Capıtulo 3 - 3.3 Exercıcios de RevisaoExercıcio 2(a) Usando o desenvolvimento de Laplace escolhendo a linha 1 temos,detA = 1 ·∆11 + 2 ·∆21 + 1 ·∆31

= 1 · (−1)1+1 · detA11 + 2 · (−1)2+1 · detA21 + 1 · (−1)3+1 · detA31

= 1 · (21− 20)− 2 · (14− 10) + 1 · (8− 6)= −5.

(b) Temos,

Adj A = AT , onde A =

∆11 ∆12 ∆13

∆21 ∆22 ∆23

∆31 ∆32 ∆33

. Vamos calcular os elementos de A.

∆11 = (−1)1+1 · detA11 = 1 ·∣∣∣∣ 3 4

5 7

∣∣∣∣ = 1 · (21− 20) = 1;

∆12 = (−1)1+2 · detA12 = (−1) ·∣∣∣∣ 2 4

1 7

∣∣∣∣ = (−1) · (14− 4) = −10;

∆13 = (−1)1+3 · detA13 = ·∣∣∣∣ 2 3

1 5

∣∣∣∣ = 1 · (10− 3) = 7;

∆21 = (−1)2+1 · detA21 = (−1) ·∣∣∣∣ 2 2

5 7

∣∣∣∣ = (−1) · (14− 10) = −4;

∆22 = (−1)2+2 · detA22 = 1 ·∣∣∣∣ 1 2

1 7

∣∣∣∣ = 1 · (7− 2) = 5;

∆23 = (−1)2+3 · detA23 = (−1) ·∣∣∣∣ 1 2

1 5

∣∣∣∣ = (−1) · (5− 2) = −3;

∆31 = (−1)3+1 · detA31 = 1 ·∣∣∣∣ 2 2

3 4

∣∣∣∣ = 1 · (8− 6) = 2;

∆32 = (−1)3+2 · detA32 = ·∣∣∣∣ 1 2

2 4

∣∣∣∣ = (−1) · (4− 4) = 0;

∆33 = (−1)3+3 · detA33 = ·∣∣∣∣ 1 2

2 3

∣∣∣∣ = 1 · (3− 4) = −1

Assim,

A =

1 −10 7−4 5 −32 0 −1

e Adj A = AT =

1 −4 2−10 5 0

7 −3 −1

.

(c) Temos que,

A · Adj A =

1 2 22 3 41 5 7

· 1 −4 2−10 5 0

7 −3 −1

=

1 · 1 + 2 · (−10) + 2 · 7 1 · (−4) + 2 · 5 + 2 · (−3) 1 · 2 + 2 · 0 + 2 · (−1)2 · 1 + 3 · (−10) + 4 · 7 2 · (−4) + 3 · 5 + 4 · (−3) 2 · 2 + 3 · 0 + 4 · (−1)1 · 1 + 5 · (−10) + 7 · 7 1 · (−4) + 5 · 5 + 7 · (−3) 1 · 2 + 5 · 0 + 7 · (−1)

=

−5 0 00 −5 00 0 −5

e detA · I3 = (−5) ·

1 0 00 1 00 0 1

=

−5 0 00 −5 00 0 −5

.

Portanto, A · Adj A = detA · I3.(d) Como detA = −5 6= 0, segue que A possui inversa e ainda,

A−1 =1

detA· Adj A =

−1

5·

1 −4 2−10 5 0

7 −3 −1

=

−15

45

−25

2 −1 0−75

35

15

.

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