Results 1 to 2 of 2

Math Help - Cramer's Rule.

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    10

    Cramer's Rule.

    Well, it would have been nice if I'd have been taught Cramer's Rule, but our lecturer obviously doesn't work that way.

    So I've gone away and looked it up, and it seems relatively straight forward with nice values such as x=2 y=1 z=3 from the samples I've seen...

    Why don't my equations look like that?!

    V1(20+j48)-j8V2=j4000<0

    j20V1-V2(-200-j60)=-j2000<90

    Pesky j (i) numbers. '<' denotes polar angle.

    I think I'll have to break it down so that like terms read straight downwards, so in this case, there will be a 20V1 in equation 1, but there will effectively be 0V1 in equation 2. Is this thinking correct?

    Any help will be much appreciated. Thank you in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,393
    Thanks
    1327
    Quote Originally Posted by AngusBurger View Post
    Well, it would have been nice if I'd have been taught Cramer's Rule, but our lecturer obviously doesn't work that way.

    So I've gone away and looked it up, and it seems relatively straight forward with nice values such as x=2 y=1 z=3 from the samples I've seen...

    Why don't my equations look like that?!
    They do look like that. "20+ j48" is every bit as good a number as 2 or 1!

    V1(20+j48)-j8V2=j4000<0

    j20V1-V2(-200-j60)=-j2000<90

    Pesky j (i) numbers. '<' denotes polar angle.
    Are you sure about that? You can write a complex number as, say, " r<\theta" meaning r(cos(\theta)+ j sin(\theta)) but it would be strange to put that "j" in front. "4000<0" is just the real number 4000 and "2000<90" is the imaginary number 2000j. Perhaps they mean for you to multiply them by j and -j? That would give 4000j and +2000.

    In any case, Cramer's rule says that the solutions to ax+ by= e and cx+ dy= f are given by
    x= \frac{\left|\begin{array}{cc}e & b \\ f & d\end{array}\right|}{\left|\begin{array}{cc}a & b \\ c  & d\end{array}\right|}

    y=\frac{\left|\begin{array}{cc}a & f \\ c & e\end{array}\right|}{\left|\begin{array}{cc}a & b \\ c  & d\end{array}\right|}

    So, here,
    x= \frac{\left|\begin{array}{cc} 4000j & -8j \\ 2000 &-200- 60j \end{array}\right|}{\left|\begin{array}{cc}20+ 48j & -8j \\ 20j  & 200+ 60j\end{array}\right|}

    x= \frac{\left|\begin{array}{cc} 20+ 48j & 4000j \\ 20j & 2000 \end{array}\right|}{\left|\begin{array}{cc}20+ 48j & -8j \\ 20j  & 200+ 60j\end{array}\right|}

    The rest is arithmetic.
    [quote]I think I'll have to break it down so that like terms read straight downwards, so in this case, there will be a 20V1 in equation 1, but there will effectively be 0V1 in equation 2. Is this thinking correct?[quote]
    No, the coefficient of V1 in equation 2 is 20j. You cannot do the real and imaginary parts separately.

    Any help will be much appreciated. Thank you in advance.
    Last edited by Ackbeet; June 11th 2011 at 08:10 AM. Reason: Fixed LaTeX errors.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cramer's Rule Help
    Posted in the Algebra Forum
    Replies: 7
    Last Post: July 16th 2010, 12:05 PM
  2. cramer rule
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 11th 2009, 07:22 PM
  3. Cramer's rule
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: April 9th 2009, 06:04 AM
  4. Cramer's Rule
    Posted in the Algebra Forum
    Replies: 6
    Last Post: January 3rd 2009, 07:43 AM
  5. Cramer's Rule
    Posted in the Algebra Forum
    Replies: 12
    Last Post: May 30th 2006, 11:40 PM

Search Tags


/mathhelpforum @mathhelpforum