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Math Help - finding value of determinants

  1. #1
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    finding value of determinants

    \\\textup{ If } A_i=1,2,3,4..9 \textup{ are perfect odd squares }\\

    T=\begin{vmatrix}<br />
  A_1& A_2 &A_3\\<br />
  A_4 & A_5 &A_6\\<br />
A_7&A_8&A_9<br />
\end{vmatrix}\\

    \textup{then T is always a multiple of-}\\

    \textup{options are }-\\\textup{a)4  b)7  c)16  d)5}
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  2. #2
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    Multiple choice questions can be answered by trial and error.

    Let's use the squares 1 and 9 which are "perfect odd squares". Form the
    determinant

    9 1 1
    1 9 1
    1 1 9

    and evaluate. I get the determinant is 704, which factors to 64 x 11. That eliminates all but choices 'a' and 'c'. Certainly if the determinant is a multiple of 16 it is also a multiple of 4. So 'a' seems safe.

    Clearly this isn't a proof and assumes the problem relies on a true fact.
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  3. #3
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    Quote Originally Posted by banku12 View Post
    \\\textup{ If } A_i=1,2,3,4..9 \textup{ are perfect odd squares }\\

    T=\begin{vmatrix}<br />
A_1& A_2 &A_3\\<br />
A_4 & A_5 &A_6\\<br />
A_7&A_8&A_9<br />
\end{vmatrix}\\

    \textup{then T is always a multiple of-}\\

    \textup{options are }-\\\textup{a)4 b)7 c)16 d)5}
    you certainly meant A_i, \ i=1,2, \cdots , 9, not A_i=1,2, \cdots , 9. anyway, the answer is 64, which is not any of the options. so you choose 16. the proof is quite easy:

    every perfect odd square is in the form 8k+1, for some integer k. thus: T=\begin{vmatrix}<br />
8a_{11} + 1 & 8a_{12} + 1 & 8a_{13} + 1 \\<br />
8a_{21} + 1 & 8a_{22} + 1 & 8a_{23} + 1\\<br />
8a_{31} + 1 & 8a_{32} + 1& 8a_{33} + 1<br />
\end{vmatrix}= 64 \begin{vmatrix}<br />
a_{11} - a_{21} & a_{12} - a_{22} & a_{13} - a_{23}\\<br />
a_{21} - a_{31} & a_{22} - a_{32} & a_{23} - a_{33}\\<br />
8a_{31} + 1 & 8a_{32} + 1& 8a_{33} + 1<br />
\end{vmatrix}\\.
    Last edited by NonCommAlg; April 3rd 2010 at 09:20 PM.
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