If $a_1, \ldots ,a_9$ are in harmonic progression ,then find the value of the determinant $$ \begin{vmatrix} a_1 &a_2&a_3\\ 5&4&a_6\\ a_7&a_8&a_9\\ \end{vmatrix} $$

I calculated the terms as $a_1=\frac{20}{1},a_2=\frac{20}{2},a_3=\frac{20}{3},a_4=\frac{20}{4},a_5=\frac{20}{5},a_6=\frac{20}{6},a_7=\frac{20}{7},a_8=\frac{20}{8},a_9=\frac{20}{9}$ but now calculating the determinant is tedious.Is there some other elegant method possible?

  • $\begingroup$ perform a raw or column operation $\endgroup$ – Deepesh Meena Aug 9 '18 at 15:53
  • $\begingroup$ After taking out the factor $20$ in each column I fed it to Alpha and got $1/3360$. As it is non-zero I think it is tedious. Your determinant is then $20^3/3360$ $\endgroup$ – Ross Millikan Aug 9 '18 at 15:57
  • $\begingroup$ This is a duplicate of math.stackexchange.com/questions/1755160/… which did not attract an answer either. $\endgroup$ – Ross Millikan Aug 9 '18 at 16:02

after taking $20^3$ common and performing the following row operations

replace $R_2$ with $R_2-\frac{R1}{4}$ replace $R_3$ with $R_3-\frac{R1}{7}$ I got this \begin{vmatrix} 1 & 1/2& 1/3\\ 0 & 3/40& 1/12\\ 0 & 3/56& 4/63\\ \end{vmatrix}

which didn't take much time now calculate the determinant about column 1

I got $det(A)=\frac{20^3}{3360}$


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