

m

A

Z26 26

det(A)

26

2

13

A =

1 3

0 1

D =

(A) = 1

D−1 = 1

26

A−1 = 1

1 −3

0 1

=

1 −3

0 1

=

1 23

0 1

26.

n

n


28/70

A

p1 p2

p =

p1

p2

Ap

p

Ap

2 × 2

3 × 3

n

n

n × n

1 3

2 1

1 3

2 1

.

20

21

=

5

9

mod26

1 3

2 1

.

4

15

=

23

23

mod26

1 3

2 1

.

5

16

=

1

0

mod26

1 3

2 1

.

15

19

=

20

23

mod26


29/70

1 3

2 1

.

19

9

=

20

21

mod26

1 3

2 1

.

22

5

=

11

23

mod26

1 3

2 1

.

12

1

=

15

25

mod26

1 3

2 1 . 15

17 = 14

21 mod26

1 3

2 1

.

21

5

=

10

21

mod26

1 3

2 1

.

3

18

=

5

24

mod26

1 3

2 1 .

5

1 =

8

11mod26

EIWWAZ TWT U KWOY N U JU EX HK

NMITRMITRCPEQEOA

A

p =

p1

p2

c = Ap

p = A−1c

1 4 1 3 9 2 0 1 8 1 3 9 2 0 1 8 3 1 6 5 1 7 5 1 5 1

p = A−1c


30/70

1 230 1

1413

=

31313

=

113

26

1 23

0 1

9

20

=

469

20

=

1

20

26

1 23

0 1

18

13

=

317

13

=

5

13

26

1 23

0 1

9

20

=

469

20

=

1

20

26

1 23

0 1

18

3

=

87

3

=

9

3

26

1 23

0 1

16

5

=

131

5

=

1

5

26

1 23

0 1

17

5

=

132

5

=

2

5

26

1 23

0 1

15

1

=

38

1

=

12

1

26

AM AT EM AT IC AE BE LA


31/70

•

0 • ∗ ∗ ∗ ∗ ∗

0 0 0 • ∗ ∗ ∗

0 0 0 0 • ∗ ∗

0 0 0 0 0 0 •

0 0 0 0 0 0 0

Li ↔ L j

Li → kLi

Li → Li + kL j

k

1/k

1 ∗ ∗ ∗ ∗

0 0 1 ∗ ∗0 0 0 0 1

→

1 ∗ 0 ∗ 0

0 0 1 ∗ 00 0 0 0 1

.


32/70

x + y + 2z = 9

2x + 4y − 3z = 1

3x + 6y − 5z = 0

A

A =

1 1 2

2 4 −3

3 6 −5

9

1

0

.

1 1 2

0 2 −7

0 3 −11

9

−17

−27

.

−2 0 −110 2 −70 0 −1

−35−17

−3

.

−2 0 0

0 2 0

0 0 −1

−2

4

−3

.

(−1

2

)

1

2

1 0 0

0 1 0

0 0 1

1

2

3

.

x = 1,

y = 2,

z = 3.

x = 1, y = 2 z = 3


33/70

A =

1 2 3

2 5 3

1 0 8

A

I

A−1

A

[A | I ].

I

A−1

[I |A−1] =

1 2 3

2 5 3

1 0 8

1 0 0

0 1 0

0 0 1

.

1 2 3

0 1 −3

0 −2 5

1 0 0

−2 1 0

−1 0 1

.

1 2 3

0 1 −3

0 0 −1

1 0 0

−2 1 0

−5 2 1

.

1 2 3

0 1 −3

0 0 1

1 0 0

−2 1 0

5 −2 −1

.

1 2 0

0 1 00 0 1

−14 6 3

−2 1 05 −2 −1

.

1 0 0

0 1 0

0 0 1

−40 16 9

13 −5 −3

5 −2 −1

.

A

−1

=

−40 16 9

13 −5 −35 −2 −1

.


34/70

E

E

E 1 =

1 0 0

0 0 1

0 1 0

E 2 =

1 0 0

0 7 0

0 0 1

E 3 =

1 0 −3

0 1 0

0 0 1

E

E −1

E

E 1 =

0 1

1 0

, E 2 =

1 0

0 5

, E 3 =

1 3

0 1

.

A =

a11 a12 a13

a21 a22 a23

E 1A =

a21 a22 a23

a11 a12 a13

(L1 ↔ L2)

E 2A = a11 a12 a13

5a21 5a22 5a23 (L2 → 5L2)

E 3A =

a11 + 3a12 a12 + 3a22 a13 + 3a23

a21 a22 a23

(L1 → L1 + 3L2)

E iA A E i

k

A E i

i


35/70

i

E i

A → E 1A → E 2E 1A → · · · → E k · · · E 2E 1A,

A → U A,

U = E k · · · E 2E 1.

A

U A = I

U = A−1

A B → I A−1 .

p1, p2,...,pn

c1, c2,...,cn n

n × n

A

P =

pT 1

pT 2

pT n

n × n

pT 1 , p

T 2 ,...,p

T n

C =

cT 1

cT 2

cT n

n × n cT 1 , c

T 2 ,...,c

T n

C

I

P

(A−1)T

P

C

C = P AT

A

p1, p2,...,pn C

E 1, · · · , E k

C

I

E k · · · E 1C = I C = P A

T

E k · · · E 1P AT = I ,


36/70

E k · · · E 1P = (A

−1)T

C

I

P

(A−1)T

A−1

C

I

P

IOSBTGXESPXHOPDE

p1 =

45

↔ c1 =

915

p2 =

1

18

↔ c2 =

19

2

.

C =

cT 1

cT 2

=

9 15

19 2

I

P =

pT 1

pT 2

=

4 5

1 18

(A−1)T

9 15

19 2

4 51 18

.

9−1 = 3

1 45

19 2

12 151 18

.

45

26

1 19

19 2

12 151 18

.


37/70

−19

1 190 −359

12 15−227 −267

.

26

1 19

0 5

12 157 19

.

5−1 = 21

1 190 1 12 1517 9 .

−19 1 0

0 1

−311 −15617 9

.

1 0

0 1

1 0

17 9 .

(A−1)T =

1 0

17 9

A−1 =

1 17

0 19

1 17

0 9 .

9

15 =

4

5 26

1 17

0 9

.

19

2

=

1

18

26

1 17

0 9

.

20

7

=

9

11

26

1 17

0 9

.

24

5

=

5

19

26

1 17

0 9

.

19

16

=

5

14

26


38/70

1 17

0 9

.

24

8

=

4

20

26

1 17

0 9

.

15

16

=

1

14

26

1 17

0 9

.

4

5

=

11

19

26

DEAR IKE SEND T ANKS

{1, 2,...,k}

i

j

pij

j

i

P = [ pij]

p11 p12 p13

p21 p22 p23

p31 p32 p33

p23 3 2 p12

2

1


39/70

0, 8 0, 3 0, 2

0, 1 0, 2 0, 6

0, 1 0, 5 0, 2

1

x =

x1

x2

x3

,

x1 1 x2

2

x3 3

k

x

xi

1

P xn

xn+1 = P xn = P n+1x0


40/70

x0 =

x01x02

x0k

k

x0i

i

i ∈ {1, · · · , k} pij

j

i

k

x1 =

p11x01 + p12x02 + · · · + p1kx0k

pk1x01 + pk2x02 + · · · + pkkx0k

.

x1 = P x0

n + 1

k

xn+1 = P xn = P 2xn−1 = · · · = P n+1x0.

x0

x1, x2,...,xn,...

0, 8 0, 3 0, 2

0, 1 0, 2 0, 60, 1 0, 5 0, 2

x0 =

0

1

0

.

xn =

xn1

xn2

xn3

n ≥ 11 x11


41/70

n

xn1

xn2

xn3

n

P =

0 1

1 0

x0 =

1

0

.

P 2 = I

P 3 = P

x0 = x2 = x4 = · · · =

1

0

x1 = x3 = x5 = · · · =

0

1

.

P

N ∈ N P N

P

r × r

ε

P

x

r

M 0 m0

x

M 1 m1

xT P

M 1 ≤ M 0, m1 ≥ m0 M 1 − m1 ≤ (1 − 2ε)(M 0 − m0).

x

x

m0 M 0 xi ≤ x

i

i = 1, · · · , r P xT P

am0 + (1 − a)M 0 = M 0 − a(M 0 − m0),


42/70

a ≥ ε xT P M 0 − ε(M 0 − m0)

xi ≤ x

i i = 1, · · · , r

M 1 ≤ M 0 − ε(M 0 − m0).

−x

−m1 ≤ m0 − ε(−m0 + M 0).

M 1 − m1 ≤ M 0 − m0 − 2ε(M 0 − m0) = (1 − 2ε)(M 0 − m0).

P

r × r

P n → Q =

q 1 q 1 ... q 1

q 2 q 2 ... q 2

q k q k ... q k

n → +∞ q i q 1 + q 2 + ... + q k = 1

P ε P ρ j

j

0

M n mn

ρT j P

n

ρT j P n = ρT j P

n−1P,

M 1 ≥ M 2 ≥ M 3 ≥ · · ·

m1 ≤ m2 ≤ m3 ≤ · · ·

M n − mn ≤ (1 − 2ε)(M n−1 − mn−1), n ≥ 1.

M n mn

dn = M n − mn

dn ≤ (1 − 2ε)nd0 = (1 − 2ε)

n.

limn→+∞

dn = 0

M n mn ρ

T j P

n

q j ρT j P n

j

P n

j

P n


43/70

q j P

n

Q

q =

q 1

q 2

q r

P

P n

P n

Q

P

P

N

P N

ε

P N

dkN ≤ (1 − 2ε)k,

k ≥ 1.

(dn)

dn → 0 n → +∞

q

x0 P

nx0 q n → ∞

P

x

P nx →

q 1

q 2

q k

= q

n → +∞ q n

P n → Q n → +∞ P nx → Qx = q

n → +∞

q

q

q

P

P q = q

P P n = P n+1

P n

P n+1

Q n → +∞ P Q = Q

P q = q

q


44/70

r

P r = r

P nr = r

n = 1, 2, 3,...

n → ∞

q = r

(I − P ) p = 0

q

q 1+· · ·+q k = 1

0, 2 0, 1 0, 7

0, 6 0, 4 0, 2

0, 2 0, 5 0, 1

I 3 q =

x1

x2

x3

(I − P )q = 0.

1 0 0

0 1 0

0 0 1

−

0, 2 0, 1 0, 7

0, 6 0, 4 0, 2

0, 2 0, 5 0, 1

x1

x2

x3

=

0

0

0

0, 8 −0, 1 −0, 7

−0, 6 0, 6 −0, 2

−0, 2 −0, 5 0, 9

x1

x2

x3

=

00

0

1 0 −22210 1 −29210 0 0

x1

= 2221

x3

x

2 = 29

21x3

x3 = s

q = s

22212921

1

x1 + x2 + x3 = 1 x1 =

1136 x2 =

2972 x3 =

2172

q =

113629

722172

.


45/70

5

1

2

5

3

2

15

1

10

3

3

10

1

5

2

P =

90% 15% 10%

5% 75% 5%

5% 10% 85%

=

0, 9 0, 15 0, 1

0, 05 0, 75 0, 05

0, 05 0, 1 0, 85

I 3 q =

x1

x2

x3

(I − P )q = 0.

1 0 0

0 1 0

0 0 1

−

0, 9 0, 15 0, 1

0, 05 0, 75 0, 05

0, 05 0, 1 0, 85

x1

x2

x3

=

0

0

0

0, 1 −0, 15 −0, 1

−0, 5 0, 25 −0, 05

−0, 05 −0, 1 0, 15

x1

x2

x3

=

0

0

0

1 0 −1370 1 −470 0 0

x1 =

137 x3 x2 =

47x3

x3 = s

q = s.

13747

1

x1 + x2 + x3 = 1 x1 =

1324 x2 =

424 x3 =

724

54, 2 1 16, 7

2

29, 1

3


46/70

A

n × n λ A

Ax = λx

x

x

A

λ

A = 5 −2

4 −1

λ = 3 A

x =

1

1

x =

0

0

Ax =

5 −2

4 −1

1

1

=

3

3

= 3

1

1

= 3x

A

λ = 3

n × n

λ

A

Ax = λx

x = 0

I

A

(λI − A)x = 0 x = 0.

|λI − A| = 0.

A =

5 −2

4 −1

λI −A =

λ − 5 2−4 λ + 1

|λI −A| = (λ−5)(λ+1)+8 = (λ−3)(λ−1)

λ1 = 3 λ2 = 1 A λ1 = 3

λ2 = 1

λ2I − A =

λ2 − 5 2

−4 λ2 + 1

=

−4 2

−4 2

(λ2I − A)x = 0 x = t

1/2

1

t

x

λ2 x = t

1/2

1

t = 0


47/70

t = 2

x =

1

2

λ2 λ1 = 3

x = t

1

1

t = 0

n × n

µ1 0 · · · 0

0 µ2 · · · 0

0 0 · · · µn

A

P

P −1AP

P

A

λ1 = 3 λ2 = 1

P = 1 12 1 .

P

P −1AP = D

D =

3 0

0 1

,

A =

5 −2

4 −1

n × n

λ

A

λ

(λI − A)x = 0

A

x1, · · · , xn

P = [x1x2 · · · xn]

A

P −1AP

P

P A =

1 12 0

0 12 1

0 0 0


48/70

λ1 = 1 λ2 =

12 λ3 = 0

v1 =

1

0

0

, v2 = 1

−1

0

, v3 = 1

−2

1

P = 1 1 1

0 −1 −2

0 0 1

P −1AP = D =

1 0 0

0 12 0

0 0 0

.

A

n × n

Ak

k

P

D

P −1AP = D

A = P DP −1.

k = 1, 2, · · ·

Ak = AA · · · A = (P DP −1)(P DP −1) · · · (P DP −1)

= P D(P −1P )DP −1 · · · P DP −1

= P D(I )D(I ) · · · (I )DP −1

= P D · · · DP −1

= P DkP −1.

Ak = P DkP −1

k = 1, 2, · · ·

Dk =

λ1 0 · · · 0

0 λ2 · · · 0

0 0 · · · λn

k

=

λ

k

1 0 · · · 00 λk2 · · · 0

0 0 · · · λkn

.

A

An = P DnP −1 =

1 1 1

0 −1 −2

0 0 1

1 0 0

0

12

n

0

0 0 0

1 1 1

0 −1 −2

0 0 1

=

1 1 −12

n1 −12

n−10

12

n

12

n−1

0 0 0

.

A


49/70

A

a

AA

Aa

aa

AA

Aa

aa

A

a

a

A

A

a

AA

Aa

aa

Aa

A

a

aa

Aa

a

aa

A

a

Aa

Aa

aa

AA Aa aa

AA


50/70

n ∈ N

an AA n

bn Aa n

cn aa n

a0 b0 c0

an + bn + cn = 1 n = 0, 1, 2,...

AA

an = an−1 + 1

2bn−1,

bn = cn−1 + 1

2bn−1,

cn = 0.

AA

AA

Aa

AA

aa

Aa

Aa

Aa aa

AA

xn = Axn−1,

xn =

an

bn

cn

, xn−1 =

an−1

bn−1

cn−1

A =

1 12 0

0 12 1

0 0 0

.

A

xn = Axn−1 = A2xn−2 = ... = A

nx0

An

An = P DnP −1 =

1 1 −12

n1 −12

n−1012

n 12

n−10 0 0

.

xn = P DnP −

1x0 =

1 1 − (12)n 1 − (12)

n−1

0 (12)n (12)n−1

0 0 0

a0

b0

c0


51/70

xn =

anbncn

= a0 + b0 + c0 − (

1

2)n

b0 − (1

2)n−1

c0( 12)

nb0 + (12)

n−1c0

0

.

a0 + b0 + c0 = 1

an = 1 −

1

2

nb0 −

1

2

n−1c0

bn =

1

2

nb0 +

1

2

n−1c0

cn = 0

n

12

n

n

an → 1,

bn → 0.

cn → 0 cn

AA

AA

xn = An.x0,

A =

1 14 0

0 12 0

0 14 1

.

A

λ1 = 1 λ2 = 1 λ3 = 12

v1 =

1

0

0

v2 =

0

0

1

v3 =

1

−2

1

an = a0 + 1

2b0 −

1

2

n+1b0

bn = 1

2n

b0

cn = c0 + 1

2b0 −

1

2

n+1

b0


52/70

n

12n n

an → a0 + 1

2b0,

bn → 0,

cn → c0 + 1

2b0.

AA

aa


53/70


54/70

A

• detA

• A detA = 0

• det(AB) = detAdetB

•

detA

A

A = [aij]

A

detA

det[aij]

|A|

•

1 × 1 |a| = a

• 2 × 2 a11 a12a21 a22 = a11a22 − a12a21


55/70

• 3 × 3

a11 a12 a13a21 a22 a23

a31 a32 a33

= a11a22a33 − a11a23a32 − a12a21a33 + a12a23a31 + a13a21a32 − a13a22a31.

2 × 2

3 × 3

3 × 3

(a11a22a33 − a11a23a32) − (a12a21a33 − a12a23a31) + (a13a21a32 − a13a22a31),

a11 a12 a13

a21 a22 a23

a31 a32 a33

= a11

a22 a23a32 a33− a12 a21 a23a31 a33+ a13 a21 a22a31 a32

a11 a12 a13

a21 a22 a23

a31 a32 a33

= a11|A11| − a12|A12| + a13|A13|,

A11 A12 A13 2 × 2 A

A

3 × 3

−(a12a21a33 − a13a21a32) + (a11a22a33 − a13a22a31) − (a11a23a32 − a12a23a31),

a11 a12 a13

a21 a22 a23

a31 a32 a33

= −a21|A21| + a22|A22| − a23|A23|.

n × n

(n − 1) × (n − 1)

n ≥ 2

n × n A = [aij]

|A| = ai1|Ai1| − ai2|Ai2| + · · · + (−1)1+nain|Ain| =

n j=1

(−1)i+ jaij|Aij|,

i, j ∈ {1, . . . , n} Aij (n − 1) × (n − 1) A

i

j

A


56/70

i

A =

3 −7 8 9 −6

0 2 −5 7 3

0 0 1 5 0

0 0 2 4 −1

0 0 0 −2 0

,

|A| = 3

2 −5 7 3

0 1 5 0

0 2 4 −1

0 0 −2 0

.

4×4

|A| = 3 · 2

1 5 0

2 4 −1

0 −2 0

= 3 · 2 · (−2) = −12,

3 × 3

A

|A|

A


57/70

A

n × n

B

13 may - aplicações de matrizes no ensino médio

Documents