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Corrigé des exercices - Dunod

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111<br />

Donc E 3 = Vect((1,3,9)).<br />

A possédant trois valeurs propres, elle est diagonalisable. On sait alors qu’en posant<br />

⎛<br />

1 1<br />

⎞<br />

1<br />

P = ⎝ 1 2 3 ⎠<br />

1 4 9<br />

⎛<br />

1 0<br />

⎞<br />

0<br />

on a P −1 AP = ⎝ 0 2 0 ⎠.<br />

0 0 3<br />

3. On inverse d’abord P en résolvant le système associé<br />

⎧<br />

⎧<br />

⎨<br />

⎨<br />

⎧<br />

⎨<br />

⇔<br />

⎩<br />

⎩<br />

x + y + z = a<br />

x + 2y + 3z = b<br />

x + 4y + 9z = c<br />

⇔<br />

⎩<br />

x + y + z = a<br />

y + 2z = −a + b<br />

3y + 8z = −a + c<br />

x + y + z = a<br />

y + 2z = −a + b<br />

⇔<br />

2z = 2a − 3b + c L 3 ← L 3 − 3L 2<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

⎧<br />

x = 3a − 5 2 ⎪⎨<br />

b + 1 2 c<br />

⇔ y = −3a + 4b − c<br />

L 2 ← L 2 − L 1<br />

L 3 ← L 3 − L 1<br />

x + y + z = a<br />

y = −3a + 4b − c<br />

z = a − 3 2 b + 1 2 c<br />

⎪⎩<br />

z = a − 3 2 b + 1 2 c<br />

D’où<br />

On en déduit que<br />

⎛<br />

1 0<br />

⎞n<br />

0<br />

A n = P ⎝ 0 2 0 ⎠ P −1<br />

0 0 3<br />

⎛ ⎞ ⎛<br />

= 1 1 1 1<br />

⎝ 1 2 3 ⎠ ⎝<br />

2<br />

1 4 9<br />

⎛<br />

⎞<br />

1 2 n 3 n<br />

1 2 n+1 3 n+1 ⎠<br />

1 2 n+2 3 n+2<br />

= 1 ⎝<br />

2<br />

⎛<br />

= 1 ⎝<br />

2<br />

⎛<br />

P −1 =<br />

⎜<br />

⎝<br />

⎞<br />

1 0 0<br />

0 2 n 0 ⎠<br />

0 0 3 n<br />

⎛<br />

⎝<br />

3 − 5 1<br />

⎞<br />

2 2<br />

−3 4 −1<br />

⎟<br />

1 − 3 ⎠<br />

1<br />

2 2<br />

⎛<br />

⎝<br />

6 −5 1<br />

−6 8 −2<br />

2 −3 1<br />

6 −5 1<br />

−6 8 −2<br />

2 −3 1<br />

⎞<br />

6 − 6 · 2 n + 2 · 3 n −5 + 8 · 2 n − 3 · 3 n 1 − 2 · 2 n + 3 n<br />

6 − 6 · 2 n+1 + 2 · 3 n+1 −5 + 8 · 2 n+1 − 3 · 3 n+1 1 − 2 · 2 n+1 + 3 n+1 ⎠<br />

6 − 6 · 2 n+2 + 2 · 3 n+2 −5 + 8 · 2 n+2 − 3 · 3 n+2 1 − 2 · 2 n+2 + 3 n+2<br />

⎞<br />

⎠<br />

⎞<br />

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