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TIM4_Rizka Putri

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DETERMINAN FUNGSI

Dosen Pengampu :

Ngatini, S.Si., M.Si.

Nama Kelompok :

1. Bayu Inthan Putri P. (3012010006)

2. Pradipta Shofianisa (3012010027)

3. Rizka Putri (3012010028)

UNIVERSITAS INTERNASIONAL SEMEN INDONESIA

GRESIK

2021


Dapatkan determinan dari matriks berikut dengan metode kofaktor.

Sederhanakan dengan di lakukan OBE

1 2 3 4

8 7 6 5

[

]

9 −1 −2 −3

−4 −5 −5 −4

Penyelesaian 1 :

1. Metode OBE

1 2 3 4

0 −3 −4 −3

[

]B2 = B2 + (2)B4

9 −1 −2 −3

−4 −5 −5 −4

1 2 3 4

0 −3 −4 −3

[

]B3 = B3 + (-9)B1

0 −19 −29 −39

−4 −5 −5 −4

1 2 3 4

0 −3 −4 −3

[

]B4 = B4 + (4)B1

0 −19 −29 −39

0 3 7 12

8 7 6 5

-8 -10 -10 -8

0 -3 -4 -3

9 -1 -2 -3

-9 -18 -27 -36

0 -19 -29 -39 +

-4 -5 -5 -4

4 8 12 16

0 3 7 12

+

+

2. Metode Kofaktor

+ 1 2 3 4

−3 − 4 − 3

0 −3 −4 −3

- [

] = (1) · [ −19

0 −19 −29 −39

− 29 −39]

+

3 7 12

0 3 7 12

-


Nomor 1, 2, dan 3 dengan Ekspansi Kolom Pertama

+ −3 −4 − 3

1. - [ −19 −29 −39]

3 7 12

+

+ −3 −4 − 3

2. - [ −19 −29 −39]

3 7 12

+

+ −3 −4 − 3

3. - [ −19 −29 −39]

3

+

7 12

−29 −39

= (1) (-3) |

7 12 |

= (-3) ((-29 ·12 – (-39) · 7)

= (-3) (-348 – (-273))

= (-3) (-75)

= 225

−4 −3

= (1) (19) |

7 12 |

= (19) ((-4) · 12 – (-3) · 7)

= (19) (-48 – (-21))

= (19) (-27)

= -513

−4 −3

= (1) (3) |

−29 −39 |

= (3) ((-4) · (-39) – (-3) · (-29)

= (3) (156 – 87)

= (3) (69)

= 207

1 2 3 4

8 7 6 5

det[

] = 225 – 513 + 207 = -81

9 −1 −2 −3

−4 −5 −5 −4


Penyelesaian 2 :

1 2 3 4

8 7 6 5

[

]

9 −1 −2 −3

−4 −5 −5 −4

1 2 3 4

8 7 6 5

1. [

]B3 = B3 – B2

1 −8 −8 −8

−4 −5 −5 −4

1 2 3 4

0 −3 −4 −3

2. [

]B2 = B2 +(2)B4

1 −8 −8 −8

−4 −5 −5 −4

1 2 3 4

0 −3 −4 −3

3. [

]B3 = B3 – B1

0 −10 −11 −12

−4 −5 −5 −4

4. [

1 2 3 4

0 −3 −4 −3

]B4 = B4 + (4)B1

0 −10 −11 −12

0 3 7 12

+

-

+

-

[

1 2 3 4

−3 − 4 − 3

0 −3 −4 −3

] = (1) · [ −10

0 −10 −11 −12

− 11 −12]

3 7 12

0 3 7 12


−3 −4 − 3

1. (1) · [ 0 37 84] B2 = (3)B2 + (10)B3

3 7 12

−3 −4 − 3

2. (1) · [ 0 37 84] B3 = B3 + B1

0 3 9

−1 − 4 − 1

3

3. (1) · [ 0 37 84

] B1 = (1/3)B1

0 3 9

−1 − 4 − 1

3

(1) · [ 0 37 84

]

0 3 9

+

-

+

37 84

(1) [

3 9 ]

= (1)(-1)(37·9 – 84·3)

= - (333 – 252)

= - 81

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