solve for x by cramer's rule

Kiran · 2006-12-18 06:50:53

x + 2y + x = 0
3x - y - z = 5
5x - 2y - 3z = 12

-------------------------------------------------------------------------
i already know the rules:

RULES:
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3

where the determinant D =
| a1 b1 c1 |
| a2 b2 c2 | = / 0
| a3 b3 c3 |

then

Dx =
| d1 b1 c1 |
| d2 b2 c2 |
| d3 b3 c3 |

Dy =
| a1 d1 c1 |
| a2 d2 c2 |
| a3 d3 c3 |

Dz =
| a1 b1 d1 |
| a2 b2 d2 |
| a3 b3 d3 |

x = Dx/D
y = Dy/D
z = Dz/D

---------------------------------------------------------------------
now i did ::::::>>

x + 2y + x = 0
3x - y - z = 5
5x - 2y - 3z = 12

i did but not finished please help!

| 1 2 1 |
| 3 -1 -1 | = / what ? not zero right
| 5 -2 -3 |

Dx =
| 0 2 1 |
| 5 -1 -1 |
|12 -2 -3 |

Dy =
| 1 0 1 |
| 3 5 -1 |
| 5 12 -3 |

Dz =
| 1 2 0 |
| 3 -1 5 |
| 5 -2 12|

x = Dx/D
y = Dy/D
z = Dz/D

HELP

fgarb · 2006-12-19 06:26:49

Hi, it's been a while since I've done this kind of thing, and I'm not sure what background you have, so please ask if you don't understand something in my explanation.

First off, in your post you are forming several different matrices and taking their determinants. As far as I know, there is no use for the determinants of the matrices you call Dx, Dy, and Dz, only of the original matrix of a's, b's, and c's.

There are two ways I remember that you can approach the problem. You can try to invert the matrix, or you can use elementary row operations to solve for each of the variables. I'm not sure which of these you've been exposed to. But either way you should try to understand why the determinants are important before you try to use them.

Let's look at your problem:
Solve:
x + 2y + x = 0
3x - y - z = 5
5x - 2y - 3z = 12

Hopefully you've seen matrix multiplication. It allows me to rewrite this:
( 1 2 1 ) (x) (0)
( 3 -1 -1 ) x (y) = (5)
( 5 -2 -3 ) (z) (12)

If you multiply out the matrices, this one is exactly the same as the three previous equations. So why do we care about the determinants? Because if a determinant is zero then you can multiply by the inverse on both sides. It cancels the matrix on the left, and you are left with:

(x) (0)
(y) = (inverse) * (5)
(z) (12)

If the inverse exsits, you can solve the problem by finding the inverse and plugging it into this equation. If it doesn't, you can't solve the problem no matter how you approach it. But the determinant can never be directly used to get the solution like you're trying to do.

Have you been taught how to find inverses of matrices?

basmah · 2006-12-19 06:37:43

Last edited by basmah (2006-12-19 06:44:28)

fgarb · 2006-12-19 07:00:16

Wow, I vaguely remember something about that. That's kind of cool actually - I wonder why that works. Already spent way too long messing around here today though, got to get back to work!

(please ignore my previous post )

Kiran · 2006-12-19 10:53:58

so i was close thanks

Math Is Fun Forum

#1 2006-12-18 06:50:53

solve for x by cramer's rule

#2 2006-12-19 06:26:49

Re: solve for x by cramer's rule

#3 2006-12-19 06:37:43

Re: solve for x by cramer's rule

#4 2006-12-19 07:00:16

Re: solve for x by cramer's rule

#5 2006-12-19 10:53:58

Re: solve for x by cramer's rule

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