Matrix 唔識做啊><要有步驟.thankyou

a) Expanding it, we have
-a+3b-c=1
a-b+3c=0
2a+b+4c=0
.....
2g+h+4i=0
The augmented matrix is
-1 3 -1 0 0 0 0 0 0 | 1
1 -1 3 0 0 0 0 0 0 | 0
2 1 4 0 0 0 0 0 0 | 0
0 0 0 -1 3 -1 0 0 0 | 0
0 0 0 1 -1 3 0 0 0 | 1
0 0 0 2 1 4 0 0 0 | 0
0 0 0 0 0 0 -1 3 -1 | 0
0 0 0 0 0 0 1 -1 3 | 0
0 0 0 0 0 0 2 1 4 | 1
___________________________

1 -3 1 0 0 0 0 0 0 | -1
0 2 2 0 0 0 0 0 0 | 1
0 7 2 0 0 0 0 0 0 | 2
0 0 0 1 -3 1 0 0 0 | 0
0 0 0 0 2 2 0 0 0 | 1
0 0 0 0 7 2 0 0 0 | 0
0 0 0 0 0 0 1 -3 1 | 0
0 0 0 0 0 0 0 2 2 | 0
0 0 0 0 0 0 0 7 2 | 1
___________________________

1 0 4 0 0 0 0 0 0 | -1+3/2=1/2
0 1 1 0 0 0 0 0 0 | 1/2
0 0 -5 0 0 0 0 0 0 | 2-7/2=-3/2
0 0 0 1 0 4 0 0 0 | 3/2
0 0 0 0 1 1 0 0 0 | 1/2
0 0 0 0 0 -5 0 0 0 | -7/2
0 0 0 0 0 0 1 0 4 | 0
0 0 0 0 0 0 0 1 1 | 0
0 0 0 0 0 0 0 0 -5 | 1
___________________________

1 0 0 0 0 0 0 0 0 | 1/2-12/10=-7/10
0 1 0 0 0 0 0 0 0 | 1/2-3/10=2/10
0 0 1 0 0 0 0 0 0 | 3/10
0 0 0 1 0 0 0 0 0 | 3/2-28/10=-14/10
0 0 0 0 1 0 0 0 0 | 1/2-7/10=-2/10
0 0 0 0 0 1 0 0 0 | 7/10
0 0 0 0 0 0 1 0 4 | 4/5
0 0 0 0 0 0 0 1 1 | 1/5
0 0 0 0 0 0 0 0 1 | -1/5
______________________________

so the inverse is
( -7 2 3)
1/10 (-13-2 7)
( 8 2 -2)

Note: This is actually a standard way to find the inverse of a given martix~~
but usually we perform it like this:

______________
-1 3 -1 | 1 0 0
1 -1 3 | 0 1 0
2 1 4 | 0 0 1
_______________

and then perform guassian elimination until the left hand side become identity
then the right hand side will be the inverse.

2008-07-31 19:04:12 補充：
Not enough space for me......words limitation.........

2008-07-31 19:05:00 補充：
(b)(i) As you may see, the ysytem is equal to
(-1 3 -1) (Y1) (2)
( 1 -1 3) (Y2) = (6)
( 2 1 4) (Y3) (4)

2008-07-31 19:05:27 補充：
Determinant Method should means using Crammar's rule:
| 2 3 -1 | |-1 3 -1 |
Y1 = | 6 -1 3 | / | 1 -1 3 | = 1
| 4 1 4 | | 2 1 4 |

2008-07-31 19:05:39 補充：
|-1 2 -1 | |-1 3 -1 |
Y2 = | 1 6 3 | / | 1 -1 3 | = -1
| 2 4 4 | | 2 1 4 |

2008-07-31 19:07:28 補充：
Y3 =

|-1 3 2 |
| 1 -1 6 |
| 2 1 4 |
________
|-1 3 -1 |
| 1 -1 3 |
| 2 1 4 |

=2

2008-07-31 19:07:52 補充：
(ii) the matrix method should means using the inverse you find in (a),
you will have:

(Y1) (-1 3 -1) ^ -1 (2)
(Y2) = ( 1 -1 3) (6)
(Y3) ( 2 1 4) (4)

( -7 2 3) (2)
= 1/10 (-13-2 7) (6)
( 8 2 -2) (4)

( 1)
= (-1)
( 2)