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Determinan

det A =

-1 -1 1

-1 2 2

2 -1 -2

=

= (-1) ·2 2

-1 -2- (-1) ·

-1 2

2 -2+ 1 ·

-1 2

2 -1=

 = -3

Soal : K | -1 -1 1 |

  | -1 2 2 |

  | 2 -1 -2 |

 Jawab : invers

Solution:

To find the inverse matrix write the matrix А and added to it on the right identity matrix:

  -1 -1 1 1 0 0

-1 2 2 0 1 0

2 -1 -2 0 0 1

devide the 1-th row by -1

  1 1 -1 -1 0 0

-1 2 2 0 1 02 -1 -2 0 0 1

from 2 3 rows we s!btra"t the 1-th row# m!$ti%$ied res%e"tive$y by -1 2

  1 1 -1 -1 0 0

0 3 1 -1 1 0

0 -3 0 2 0 1

devide the 2-th row by 3

  1 1 -1 -1 0 0

0 1 1&3 -1&3 1&3 00 -3 0 2 0 1

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from 1 3 rows we s!btra"t the 2-th row# m!$ti%$ied res%e"tive$y by 1 -3

  1 0 -'&3 -2&3 -1&3 0

0 1 1&3 -1&3 1&3 0

0 0 1 1 1 1

from 1 2 rows we s!btra"t the 3-th row# m!$ti%$ied res%e"tive$y by -'&3 1&3

  1 0 0 2&3 1 '&3

0 1 0 -2&3 0 -1&3

0 0 1 1 1 1

Answer:

A-1 =

  2&3 1 '&3

-2&3 0 -1&3

1 1 1

Metode Kofaktor :

k  =

 Jawab :

Solution:

A=

  -1 -1 1

-1 2 2

2 -1 -2

inding the determinant of the matrix А

det A = -3

The determinant of A is non-ero# hen"e the inverse matrix A-1 exists* To find the inverse

matrix was "a$"!$ated the minors and "ofa"tors of the matrix А

• inding the minor 1#1 and "ofa"tor ,1#1* The matrix A  "ross o!t the row 1 and

"o$!mn 1*

1#1 =

2 2

-1 -2 = -2

-1 -1 1

-1 2 2

2 -1 -2

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•

1#1 = (-1)1+11#1 = -2

•

• inding the minor 1#2 and "ofa"tor ,1#2* The matrix A  "ross o!t the row 1 and

"o$!mn 2*

1#2 =-1 2

2 -2= -2

•

1#2 = (-1)1+21#2 = 2

•

• inding the minor 1#3 and "ofa"tor ,1#3* The matrix A  "ross o!t the row 1 and

"o$!mn 3*

1#3 =-1 2

2 -1

= -3

•

1#3 = (-1)1+31#3 = -3

•

• inding the minor 2#1 and "ofa"tor ,2#1* The matrix A  "ross o!t the row 2 and

"o$!mn 1*

2#1 =-1 1

-1 -2= 3

•

2#1 = (-1)2+12#1 = -3

•

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• inding the minor 2#2 and "ofa"tor ,2#2* The matrix A  "ross o!t the row 2 and

"o$!mn 2*

2#2 =-1 1

2 -2= 0

•

2#2 = (-1)2+22#2 = 0

•

• inding the minor 2#3 and "ofa"tor ,2#3* The matrix A  "ross o!t the row 2 and

"o$!mn 3*

2#3 =-1 -1

2 -1= 3

•

2#3 = (-1)2+32#3 = -3

•

• inding the minor 3#1 and "ofa"tor ,3#1* The matrix A  "ross o!t the row 3 and

"o$!mn 1*

3#1 =-1 1

2 2= -'

•

3#1 = (-1)3+13#1 = -'

•

• inding the minor 3#2 and "ofa"tor ,3#2* The matrix A  "ross o!t the row 3 and

"o$!mn 2*

3#2 =-1 1

-1 2= -1

•

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3#2 = (-1)3+23#2 = 1

•

• inding the minor 3#3 and "ofa"tor ,3#3* The matrix A  "ross o!t the row 3 and

"o$!mn 3*

3#3 =-1 -1

-1 2= -3

•

3#3 = (-1)3+33#3 = -3

•

atrix of "ofa"tors:

C =

  -2 2 -3

-3 0 -3

-' 1 -3

Trans%ose of "ofa"tor matrix:

CT =

  -2 -3 -'

2 0 1

-3 -3 -3

inding the inverse matrix*

A-1 =

CT

 =det A

  2&3 1 '&3

-2&3 0 -1&3

1 1 1Metode Gauss jordan

Divide row1 by -1

1 1 -1

-1 2 2

2 -1 -2

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 Add (1 * row1) to row2

1 1 -1

0 3 1

2 -1 -2

 Add (-2 * row1) to row3

1 1 -1

0 3 1

0 -3 0

Divide row2 by 3

1 1 -1

0 1 1/3

0 -3 0

 Add (3 * row2) to row3

1 1 -1

0 1 1/3

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0 0 1

 Add (-1/3 * row3) to row2

1 1 -1

0 1 0

0 0 1

 Add (1 * row3) to row1

1 1 0

0 1 0

0 0 1

 Add (-1 * row2) to row1

1 0 0

0 1 0

0 0 1

Metodekaidah sarrus :

-x + .y + 2 = 3

2x + 3y + 3 = -2

/ 2y = .

=

-1 . 2

2 3 3

1 -2 -1

=

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1 =

3 . 2

-2 3 3

. -2 -1

= .2

2 =

-1 3 2

2 -2 3

1 . -1

= .2

3 =

-1 . 3

2 3 -2

1 -2 .

= -2

1 =

1 =

.2 =

13

2

2 =

2 =

.2 =

13

2

3 =

3 =

-2 = -

23

2

Gauss jordan

Solusi:

en!$is !$ang sistem %ersamaan da$am bent!4 matri4s dan menye$esai4annya dengan

metode 5a!ss-6ordan

  -1 . 2 3

2 3 3 -2

1 -2 -1 .

7agi$ah dengan 1-th -1

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  1 -. -2 -3

2 3 3 -2

1 -2 -1 .

dari 2 3 baris s!straigamos 1 baris# masing-masing di4a$i4an dengan 2 1

  1 -. -2 -3

0 13 8 '

0 3 1

7agi$ah dengan 2-13 th

  1 -. -2 -3

0 1 8-13 '&13

0 3 1

dari 1 3 baris 2 s!straigamos baris di4a$i4an masing-masing sebesar -. 3

  1 0 -13 -1&13

0 1 8-13 '&13

0 0 -&13 2&13

7agi$ah dengan 3-yo th -&13

  1 0 -13 -1&13

0 1 8-13 '&13

0 0 1 -11#.

dari 1 2 baris s!straigamos baris 3 di4a$i4an dengan &13 masing-masing 8-13

  1 0 0 9*.0 1 0 9*.

0 0 1 -11#.

Hasil:

x 1 = 9*.

x 2 = 9*.

x 3 = -11#.

2w ! ! " ! #$ % 2

#w ! & "- % 1

2w & 2 & " ! ' $ % (

(w & & 2" -#$ % -2

 Jawab:

Metode )auss jordan

Solusi:

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en!$is !$ang sistem %ersamaan da$am bent!4 matri4s dan menye$esai4annya dengan

metode 5a!ss-6ordan

  2 1 1 3 2

3 1 -1 0 1

2 -2 -1 9 '

' -1 -2 -3 -2

ivide 1-th 2

  1 0#. 0#. 1*. 1

3 1 -1 0 1

2 -2 -1 9 '

' -1 -2 -3 -2

dari 2 3 ' baris s!straigamos 1 baris# masing-masing di4a$i4an dengan 3 2 '

  1 0#. 0#. 1*. 10 -0*. -2*. -'*. -2

0 -3 -2 3 2

0 -3 -' - -9

7agi$ah dengan -0*. 2-th

  1 0#. 0#. 1*. 1

0 1 . '

0 -3 -2 3 2

0 -3 -' - -9

dari 1 3 ' baris 2 s!straigamos baris di4a$i4an masing-masing sebesar 0#. -3 -3

  1 0 -2 -3 -1

0 1 . '

0 0 13 30 1'

0 0 11 1 9

7agi$ah 3 o$eh 13 th-;

  1 0 -2 -3 -1

0 1 . '

0 0 1 30&13 1'&130 0 11 1 9

dari 1 2 ' baris s!straigamos baris 3 di4a$i4an masing-masing o$eh -2 ei 11

  1 0 0 21&13 1.&13

0 1 0 -33&13 -1&13

0 0 1 30&13 1'&13

0 0 0 -9&13 -89&13

7agi$ah dengan '-yo th -9&13

  1 0 0 21&13 1.&130 1 0 -33&13 -1&13

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  0 0 1 30&13 1'&13

0 0 0 1 1-2'

dari 1 2 3 baris s!straigamos ' baris# di4a$i4an dengan 21&13 masing-masing -33&13 30&13  1 0 0 0 -0#12.

0 1 0 0 0*92.

0 0 1 0 -0*8.

0 0 0 1 1-2'

Hasil:

x 1 = -0#12.

x 2 = 0*92.

x 3 = -0*8.

x ' = 1&2'

Metode kaidah sarrus

< =

2 1 1 3

3 1 -1 0

2 -2 -1 9

' -1 -2 -3

= 9

< 1  =

2 1 1 3

1 1 -1 0

' -2 -1 9

-2 -1 -2 -3

= -12

< 2  =

2 2 1 3

3 1 -1 0

2 ' -1 9

' -2 -2 -3

= 90

< 3  =

2 1 2 3

3 1 1 0

2 -2 ' 9

' -1 -2 -3

= -82

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< '  =

2 1 1 2

3 1 -1 1

2 -2 -1 '' -1 -2 -2

= 89

1  =

< 1 =

-12 = -

1

< 9

2  =

< 2 =

90 =

.

< 9

3  =

< 3 =

-82 = -

3

'< 9

'  =

< ' =

89 =

1

2'< 9