Question

$\left|\begin{array}{ccc}125& 5& 25\\ 343& 7& 49\\ 729& 9& 81\end{array}\right|$ =

• A. 9!
• B. 6!
• C. 8!
• D. 7!

Answer 1

Answer: (D)

$\left|\begin{array}{ccc}125& 5& 25\\ 343& 7& 49\\ 729& 9& 81\end{array}\right|$
Applying $R_1\to \frac{1}{5}{R}_1,R_2\to \frac{1}{7}{R}_2,R_3\to \frac{1}{9}{R}_3$
$=5\cdot 7\cdot 9\left|\begin{array}{ccc}25& 1& 5\\ 49& 1& 7\\ 81& 1& 9\end{array}\right|$
Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$
$=5\cdot 7\cdot 9\left|\begin{array}{ccc}25& 1& 5\\ 24& 0& 2\\ 56& 0& 4\end{array}\right|$
expanding for $a_{12}$
$=-5\cdot 7\cdot 9(96-112)$
$=\mathrm{5.7.9.16}$
$=7!$