Solving an inverse matrice

neon_glows · 2005-05-25 15:38:28

This the what i believe is the answer, would appreciate if some one cud verify it to be right or wrong. 3x + 2y = 16 x + 4y = 22 |3 2| | x | = | 16 | |1 4| | y | = | 22 | A X = B x = A-1 B 1/ad-bc = 1/(3)(4)-(2)(1) = 1/9 1/9 |4 -2| | x | = | 16 | |-1 3| | y | = | 22 | = 1/9 |64 - 44| |-16 + 66| = 1/9 (20) = 2.2 (50) = 5.5…

25-05-2005 04:38PM

#1

neon_glows

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This the what i believe is the answer, would appreciate if some one cud verify it to be right or wrong.

3x + 2y = 16
x + 4y = 22

|3 2| | x | = | 16 |
|1 4| | y | = | 22 |
A X = B

x = A-1 B

1/ad-bc = 1/(3)(4)-(2)(1) = 1/9

1/9 |4 -2| | x | = | 16 |
|-1 3| | y | = | 22 |

=

1/9 |64 - 44|
|-16 + 66|

= 1/9 (20) = 2.2
(50) = 5.5

1/9

0

Comments

#2 25-05-2005 06:31PM

D-Generate

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Join Date: April 2000

Posts: 2859

The detirminent is 1/10

1/(ad-bc)
= 1/[(4.3)-(2.1)]
= 1/[12-2]
= 1/10

Rest is all right final line should be

1/10 [20]
[50]

Giving you
x= 2 and y = 5

0