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 )

(

)