# Thread: finding value of determinants

1. ## finding value of determinants

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

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

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

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

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

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

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

$\textup{options are }-\\\textup{a)4 b)7 c)16 d)5}$
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: $T=\begin{vmatrix}
8a_{11} + 1 & 8a_{12} + 1 & 8a_{13} + 1 \\
8a_{21} + 1 & 8a_{22} + 1 & 8a_{23} + 1\\
8a_{31} + 1 & 8a_{32} + 1& 8a_{33} + 1
\end{vmatrix}= 64 \begin{vmatrix}
a_{11} - a_{21} & a_{12} - a_{22} & a_{13} - a_{23}\\
a_{21} - a_{31} & a_{22} - a_{32} & a_{23} - a_{33}\\
8a_{31} + 1 & 8a_{32} + 1& 8a_{33} + 1
\end{vmatrix}\\$
.