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Question: Answered & Verified by Expert
$\left|\begin{array}{lll}125 & 5 & 25 \\ 343 & 7 & 49 \\ 729 & 9 & 81\end{array}\right|=$
MathematicsDeterminantsAP EAMCETAP EAMCET 2018 (24 Apr Shift 1)
Options:
  • A $9 !$
  • B $6 !$
  • C $8 !$
  • D $7 !$
Solution:
2401 Upvotes Verified Answer
The correct answer is: $7 !$
$\left|\begin{array}{lll}125 & 5 & 25 \\ 343 & 7 & 49 \\ 729 & 9 & 81\end{array}\right|$

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 $a_{12}$
$$
\begin{aligned}
& =-5 \cdot 7 \cdot 9(96-112) \\
& =5.7 .9 .16 \\
& =7 !
\end{aligned}
$$

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