1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Linear Algebra Cramer's Rule

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve the system:
    ax1+3x2+x3= y1
    ax2= y2
    -5x1+ax2-x3=y3

    using cramer's rule for those a such that the det of A is not 0


    2. Relevant equations

    x1= det(A1)/det(A)
    det(A1)=a11+c11+a21c21+a31c31
    and A1=

    y1 3 1
    y2 a 0
    y3 a 1



    3. The attempt at a solution
    By using those equations I got the det(A1) to go ay1+(-3+a)y2-ay3
    and the det(A) was a[itex]^{}2[/itex]-5a
    and x1=(ay1+(-3+a)y2-ay3)/a[itex]^{}2[/itex]-5a

    But that was counted wrong. I don't understand where I went wrong since I have doublechecked my calculations
     
  2. jcsd
  3. Sep 26, 2011 #2

    I like Serena

    User Avatar
    Homework Helper

    Hi lina29! :smile:

    You appear to have made a few mistakes with signs.
    For starters, det(A) is not a2-5a, but -a2+5a.
    How did you calculate it?

    And your A1 should be (note the -1 in the lower right corner):
    y1 3 1
    y2 a 0
    y3 a -1
     
  4. Sep 26, 2011 #3
    I'm sorry I typed the equation wrong it was
    -5x1+ ax2+x3. Even using -a2+5a as the denominator I got it wrong
     
  5. Sep 26, 2011 #4

    I like Serena

    User Avatar
    Homework Helper

    Ah, but with your correction your det(A) is still wrong.
    Can you recalculate it?
     
  6. Sep 26, 2011 #5
    and when doing x2=(a-5)y2/(-a2+5a) I also got it counted wrong. Is there something I'm missing out?
     
  7. Sep 26, 2011 #6
    I'm still getting a2-5a. Using the equation a11c11+a12c12+a13c13.
    a11=a
    a12=3
    a13=1
    c11=a
    c12=0
    c13=-5a

    and combining them together I get a2-5a
     
  8. Sep 26, 2011 #7

    I like Serena

    User Avatar
    Homework Helper

    Yes, your det(A) is wrong, so x2 is wrong too.
    You appear to have made the same mistake calculating det(A2).
    Strangely you did calculate det(A1) properly.
     
  9. Sep 26, 2011 #8
    Could you explain to me where I'm going wrong?
     
  10. Sep 26, 2011 #9

    I like Serena

    User Avatar
    Homework Helper

    I don't know this equation for a determinant, but it is not right.

    Here's for instance the formula from wikipedia:
    b9845d679f1ad3b093e3572a8120f8d6.png
     
  11. Sep 26, 2011 #10

    I like Serena

    User Avatar
    Homework Helper

    Oh, I think your c11,c12,c13 are the determinants of the sub matrices in the lower 2 rows.
    In that case you miscalculated c13.
     
  12. Sep 26, 2011 #11
    c11, c12, and c13 are determinants of the submatrices. In class we were taught to calculate the determinant using cofactors. Using your method though I had det(A2)= (a-5)y2 -ay3 and det(A)=a2+5a
     
  13. Sep 26, 2011 #12

    I like Serena

    User Avatar
    Homework Helper

    Yep, I got that at second sight.
    See my previous post.

    So here you have the right det(A), but you still miscalculated det(A2).
     
  14. Sep 26, 2011 #13
    Got it! (a+5)y2
    Thank you so much!
     
  15. Sep 26, 2011 #14

    I like Serena

    User Avatar
    Homework Helper

    You're welcome! :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linear Algebra Cramer's Rule
Loading...