1. Not finding help here? Sign up for a free 30min 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!

Determinant of a 3x3 matrix via row reduction

  1. Sep 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that the determinant of
    upload_2016-9-12_18-40-40.png
    is (a-b)(b-c)(c-a)

    2. Relevant equations
    Row reduction, determinants

    3. The attempt at a solution
    upload_2016-9-12_18-41-32.png
    upload_2016-9-12_18-41-45.png

    Apparently I got a (a-b)^2 instead of (a-b) when I multiplied them up. It would be grateful if someone can point me out where the mistakes are. upload_2016-9-12_18-40-40.png upload_2016-9-12_18-41-32.png upload_2016-9-12_18-41-45.png upload_2016-9-12_18-40-40.png upload_2016-9-12_18-41-32.png upload_2016-9-12_18-41-45.png
     
  2. jcsd
  3. Sep 12, 2016 #2

    Math_QED

    User Avatar
    Homework Helper

    The problem is this. You seem to believe that when you perform a row operation: ##(a-c)R_2 - (a-b)R_3 \rightarrow R3##, the determinant remains unchanged.

    To show you an easy example that this is not true:

    ## A = \begin{pmatrix}
    1 & 2 \\
    2 & 3
    \end{pmatrix}##

    It's obvious that this matrix has a determinant equal to ##-1##

    Now perform: ##2R_1 + 3R_2 \rightarrow R_2##

    Then, we obtain a new matrix A':

    ## A' = \begin{pmatrix}
    1 & 2 \\
    8 & 13
    \end{pmatrix}##

    And this matrix has a determinant equal to ##-3##
     
    Last edited: Sep 12, 2016
  4. Sep 12, 2016 #3
    So any ideas to work it out then?
     
  5. Sep 12, 2016 #4

    Math_QED

    User Avatar
    Homework Helper

    You were on the right track. I will give you this hint:

    When you perform a row (kolom) operation:

    ##R_a + k*R_b \rightarrow R_a##, the determinant remains unchanged.
    ##l*R_a + R_b \rightarrow R _a##, the determinant is multiplied by ##l##.
    ##l*R_a + k*R_b \rightarrow R_a##, the determinant is multiplied by ##l##.

    Where ##R_a## and ##R_b## are the a'th and the b'th row and ##k,l \in \mathbb{R}##

    Now, keep in mind that you had something of the form ##l*R_a + k*R_b \rightarrow R_a##, so your determinant is multiplied with ##-(a-b)##. To make sure that the equality will still hold, multiply the determinant with the factor ##\frac{-1}{a-b}##.
     
    Last edited: Sep 12, 2016
  6. Sep 12, 2016 #5

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Why bother with row operations? Why not evaluate the determinant as it is and simplify?
     
  7. Sep 12, 2016 #6

    Math_QED

    User Avatar
    Homework Helper

    I supppose that's the exercise since it's in the title...
     
  8. Sep 12, 2016 #7

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you got this far, why not just take out the factors of ##(a-b)## and ##(c-a)## and you're nearly finished.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Determinant of a 3x3 matrix via row reduction
Loading...