MATEMATIKA 1 skripta studij: Biotehnologija i Prehrambena ... - PBF

A 2,1 = (−1) 3 ∣ ∣∣∣∣
0 1
3 1
A 3,1 = (−1) 4 ∣ ∣∣∣∣
0 1
0 2
Odavdje je
A −1 = 1 −6
∣ ∣ ∣∣∣∣ ∣ = 3, A 2,2 = (−1) 4 1 1
∣∣∣∣ −1 1 ∣ = 2, A 2,3 = (−1) 5 1 0
−1 3 ∣ = −3,
∣ ∣ ∣∣∣∣ ∣ = 0, A 3,2 = (−1) 5 1 1
∣∣∣∣ 0 2 ∣ = −2, A 3,3 = (−1) 6 1 0
0 0 ∣ = 0.
⎡
⎢
⎣
−6 −2 0
3 2 −3
0 −2 0
⎤
⎥
⎦
T
=
⎡
⎢
⎣
1 −1/2 0
1/3 −1/3 1/3
0 1/2 0
b)
⎡ ⎤ ⎡ ⎤ ⎡ ⎤
1 0 1 1 0 0 1 0 1 1 0 0 1 0 1 1 0 0
[A|I] = ⎢
⎣ 0 0 2 0 1 0 ⎥
⎦ ∼ ⎢
⎣ 0 0 2 0 1 0 ⎥
⎦ ∼ ⎢
⎣ 0 3 2 1 0 1 ⎥
⎦
−1 3 1 0 0 1 0 3 2 1 0 1 0 0 2 0 1 0
⎡ ⎤ ⎡
⎤ ⎡
⎤
1 0 1 1 0 0 1 0 1 1 0 0 1 0 0 1 −1/2 0
∼ ⎢
⎣ 0 3 0 1 −1 1 ⎥
⎦ ∼ ⎢
⎣ 0 1 0 1/3 −1/3 1/3 ⎥
⎦ ∼ ⎢
⎣ 0 1 0 1/3 −1/3 1/3 ⎥
⎦ .
0 0 2 0 1 0 0 0 1 0 1/2 0 0 0 1 0 1/2 0
Odavde čitamo:
A −1 =
⎡
⎢
⎣
1 −1/2 0
1/3 −1/3 1/3
0 1/2 0
⎤
⎥
⎦ .
⎤
⎥
⎦ .
Zadatak 20⎡Za koje vrijednosti ⎤ parametra k ∈ R postoji inverzna matrica
2 −k 3
matrice A = ⎢
⎣ −2 1 −3 ⎥
⎦ .
2 7 5
Rješenje: Kako je
detA =
∣
2 −k 3
−2 1 −3
2 7 5
=
∣ ∣
2 −k 3
0 1−k 0
0 7+k 2
= 4(1−k),
∣
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