Cap Tulo3 Algebra

Cap´ıtulo 3 - 3.3 Exerc´ıcios de Revis˜ ao Exerc´ıcio 2
(a) Usando o desenvolvimento de Laplace escolhendo a linha 1 temos, det A = 1 · ∆11 + 2 · ∆21 + 1 · ∆31
= 1 · (−1)1+1 · det A11 + 2 · (−1)2+1 · det A21 + 1 · (−1)3+1 · det A31
= 1 · (21 − 20) − 2 · (14 − 10) + 1 · (8 − 6)
= −5.
(b) Temos,


∆11 ∆12 ∆13
¯
Adj A = A¯T , onde A¯ =  ∆21 ∆22 ∆23  . Vamos calcular os elementos de A.
∆31 ∆32 ∆33
3 4
∆11 = (−1)1+1 · det A11 = 1 ·
= 1 · (21 − 20) = 1;
5 7
2 4
∆12 = (−1)1+2 · det A12 = (−1) ·
= (−1) · (14 − 4) = −10;
1 7
2 3
∆13 = (−1)1+3 · det A13 = ·
= 1 · (10 − 3) = 7;
1 5
2 2
∆21 = (−1)2+1 · det A21 = (−1) ·
= (−1) · (14 − 10) = −4;
5 7
1 2
∆22 = (−1)2+2 · det A22 = 1 ·
= 1 · (7 − 2) = 5;
1 7
1 2
∆23 = (−1)2+3 · det A23 = (−1) ·
= (−1) · (5 − 2) = −3;
1 5
2 2
= 1 · (8 − 6) = 2;
∆31 = (−1)3+1 · det A31 = 1 ·
3 4
1 2
= (−1) · (4 − 4) = 0;
∆32 = (−1)3+2 · det A32 = ·
2 4
1 2
∆33 = (−1)3+3 · det A33 = ·
= 1 · (3 − 4) = −1
2 3
Assim,




1 −10 7
1 −4 2
0 .
A¯ =  −4 5 −3  e Adj A = A¯T =  −10 5
2
0 −1
7 −3 −1
(c) Temos que,

 

1 2 2
1 −4 2
0 
A · Adj A =  2 3 4  ·  −10 5
7 −3 −1
1 5 7

 

1 · 1 + 2 · (−10) + 2 · 7 1 · (−4) + 2 · 5 + 2 · (−3) 1 · 2 + 2 · 0 + 2 · (−1)
−5 0
0
=  2 · 1 + 3 · (−10) + 4 · 7 2 · (−4) + 3 · 5 + 4 · (−3) 2 · 2 + 3 · 0 + 4 · (−1)  =  0 −5 0 
1 · 1 + 5 · (−10)+ 7 · 7 1 ·(−4)
0
0 −5
 + 5 · 5 + 7 · (−3)
 1 · 2 + 5 · 0 + 7 · (−1)
1 0 0
−5 0
0



0 −5 0  . e det A · I3 = (−5) · 0 1 0 =
0 0 1
0
0 −5
Portanto, A · Adj A = det A · I3 .
(d) Como det A = −5 = 0, segue
 que A possuiinversa
 −1e ainda,

4
−2
1 −4 2
5
5
5
−1 
1
0  =  2 −1 0  .
· Adj A =
· −10 5
A−1 = det A
5
−7
3
1
7 −3 −1
5
5
5

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