a −1
a +1
a+2
a −1 a +1 a a+2 F2 = F2 − F1 − 3 2 − 2a − 2 C 2 = C 2 + 2C1 a −1 a − 2 a + 2 − a 0 = F3 = F3 − F1 = = = C 4 = C 4 + 4C1 −2 0 −4 a a −1 a a−2 1 F4 = F4 − F1 3 −1 −1 −1 a+2 a a −1 a +1 a
a − 1 3a − 1 =
0 1
−3 0
3
5
8a − 1 = 7 0
5a − 2
a
3a − 1 a 5a − 2 − 2a − 2 C1 = C1 + 5C 2 = +1 ⋅ − 3 2 − 2a − 2 = = C 3 = C 3 + 11C 2 0 5 −1 11 11 −1 2 0
16a − 2 8a − 1 16a − 2 − 2a + 20 = −(− 1) ⋅ = (8a − 1) ⋅ (− 2a + 20) − 7 ⋅ (16a − 2 ) = 7 − 2a + 20 −1 0 a 2
(
)
= −16a 2 + 160a + 2a − 20 − (112a − 14 ) = −16a 2 + 50a − 6 = −2 ⋅ 8a 2 − 25a + 3
a 1 0 0 0 4 a 2 0 0 0 3 a 3 0 0 0 2 a 4 0 0 0 1 a
Observando el determinante me doy cuenta que la suma de las columnas es constante: a 1 0 0 0 4 a 2 0 0
0 3 a 3 0 = {F1 = F1 + F2 + F3 + F4 + F5 } = 0 0 2 a 4 0 0 0 1 a 1 1 1 1 1 4 a 2 0 0
C 2 C 3 ( ) = a+4 ⋅ 0 3 a 3 0 = C4 0 0 2 a 4 C 5 0 0 0 1 a
a+4 a+4 a+4 a+4 a+4 4 a 2 0 0 0 0
3 0
a 2
3 a
0 4
0
0
0
1
a
= C 2 − C1 = C 3 − C1 = (a + 4) ⋅ = C 4 − C1 = C 5 − C1
1 0 0 0 0 4 a−4 −2 −4 −4 0 0
3 0
a 2
3 a
0 = 4
0
0
0
1
a
a−4 −2 −4 −4 = (a + 4) ⋅
=
a − 4 − 2 − 4 − 4 + 4a
3 0
a 2
3 a
0 = {C 4 = C 4 − aC 3 } = (a + 4) ⋅ 4
3 0
a 2
3 a
0
0
1
a
0
0
1
− 3a = 4 − a2 0
a − 4 − 2 − 4 + 4a = (− 1) ⋅ (a + 4 ) ⋅ 3 a − 3a = (− 1) ⋅ (a + 4) ⋅ a ⋅ (4 − a ) ⋅ a 2 − 4 = 0
2
4−a
2
Sarrus
= a ⋅ (a + 4 ) ⋅ (a − 4) ⋅ (a + 2) ⋅ (a − 2 )
(
)