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Math Help - Are these right? (Determinants)

  1. #1
    Junior Member StonerPenguin's Avatar
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    Are these right? (Determinants)

    \left|\begin{array}{ccc}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|
    Hello all! I'd like to check these as I'm not very confident that I got them right, and I'd really appreciate your input, especially as I've scrambled my brains with trilingual madness! So much fun

    Anywho here's the questions with my answers, are they right?

    Evaluate the following determinants.
    \left|\begin{array}{ccc}-5 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 4 & 0 \end{array}\right|
    and
    3\left|\begin{array}{ccc}-2 & -1 & 0 \\ -3 & 5 & -2 \\ 0 & 8 & -1 \end{array}\right|

    My answers;
    \left|\begin{array}{ccc}5 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 4 & 0 \end{array}\right|
    =(5)(3)+(4)+(-2)-(3)-(-5)(4)(-2)=-56
    and
    3\left|\begin{array}{ccc}-2 & -1 & 0 \\ -3 & 5 & -2 \\ 0 & 8 & -1 \end{array}\right|
    =3(-2\left|\begin{array}{cc}5 & -2 \\ 8 & -1 \end{array}\right| +1\left|\begin{array}{cc}-3 & -2 \\ 0 & -1 \end{array}\right| +0\left|\begin{array}{cc}-3 & 5 \\ 0 & 8 \end{array}\right|)
    =3(-22 + 3 + 0) = -57

    And one more question;
    Expand by minors
    \left|\begin{array}{ccc}3 & 4 & -2 \\ 6 & 4 & 3 \\ 0 & 4 & 2 \end{array}\right|
    This question really bothers me considering that supposedly it doesn't matter which row or column you expand- you're supposed to get the same answer everytime, I'm getting a different answer each time.

    Example, expanding the first row;
    =3\left|\begin{array}{cc}4 & 3 \\ 4 & 2 \end{array}\right| -4\left|\begin{array}{cc}6 & 3 \\ 0 & 2 \end{array}\right| -2\left|\begin{array}{cc}6 & 4 \\ 0 & 4 \end{array}\right|
    =3(8-12)-4(12-0)-2(24-0)=-108

    And if I expand the middle column;
    Example, expanding the first row;
    =-4\left|\begin{array}{cc}6 & 3 \\ 0 & 2 \end{array}\right| +4\left|\begin{array}{cc}3 & -2 \\ 0 & 2 \end{array}\right| -4\left|\begin{array}{cc}3 & -2 \\ 6 & 3 \end{array}\right|
    =-4(12-0)+4(6-0)-4(9-12)=-12

    So... what am I doing wrong? D:

    Thanks in advance for any help
    Last edited by StonerPenguin; February 26th 2010 at 01:55 PM. Reason: Fixing my stupidity
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  2. #2
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    Hi

    Quote Originally Posted by StonerPenguin View Post
    \left|\begin{array}{ccc}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right|
    Hello all! I'd like to check these as I'm not very confident that I got them right, and I'd really appreciate your input, especially as I've scrambled my brains with trilingual madness! So much fun

    Anywho here's the questions with my answers, are they right?

    Evaluate the following determinants.
    \left|\begin{array}{ccc}5 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 4 & 0 \end{array}\right|
    and
    3\left|\begin{array}{ccc}-2 & -1 & 0 \\ -3 & 5 & -2 \\ 0 & 8 & -1 \end{array}\right|

    My answers;
    \left|\begin{array}{ccc}5 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 4 & 0 \end{array}\right|
    =(5)(3)+(4)+(-2)-(3)-(-5)(4)(-2)=-56
    No, how did you calculate it anyway?

    Have you ever heard of "rule of Sarrus". I get : det() = +40

    Quote Originally Posted by StonerPenguin View Post
    and

    3\left|\begin{array}{ccc}-2 & -1 & 0 \\ -3 & 5 & -2 \\ 0 & 8 & -1 \end{array}\right|
    =3(-2\left|\begin{array}{cc}5 & -2 \\ 8 & -1 \end{array}\right| +1\left|\begin{array}{cc}-3 & -2 \\ 0 & -1 \end{array}\right| +0\left|\begin{array}{cc}-3 & 5 \\ 0 & 8 \end{array}\right|)
    =3(-22 + 3 + 0) = -57

    This is correct, well done

    Quote Originally Posted by StonerPenguin View Post
    And one more question;
    Expand by minors
    \left|\begin{array}{ccc}3 & 4 & -2 \\ 6 & 4 & 3 \\ 0 & 4 & 2 \end{array}\right|
    This question really bothers me since supposedly it doesn't matter with row or column you expan you're supposed to get the same answer everytime and I get a different answer everytime.

    Example, expanding the first row;
    =3\left|\begin{array}{cc}4 & 3 \\ 4 & 2 \end{array}\right| -4\left|\begin{array}{cc}6 & 3 \\ 0 & 2 \end{array}\right| -2\left|\begin{array}{cc}6 & 4 \\ 0 & 4 \end{array}\right|
    =3(8-12)-4(12-0)-2(24-0)=-108
    This is correct, too

    Quote Originally Posted by StonerPenguin View Post
    And if I expand the middle column;
    Example, expanding the first row;
    =-4\left|\begin{array}{cc}6 & 3 \\ 0 & 2 \end{array}\right| +4\left|\begin{array}{cc}3 & -2 \\ 0 & 2 \end{array}\right| -4\left|\begin{array}{cc}3 & -2 \\ 6 & 3 \end{array}\right|
    =-4(12-0)+4(6-0)-4(9-12)
    No, it is -4\left|\begin{array}{cc}3 & -2 \\ 6 & 3 \end{array}\right| =  -4(9-(-12)) = -4(9+12)

    So it should be -4*21+24-48 = -108



    Quote Originally Posted by StonerPenguin View Post
    So... what am I doing wrong? D:

    Thanks in advance for any help
    By the way, nice solution!

    Rapha
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  3. #3
    Junior Member StonerPenguin's Avatar
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    Ah thank you so much! Yeah, I had a serious case of brainfart here, how embarassing O: For the first question I used a_1 b_2 c_3 + a_2 b_3 c_1 + a_3 b_1 c_2 - a_3 b_2 c_1 - a_2 b_1 c_3 - a_1 b_3 c_2
    but like an idiot I cast out some of the zeros D: e.g. (5)(3)(0)=0 not 15. However re-evaluating it, I get NEG 40, because
    (5)(3)(0) + (0)(4)(0) + (0)(0)(-2) - (0)(3)(0) - (0)(0)(0) - (-5)(4)(-2)

    So then det = - (-5)(4)(-2)= -40 Right? Or am I being stupid again?
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  4. #4
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    Quote Originally Posted by StonerPenguin View Post
    Ah thank you so much! Yeah, I had a serious case of brainfart here, how embarassing O: For the first question I used a_1 b_2 c_3 + a_2 b_3 c_1 + a_3 b_1 c_2 - a_3 b_2 c_1 - a_2 b_1 c_3 - a_1 b_3 c_2
    but like an idiot I cast out some of the zeros D: e.g. (5)(3)(0)=0 not 15. However re-evaluating it, I get NEG 40, because
    (5)(3)(0) + (0)(4)(0) + (0)(0)(-2) - (0)(3)(0) - (0)(0)(0) - (-5)(4)(-2)

    So then det = - (-5)(4)(-2)= -40 Right? Or am I being stupid again?
    It is actually
    (5)(3)(0) + (0)(4)(0) + (0)(0)(-2) - (0)(3)(0) - (0)(0)(0) - (5)(4)(-2)
    because a_1=5
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  5. #5
    Junior Member StonerPenguin's Avatar
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    Quote Originally Posted by running-gag View Post
    It is actually
    (5)(3)(0) + (0)(4)(0) + (0)(0)(-2) - (0)(3)(0) - (0)(0)(0) - (5)(4)(-2)
    because a_1=5
    Ah crap! I wrote my original question wrong! The question was \left|\begin{array}{ccc}-5 & 0 & 0 \\ 0 & 3 & -2 \\ 0 & 4 & 0 \end{array}\right| so a_1 = -5 I'm so sorry you guys, I'm usually not this bad about making mistakes D: Like I said, I think I scrambled my brains a bit, I stayed up 22 hours straight after only 4 hours of sleep (Nice excuse amirite? I'm retarded BAAWWW)

    So it is -40 then right? Thank you so much for helping my stupid ass D':
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  6. #6
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    Yes -40 is the right result
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  7. #7
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    Looks like you need to lay off that reefer, son ^_^
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