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Cramer's rule

  • Thread starter Nylex
  • Start date
  • #1
551
1
It won't work and I don't see what I'm doing wrong.

Find the solutions of the following system of linear equations:

x + 3y - z = 1
2x + y + 2z = 3
5x + z = 2

I put these into the form Ax = b, where

A = (1 3 -1)
(2 1 2 )
(5 0 1 )

x = (x)
(y)
(z)

b = (1)
(3)
(2)

I worked out det A = 20.

Cramer's rule says the solutions are given by:

x = (1/det A) | 1 3 -1 | => x = 1/10
| 3 1 2 |
| 2 0 1 |

y = (1/det A) | 1 1 -1 | => y = 9/10
| 2 3 2 |
| 5 2 1 |

z = (1/det A) | 1 3 1 | => z = 11/10
| 2 1 3 |
| 5 0 2 |

These solutions are wrong, where have I gone wrong?? Grr.

When I work out the answers algebraically, I get x = 1/5, y = 3/5 and z = 1. These are correct.
 

Answers and Replies

  • #2
3,763
8
det A is 30 not 20

marlon
 
  • #3
If you are not limited to using Cramer's Rule, I always find that just computing the cofactors of the matrix is much easier... and also a good way to check if your Cramer's Rule method is correct.
 
  • #4
551
1
marlon said:
det A is 30 not 20

marlon
ARGH, thank you.

Theelectricchild said:
If you are not limited to using Cramer's Rule, I always find that just computing the cofactors of the matrix is much easier... and also a good way to check if your Cramer's Rule method is correct.
I didn't understand cofactors :(.
 

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