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Math Help - getting wrong answer for determinant

  1. #1
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    Still need help! getting very confused: getting wrong answer for determinant

    I am trying to calculate the area of a triangle given the points (3,3) (-1,-1) (4,1).

    I am having trouble calculating the determinant
    <br />
\left[\begin{matrix}<br />
3 & 3 & 1\\<br />
-1 & -1 & 1\\<br />
4 & 1 & 1<br />
\end{matrix} \right]<br />
=-<br />
\left[\begin{matrix}<br />
-1 & -1 & 1\\<br />
3 & 3 & 1\\<br />
4 & 1 & 1<br />
\end{matrix} \right]<br />
=<br />
\left[\begin{matrix}<br />
1 & 1 & -1\\<br />
3 & 3 & 1\\<br />
4 & 1 & 1<br />
\end{matrix} \right]<br />
=<br />
\left[\begin{matrix}<br />
1 & 1 & -1\\<br />
0 & 0 & 4\\<br />
0 & -3 & 5<br />
\end{matrix} \right]<br />

    since there is a 0 in the very middle of the last determinant, the product along the diagonal is 0. The answer key says I should get 6 but I can't find where I'm going wrong
    Last edited by superdude; February 26th 2010 at 05:24 PM.
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  2. #2
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    Did you manage 12? That's excellent. Now add the final factor of 1/2 and you are done.
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  3. #3
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    I don't follow
    1\times 0\times 5=0 \neq 12

    this is seriously starting to bother me, why is it whenever I use this method I get a differnt answer than if I do cofactor expansion? I thought a matrix only has one determinant?
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  4. #4
    Member mybrohshi5's Avatar
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    I know this probably doesnt help but using a calculator to find the determinant of that matrix it says the answer is 12 and you said the answer on the key is 6 so maybe there is an error somewhere and thats why its not working out....
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  5. #5
    Newbie Nissplus's Avatar
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    You were on the right track with your original question, you just need a few more steps.

    The product of the diagonal elements of a matrix equals the determinant of a triangular matrix. Exchanging rows 2 and 3 in your last matrix and multiplying by the requisite -1 will give you an upper triangular matrix and the product of the diagonal (and the determinant) will equal 12.

    Then you must divide by 2. Not for the determinant (which is 12), but because the determinant calculates the area of a parallelogram one half of which is your triangle. These two wikipedia entries explain:
    Triangular matrix - Wikipedia, the free encyclopedia
    Triple product - Wikipedia, the free encyclopedia
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  6. #6
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    Quote Originally Posted by superdude View Post
    I don't follow
    1\times 0\times 5=0 \neq 12

    this is seriously starting to bother me, why is it whenever I use this method I get a differnt answer than if I do cofactor expansion? I thought a matrix only has one determinant?
    A matrix has only one determinant. You must do it correctly.

    From the beginning, the determinant can be calculated in six pieces.

    -3 - 1 + 12 + 4 - 3 + 3 = -4 + 16 + 0 = 12

    Your row operations were impeccable. The same six steps from that form:

    0 + 0 + 0 - 0 + 12 - 0 = 12

    Using expansion my minors across the second row of your reduced form:

    -4*(-3-0) = 12

    Using expansion my minors down the first column of your reduced form:

    1*(12) = 12

    I think, if you do another row operation, you will see that your triangular form was not yet accomplished. Swap rows 2 and 3 and think about it again.
    Last edited by TKHunny; February 27th 2010 at 06:47 PM. Reason: Right Word
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  7. #7
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    Quote Originally Posted by superdude View Post
    I am trying to calculate the area of a triangle given the points (3,3) (-1,-1) (4,1).

    I am having trouble calculating the determinant
    <br />
\left[\begin{matrix}<br />
3 & 3 & 1\\<br />
-1 & -1 & 1\\<br />
4 & 1 & 1<br />
\end{matrix} \right]<br />
=-<br />
\left[\begin{matrix}<br />
-1 & -1 & 1\\<br />
3 & 3 & 1\\<br />
4 & 1 & 1<br />
\end{matrix} \right]<br />
=<br />
\left[\begin{matrix}<br />
1 & 1 & -1\\<br />
3 & 3 & 1\\<br />
4 & 1 & 1<br />
\end{matrix} \right]<br />
=<br />
\left[\begin{matrix}<br />
1 & 1 & -1\\<br />
0 & 0 & 4\\<br />
0 & -3 & 5<br />
\end{matrix} \right]<br />

    since there is a 0 in the very middle of the last determinant, the product along the diagonal is 0. The answer key says I should get 6 but I can't find where I'm going wrong
    The product along the diagonal is NOT the determinant because this is not yet a triangular matrix. On more row operation, swapping the second and third rows, gives
    -\begin{bmatrix}1 & 1& -1 \\0 & -3 & 5 \\ 0 & 0 & 4\end{bmatrix}
    which is triangular and has determinant -(1)(-3)(4)= 12.
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  8. #8
    Senior Member
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    thanks guy I see where I went wrong
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