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Math Help - [SOLVED] Are determinants always positive?

  1. #1
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    [SOLVED] Are determinants always positive?

    I got the matrix A

    3 2 -1
    1 6 3
    2 -4 0

    I'm not sure if determinants are always positive, but they must be because in my textbook, the determinant is 64.

    I got this calculating the minors of row 1. I got 36+12+16. For row 2 I got -4-12-48 and row 3 I got 24-40+0.

    This leads me to believe dets are always positive, but I'm not sure. What are the conditions for the sign of determinants?
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  2. #2
    o_O
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    No matter which row you expand, you should always get the same determinant. So you must've made a mistake when expanding across row 2.

    As for matrices in general, of course you can have a non-positive determinant. Consider the matrix:

    \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
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  3. #3
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    I was going to give \left|\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right| as an example having a negative determinant.

    And, of course, \left|\begin{array}{cc}1 & 1 \\ 1 & 1\end{array}\right| has determinant 0.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by o_O View Post
    No matter which row you expand, you should always get the same determinant. So you must've made a mistake when expanding across row 2.

    As for matrices in general, of course you can have a non-positive determinant. Consider the matrix:

    \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
    well, \begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix} was a simpler example

    @OP: personally I would have expanded along the 3rd column
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  5. #5
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    I ended up with determinant = 64

    \begin{bmatrix} 3 & 2 & -1 \\ 1 & 6 & 3 \\ 2 & -4 & 0 \end{bmatrix}

    [(3 \times 6 \times 0)+(2\times3\times2)+(-1\times-4\times1)] - [(2\times6\times-1)-(-4\times3\times3)-(0\times2\times1)]

     16+48 =64

    By the way, determinants are not the identical to absolute values and they can be negative. If you end up with a negative that simply means that the shape is left-handed rather than right-handed.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Roam View Post
    I ended up with determinant = 64

    \begin{bmatrix} 3 & 2 & -1 \\ 1 & 6 & 3 \\ 2 & -4 & 0 \end{bmatrix}

    [(3 \times 6 \times 0)+(2\times3\times2)+(-1\times-4\times1)] - [(2\times6\times-1)-(-4\times3\times3)-(0\times2\times1)]

     16+48 =64

    By the way, determinants are not the identical to absolute values and they can be negative. If you end up with a negative that simply means that the shape is left-handed rather than right-handed.
    yes, i got +64 also.

    and no one said they had to be positive, we all gave examples of negative determinants. the straight bars are notation: \text{det} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix}
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  7. #7
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    Expanding row 2:

    (-1)^2+1 * 1 det (2*0)-(-1*-4) = 4

    +

    (-1)^2+2 * 6 det (3*0)-(-1*2) = 12

    +

    (-1)^2+3 * 3 det (3*-4)-(2*2) = 48

    Okay I get 64 as well. I understand my mistake. Originally I did not pay attention to the sign and sign changes depending on the location of the minors 4, 12, 48. So originally I had -4, -12 and -48 because I forgot to change the sign, or apply the correct ij in the (-1)^ij part.
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