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Cramer's rule |
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| Apr15-05, 07:34 AM | #1 |
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Cramer's rule
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. |
| Apr15-05, 07:41 AM | #2 |
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det A is 30 not 20
marlon |
| Apr16-05, 01:33 AM | #3 |
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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.
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| Apr17-05, 02:19 AM | #4 |
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Cramer's rule |
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