Question

Let $a>0,d>0.$ Find the value of the determinant
$\left|\begin{array}{ccc}\frac{1}{a}& \frac{1}{a(a+d)}& \frac{1}{(a+d)(a+2d)}\\ \frac{1}{(a+d)}& \frac{1}{(a+d)(a+2d)}& \frac{1}{(a+2d)(a+3d)}\\ \frac{1}{(a+2d)}& \frac{1}{(a+2d)(a+3d)}& \frac{1}{(a+3d)(a+4d)}\end{array}\right|$

Answer 1

Multiplying $R_1$ by $a(a+d)(a+2d)$ to remove fraction and hence divide $\Delta$ by the same.
Similarly clean fractions in $R_2$ and $R_3$
If $k=a(a+d)(a+2d)(a+d)(a+2d)(a+3d)(a+2d)(a+3d)(a+4d)$
or $k=a(a+d)(a+2d{)}^2(a+3d{)}^2(a+4d)$
then$\Delta =\frac{1}{k}\left|\begin{array}{ccc}(a+d)(a+2d)& a+2d& a\\ (a+2d)(a+3d)& a+3d& a+d\\ (a+3d)(a+4d)& a+4d& a+2d\end{array}\right|$
Apply $C_2-C_3$ and take $2d$ common from $C_2$
$\Delta =\frac{2d}{k}\left|\begin{array}{ccc}(a+d)(a+2d)& 1& a\\ (a+2d)(a+3d)& 1& a+d\\ (a+3d)(a+4d)& 1& a+2d\end{array}\right|$
Now make two zeros by $R_3-R_2$ and $R_2-R_1$
$\Delta =\frac{2d}{k}\left|\begin{array}{ccc}(a+d)(a+2d)& 1& a\\ (a+2d).2d& 0& d\\ (a+3d).2d& 0& d\end{array}\right|$
Expanding with $2nd$ column after taking $2d$ and $d$ common
$\Delta =\frac{2d}{k}\times 2d^2\left|\begin{array}{cc}a+2d& 1\\ a+3d& 1\end{array}\right|=\frac{4d^4}{k}$
$=\frac{{4d}^4}{a{(a+d)}^2{(a+2d)}^2{(a+3d)}^2(a+4d)}$