$\left|\begin{array}{lll}125 & 5 & 25 \\ 343 & 7 & 49 \\ 729 & 9 & 81\end{array}\right|=$
$\left|\begin{array}{lll}125 & 5 & 25 \\ 343 & 7 & 49 \\ 729 & 9 & 81\end{array}\right|=$
- $9 !$
- $6 !$
- $8 !$
- $7 !$
Solution
$\left|\begin{array}{lll}125 & 5 & 25 \\ 343 & 7 & 49 \\ 729 & 9 & 81\end{array}\right|$
Applying $R_1 \rightarrow \frac{1}{5} R_1, R_2 \rightarrow \frac{1}{7} R_2, R_3 \rightarrow \frac{1}{9} R_3$
$
=5 \cdot 7 \cdot 9\left|\begin{array}{ccc}
25 & 1 & 5 \\
49 & 1 & 7 \\
81 & 1 & 9
\end{array}\right|
$
Applying $R_2 \rightarrow R_2-R_1, R_3 \rightarrow R_3-R_1$
$
=5 \cdot 7 \cdot 9\left|\begin{array}{ccc}
25 & 1 & 5 \\
24 & 0 & 2 \\
56 & 0 & 4
\end{array}\right|
$
expanding for $a_{12}$
$
\begin{aligned}
& =-5 \cdot 7 \cdot 9(96-112) \\
& =5.7 .9 .16 \\
& =7 !
\end{aligned}
$
Asked in: AP EAMCET 2018 (24 Apr Shift 1)
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