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Cap´ıtulo 1 Matrizes e Sistemas Lineares

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28 CAPÍTULO 2. MATRIZ INVERSA E DETERMINANTES<br />

Exemplo:<br />

Seja A =<br />

Solução :<br />

⎡<br />

⎢<br />

⎣<br />

<br />

An×n<br />

1 2 3<br />

1 1 2<br />

0 1 2<br />

. In×n<br />

⎤<br />

<br />

→ Gauss − Jordan → <br />

⎥<br />

⎦ ,encontre A −1 caso exista.<br />

In×n<br />

. A −1<br />

A −1 ⎡<br />

⎤<br />

1 2 3 . 1 0 0<br />

⎢<br />

⎥<br />

⎢<br />

⎥<br />

= ⎢<br />

⎣ 1 1 2 . 0 1 0 ⎥<br />

⎦<br />

0 1 2 . 0 0 1<br />

L (1)<br />

2 = (−1)L (0)<br />

1 + L (0)<br />

⎡<br />

⎤<br />

1 2 3 . 1 0 0<br />

⎢<br />

⎥<br />

⎢<br />

⎥<br />

2 → ⎢<br />

⎣ 0 −1 −1 . −1 1 0 ⎥<br />

⎦<br />

0 1 2 . 0 0 1<br />

→ L (1)<br />

3 = L (1)<br />

2 + L (0)<br />

⎡<br />

⎤<br />

1 2 3 . 1 0 0<br />

⎢<br />

⎥<br />

⎢<br />

⎥<br />

3 = ⎢<br />

⎣ 0 −1 −1 . −1 1 0 ⎥<br />

⎦<br />

0 0 1 . −1 1 1<br />

p = 3 e n = 3 ⇐⇒ p = n.<br />

Então a inversa existe. Assim podemos prosseguir...<br />

⎡<br />

⎤<br />

1 2 3 . 1 0 0<br />

⎢<br />

⎥<br />

⎢<br />

⎥<br />

⎢<br />

⎣ 0 −1 −1 . −1 1 0 ⎥ → L(1)<br />

⎦ 1 = (−3)L<br />

0 0 1 . −1 1 1<br />

(1)<br />

3 + L (0)<br />

1 e L (2)<br />

2 = L (1)<br />

2 + L (1)<br />

3 →<br />

⎡<br />

⎤<br />

1 2 0 . 4 −3 −3<br />

⎢<br />

⎥<br />

⎢<br />

⎥<br />

⎢<br />

⎣ 0 −1 0 . −2 2 1 ⎥ → L(2)<br />

⎦ 1 = 2L<br />

0 0 1 . −1 1 1<br />

(2)<br />

2 + L (1)<br />

1 e L (3)<br />

2 = (−1)L (2)<br />

2 →

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