1. Tardigrade
  2. Question
  3. Mathematics
  4. |125 5 25 343 7 49 729 9 81| =

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. $\begin{vmatrix}125 & 5 & 25 \\343 & 7 & 49 \\729 & 9 & 81\end{vmatrix}$ =

AP EAMCETAP EAMCET 2018

Solution:

$\begin{vmatrix}125 & 5 & 25 \\343 & 7 & 49 \\729 & 9 & 81\end{vmatrix}$
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} $
$=5 \cdot 7 \cdot 9 \begin{vmatrix}25 & 1 & 5 \\49 & 1 & 7 \\81 & 1 & 9\end{vmatrix}$
Applying $R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1} $
$=5 \cdot 7 \cdot 9 \begin{vmatrix}25 & 1 & 5 \\24 & 0 & 2 \\56 & 0 & 4\end{vmatrix}$
expanding for $a_{12}$
$=-5 \cdot 7 \cdot 9(96-112) $
$=5.7 .9 .16$
$=7 !$