# Thread: finding value of determinants

1. ## finding value of determinants

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

$\displaystyle T=\begin{vmatrix} A_1& A_2 &A_3\\ A_4 & A_5 &A_6\\ A_7&A_8&A_9 \end{vmatrix}\\$

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

$\displaystyle \textup{options are }-\\\textup{a)4 b)7 c)16 d)5}$

2. 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.

3. Originally Posted by banku12
$\displaystyle \\\textup{ If } A_i=1,2,3,4..9 \textup{ are perfect odd squares }\\$

$\displaystyle T=\begin{vmatrix} A_1& A_2 &A_3\\ A_4 & A_5 &A_6\\ A_7&A_8&A_9 \end{vmatrix}\\$

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

$\displaystyle \textup{options are }-\\\textup{a)4 b)7 c)16 d)5}$
you certainly meant $\displaystyle A_i, \ i=1,2, \cdots , 9,$ not $\displaystyle A_i=1,2, \cdots , 9.$ 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 $\displaystyle 8k+1,$ for some integer $\displaystyle k.$ thus: $\displaystyle T=\begin{vmatrix} 8a_{11} + 1 & 8a_{12} + 1 & 8a_{13} + 1 \\ 8a_{21} + 1 & 8a_{22} + 1 & 8a_{23} + 1\\ 8a_{31} + 1 & 8a_{32} + 1& 8a_{33} + 1 \end{vmatrix}= 64 \begin{vmatrix} a_{11} - a_{21} & a_{12} - a_{22} & a_{13} - a_{23}\\ a_{21} - a_{31} & a_{22} - a_{32} & a_{23} - a_{33}\\ 8a_{31} + 1 & 8a_{32} + 1& 8a_{33} + 1 \end{vmatrix}\\$.