Results 1 to 8 of 8

Math Help - Cramer's Rule Help

  1. #1
    Junior Member
    Joined
    Jul 2010
    Posts
    55

    Cramer's Rule Help

    Okay so i've spent a good amount of time between yesterday evening and today trying to figure out how to solve this problem using cramer's rule:

    x + y +z = 0
    7x- y +z=-10
    -x+ 4y-z=-20

    I heard that Cramer's rule isn't the best way to do this with this many equations and the simple add, substitution and elimination way would be somewhat better but even that way i've gotten the wrong solution. I've tried to eliminate X first by multiplying the first equation by -7 to get -8y-6z=-10 and then i changed the thrid equation to get 27y-6z=-150. I can't remember my exact answer because i erased the rest of it but i solved for z first once i combined them and got a small number and after i finished back substitution everything else i got 2 big numbers that didn't come out as the solution to each of the three equation. I was wondering if anyone could make it clear to me how i should go about solving this problem. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jun 2009
    Posts
    68
    Quote Originally Posted by MajorJohnson View Post
    I was wondering if anyone could make it clear to me how i should go about solving this problem. Thanks.
    do an rref on a matrix of the form
    [1, 1, 1, 0]
    [7, -1, 1, -10]
    [-1, 4, -1, -20]
    it will output a matrix like so
    [1, 0, 0, X]
    [0, 1, 0, Y]
    [0, 0, 1, Z]
    where X, Y, and Z are the answers
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    769
    "I heard that Cramer's rule isn't the best way to do this with this many equations"

    The better way is to use Gaussian elimination to reduce the matrix where the diagonal is all one's and the members below the diagonal are all zeros.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2010
    Posts
    55
    Quote Originally Posted by 1005 View Post
    do an rref on a matrix of the form
    [1, 1, 1, 0]
    [7, -1, 1, -10]
    [-1, 4, -1, -20]
    it will output a matrix like so
    [1, 0, 0, X]
    [0, 1, 0, Y]
    [0, 0, 1, Z]

    where X, Y, and Z are the answers
    i don't get where the extra 0's in the output matrix are coming from
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member eumyang's Avatar
    Joined
    Jan 2010
    Posts
    278
    Thanks
    1
    First, find the determinant of the coefficient matrix:
    \begin{vmatrix}<br />
1 & 1 & 1 \\<br />
7 & -1 & 1 \\<br />
-1 & 4 & -1<br />
\end{vmatrix} = 30

    Set up Cramer's Rule:
    x = \frac {\begin{vmatrix}<br />
0 & 1 & 1 \\<br />
-10 & -1 & 1 \\<br />
-20 & 4 & -1<br />
\end{vmatrix}}{30} =  \frac{-90}{30} = -3

    y = \frac{\begin{vmatrix}<br />
1 & 0 & 1 \\<br />
7 & -10 & 1 \\<br />
-1  & -20 & -1<br />
\end{vmatrix}}{30} = \frac{-120}{30} = -4

    z = \frac{\begin{vmatrix}<br />
1 & 1 & 0 \\<br />
7 & -1 & -10 \\<br />
-1  & 4 & -20<br />
\end{vmatrix}}{30} = \frac{210}{30} = 7

    Hopefully you can see what I did. The numerators are all determinants of the coefficient matrix with one column replaced by the numbers on the RHS. You replace the 1st column for x, the 2nd column for y, and the 3rd column for z. Yes, Cramer's Rules isn't the best here, especially if you're required to find the determinant of 3x3 matrices by hand. TT_TT
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jun 2009
    Posts
    68
    Quote Originally Posted by MajorJohnson View Post
    i don't get where the extra 0's in the output matrix are coming from
    Are you in a matrix algebra class? Unless you're supposed to know where it comes from, just shrug it off and use its useful results. All hail the calculator. If you really want to know the process (it's related to Gaussian elimination, I believe), google reduced row echelon form(rref)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member eumyang's Avatar
    Joined
    Jan 2010
    Posts
    278
    Thanks
    1
    If you REALLY want to solve by elimination,
    x + y +z = 0
    7x- y +z=-10
    -x+ 4y-z=-20

    Multiply Eq. 1 by -7 and add to Eq. 2:
    -7x - 7y - 7z = 0
    7x - y + z = -10
    ----------------
    -8y - 6z = -10 [Eq. 4]

    Add Eq. 1 and Eq. 3:
    Hey! x AND z get eliminated!
    x + y +z = 0
    -x+ 4y-z=-20
    --------------
    5y = -20
    y = -4

    Substitute into Eq. 4:
    -8(-4) - 6z = -10
    32 - 6z = -10
    -6z = -42
    z = 7

    Substitute into Eq. 1:
    x - 4 + 7 = 0
    x + 3 = 0
    x = -3
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jul 2010
    Posts
    55
    ok i understand what you did, i messed up and got 28 for the determinate instead of 30. I must have arranged my ascending numbers on the wrong side from my decending ones. Thanks.

    Quote Originally Posted by 1005 View Post
    Are you in a matrix algebra class? Unless you're supposed to know where it comes from, just shrug it off and use its useful results. All hail the calculator. If you really want to know the process (it's related to Gaussian elimination, I believe), google reduced row echelon form(rref)
    I'm in a college algebra class, i just learned Cramer's rule yesterday, so i guess she'll teach me the other way at some point idk cause i'm not familar with that way but i might search it later on. Eumyang's method worked for me though i just made some mistakes on getting the determinate that's what though off my answer. Thanks for the input though.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cramer's Rule
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 8th 2011, 02:56 AM
  2. Cramer's Rule.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: June 11th 2011, 07:41 AM
  3. Cramer's Rule
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: August 8th 2009, 05:52 AM
  4. Cramer's rule
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: April 9th 2009, 06:04 AM
  5. Cramer's Rule Help
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 3rd 2007, 06:12 AM

Search Tags


/mathhelpforum @mathhelpforum