Q44 / 160 · AP EAMCET 2018 (24 Apr Shift 1) · Mathematics

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

  • A
    $9 !$
  • B
    $6 !$
  • C
    $8 !$
  • Correct answer
    D
    $7 !$

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

Hence, the correct answer is option: 'D'.

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