Question

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

• A. $\frac{4d}{a(a+d)(a+2d)(a+3d)(a+4d)}$
• B. $\frac{4d}{a(a+d{)}^2(a+2d{)}^3(a+3d{)}^2(a+4d)}$
• C. $\frac{4d^4}{a(a+d{)}^2(a+2d{)}^3(a+3d{)}^2(a+4d)}$
• D. 0.0

Answer 1

The correct option is C $\frac{4d^4}{a(a+d{)}^2(a+2d{)}^3(a+3d{)}^2(a+4d)}$

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