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Use determinant property to solve evaluate matrice

  1. Nov 11, 2009 #1
    Use determinant property to evaluate matrice determinant

    1. The problem statement, all variables and given/known data
    3x3:
    A =
    12 27 12
    28 18 24
    70 15 40


    What determinant property can I use to find the determinate of this matrix? I cannot see any relationship between the rows or columns. Apparently 720 comes into it somehow, but I cannot see!

    Thanks :)
    Thomas
     
    Last edited: Nov 11, 2009
  2. jcsd
  3. Nov 11, 2009 #2

    Mark44

    Staff: Mentor

    What exactly is the problem?
     
  4. Nov 11, 2009 #3
    edited it. sorry about that. I wrote it in notepad, but copied wrong version!
     
  5. Nov 11, 2009 #4

    Mark44

    Staff: Mentor

    You can make the numbers smaller and easier to work with by pulling out common factors. 3 is a factor of all the numbers in the first row, 2 is a common factor in the 2nd row, and 5 is a common factor in the 3rd row. You should have a theorem that tells you what happens to the determinant of a matrix if you replace a row by some multiple of itself.
     
  6. Nov 11, 2009 #5
    I know that if a column or row is a multiple of another then it's det is 0. If you take a multiple of one row or coloum from another, then the det is still the same.

    So I've found multiples in rows and columns, what should i do with them?

    Tom
    Thanks
     
  7. Nov 11, 2009 #6

    Mark44

    Staff: Mentor

    You are referring to two separate theorems (or two parts of one theorem). There's another theorem (or part of one) that you haven't mentioned, and it has to do with replacing a row or column by a multiple of itself.

    Also, can you go back and edit your first post? There is a lone \right tex tag that's causing the page to be really huge.
     
  8. Nov 12, 2009 #7
    If the elemnts of one row (or one column) of a determinant are multiplied by k, then the value of the resulting determinants is k times the given determinant

    in the answers, an intermeddiate step is to somehow get

    |2 3 1|
    720 |7 3 3|
    |7 1 2|

    but I can't see common factors of each row they've taken out!
     
  9. Nov 12, 2009 #8

    Mark44

    Staff: Mentor

    From the first matrix, take out a factor of 3 from the first row, a factor of 2 from the second row, and a factor of 5 from the third row. Then your first determinant is equal to 3*2*5 times this determinant:
    |4 9 4|
    |14 9 12|
    |14 3 8|

    Now take out a factor of 2 from the first column, a factor of 3 from the second column, and a factor of 4 from the third column. So your first determinant equals 30 times the determinant above, which equals what?
     
  10. Nov 12, 2009 #9
    that's gives 720! yeah! I got the impression from my work book that you could only take out a factor of all, didn't know you could take out factors of multiple lines and multiply!

    right, I could find the determinate from that, but the answers seem to simplify it even more

    Can you see how (it involves another rule). After looking at the answers, I would still find it quite that hard to spot

    Thanks alot!
    Tom
     
  11. Nov 12, 2009 #10
    the answer seems to use [tex](-2C_{3} + C_{1})[/tex] then [tex](-3C_{3} + C_{2})[/tex]

    Does that mean -2 lots of coloum 3 from all the columns in the matrix and then plus the new column 1 or the old? Then on the next set, does that mean -3 lots of the new column 3 + the last generation of column 2!?

    confusing!

    How do I spot something clever like that, experience or are there tricks?

    Thanks!
    Tom
     
  12. Nov 12, 2009 #11

    Mark44

    Staff: Mentor

    I don't understand what you are saying: "-2 lots of coloum 3"

    I didn't add any row to another row, so what you're asking seems unrelated to what I did. All I did was pull common factors from all members of a row or column.

    Basically what I did was this:
    |ka kb kc|
    |e f g|
    |h i j |
    =k*
    |a b c|
    |e f g|
    |h i j|
     
  13. Nov 12, 2009 #12
    beg your pardon. I see how you did that, but to get the factors of the matrix rows and columns is only part of the way to the answer. In my answer booklet, they have perform ed the operations I listed in my previous post to get to

    720x13 = 9360

    the 720 comes from what you've shown me to do with the factors, but how you know that using [tex]
    (-2C_{3} + C_{1})
    [/tex] then [tex]
    (-3C_{3} + C_{2})
    [/tex], is a shortcut route. They get

    | 2 3 1 |
    720 | 7 3 3 |
    | 7 1 2 |

    to
    | 0 0 1 |
    720 | 1 -6 3 |
    | 3 -5 2 |

    this allows you only to have to do one calculation of a minor to get determinate. How would I go about spotting this?

    Thanks
     
  14. Nov 12, 2009 #13

    Mark44

    Staff: Mentor

    BTW, what your are calculating is the determinant.

    Starting from
    | 2 3 1 |
    | 7 3 3 |
    | 7 1 2 |

    the -2C3 + C1 means add -2 times column 3 to 1 times column 1. That gives

    | 0 3 1 |
    | 1 3 3 |
    | 3 1 2 |

    Then -3C3 + C2 means add -3 times column 3 to 1 times column 2. That gives

    | 0 0 1 |
    | 1 -6 3 |
    | 3 -5 2 |

    The operation of adding a multiple of a row (column) to another row (column) doesn't change the value of the determinant.

    With the operations that you did before pulling out factors, what we have is
    720*
    |2 3 1|
    |7 3 3|
    |7 1 2|

    = 720*
    | 0 0 1 |
    | 1 -6 3 |
    | 3 -5 2 |

    = 720 * (-5 + 18) = 720*13=9360
     
  15. Nov 12, 2009 #14
    Thank you very very very much!
    I now understand what they've done and will do some more practice questions to gain experience now. Hopefully this method will come in very useful!

    Thanks again for your time
    Thomas
     
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