Question

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

• A. $\Delta ={4d}^4/\left[a{(a+d)}^2{(a+2d)}^2{(a+3d)}^2\right(a+4d\left)\right]$
• B. $\Delta ={4d}^2/\left[a{(a+d)}^2{(a+2d)}^2{(a+3d)}^2\right(a+4d\left)\right]$
• C. $\Delta =-{4d}^4/\left[a{(a+d)}^2{(a+2d)}^2{(a+3d)}^2\right(a+4d\left)\right]$
• D. None of these.

Answer 1

The correct option is A $\Delta ={4d}^4/\left[a{(a+d)}^2{(a+2d)}^2{(a+3d)}^2\right(a+4d\left)\right]$

Given $a>0,d>0$
Let $△=\left|\begin{array}{ccc}\frac{1}{a}& \frac{1}{a(a+d)}& \frac{1}{(a+d)(a+2d)}\\ \frac{1}{a+d}& \frac{1}{(a+d)(a+2d)}& \frac{1}{(a+2d)(a+3d)}\\ \frac{1}{a+2d}& \frac{1}{(a+2d)(a+3d)}& \frac{1}{(a+3d)(a+4d)}\end{array}\right|$
Taking $\frac{1}{a(a+d)(a+2d)}$ common from $R_1,$

$\frac{1}{(a+d)(a+2d)(a+3d)}$ from $R_2,$

$\frac{1}{(a+2d)(a+3d)(a+4d)}$ from $R_3$

$\Rightarrow △=\frac{1}{a{(a+d)}^2{(a+2d)}^3{(a+3d)}^2(a+4d)}\left|\begin{array}{ccc}(a+d)(a+2d)& a+2d& a\\ (a+2d)(a+3d)& a+3d& a+d\\ (a+2d)(a+4d)& a+4d& a+2d\end{array}\right|$
$\Rightarrow △=\frac{1}{a{(a+d)}^2{(a+2d)}^3{(a+3d)}^2(a+4d)}{△}^1$
Where ${△}^1=\left|\begin{array}{ccc}(a+d)(a+2d)& a+2d& a\\ (a+2d)(a+3d)& a+3d& a+d\\ (a+3d)(a+4d)& a+4d& a+2d\end{array}\right|$
Applying $R_2\to R_2-R_1,R_3\to R_3-R_2$
$\Rightarrow {△}^1=\left|\begin{array}{ccc}(a+d)(a+2d)& (a+2d)& a\\ (a+2d)\left(2d\right)& d& d\\ (a+3d)\left(2d\right)& d& d\end{array}\right|$
Applying $R_3\to R_3-R_2$
$\Rightarrow {△}^1=\left|\begin{array}{ccc}(a+d)(a+2d)& (a+2d)& a\\ (a+d)2d& d& d\\ {2d}^2& 0& 0\end{array}\right|$
Expanding along $R_3$, we get
${△}^1=\left|\begin{array}{cc}a+2d& d\\ {2d}^2& 0\end{array}\right|\Rightarrow {△}^1=\left({2d}^2\right)\left(d\right)(a+2d-a)={4d}^2$
$\therefore △=\frac{{4d}^3}{a{(a+d)}^2{(a+2d)}^3{(a+4d)}^2(a+4d)}$