Matrices and Determinants 2 Question 34
38. Let $a>0, d>0$. Find the value of the determinant
$ \begin{vmatrix} \frac{1}{a} & \frac{1}{a(a+d)} & \frac{1}{(a+d)(a+2 d)} \\ \frac{1}{(a+d)} & \frac{1}{(a+d)(a+2 d)} & \frac{1}{(a+2 d)(a+3 d)} \\ \frac{1}{(a+2 d)} & \frac{1}{(a+2 d)(a+3 d)} & \frac{1}{(a+3 d)(a+4 d)} \end{vmatrix} $
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Answer:
Correct Answer: 38.
$\frac{4 d^{4}}{a(a+d)^{2}(a+2 d)^{3}(a+3 d)^{2}(a+4 d)}$
Solution:
- Given, $a>0, d>0$ and let
$ \Delta=\begin{vmatrix} \frac{1}{a} & \frac{1}{a(a+d)} & \frac{1}{(a+d)(a+2 d)} \\ \frac{1}{(a+d)} & \frac{1}{(a+d)(a+2 d)} & \frac{1}{(a+2 d)(a+3 d)} \\ \frac{1}{(a+2 d)} & \frac{1}{(a+2 d)(a+3 d)} & \frac{1}{(a+3 d)(a+4 d)} \end{vmatrix} $
Taking $\frac{1}{a(a+d)(a+2 d)}$ common from $R_{1}$,
$ \begin{aligned} & \frac{1}{(a+d)(a+2 d)(a+3 d)} \text { from } R_{2}, \\ & \frac{1}{(a+2 d)(a+3 d)(a+4 d)} \text { from } R_{3} \\ & \Rightarrow \Delta=\frac{1}{a(a+d)^{2}(a+2 d)^{3}(a+3 d)^{2}(a+4 d)} \\ & (a+d)(a+2 d) \quad(a+2 d) \quad a \\ & (a+2 d)(a+3 d) \quad(a+3 d) \quad(a+d) \\ & (a+3 d)(a+4 d) \quad(a+4 d) \quad(a+2 d) \end{aligned} $
$\Rightarrow \Delta=\frac{1}{a(a+d)^{2}(a+2 d)^{3}(a+3 d)^{2}(a+4 d)} \Delta^{\prime}$
where, $\Delta^{\prime}=\begin{vmatrix}(a+d)(a+2 d) & (a+2 d) & a \\ (a+2 d)(a+3 d) & (a+3 d) & (a+d) \\ (a+3 d)(a+4 d) & (a+4 d) & (a+2 d)\end{vmatrix}$
Applying $R_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{2}$
$ \Rightarrow \quad \Delta^{\prime}=\begin{vmatrix} (a+d)(a+2 d) & (a+2 d) & a \\ (a+2 d)(2 d) & d & d \\ (a+3 d)(2 d) & d & d \end{vmatrix} $
Applying $R_{3} \rightarrow R_{3}-R_{2}$
$ \Delta^{\prime}=\begin{vmatrix} (a+d)(a+2 d) & (a+2 d) & a \\ (a+2 d) 2 d & d & d \\ 2 d^{2} & 0 & 0 \end{vmatrix} $
Expanding along $R_{3}$, we get
$ \begin{aligned} & \Delta^{\prime}=2 d^{2} \begin{vmatrix} a+2 d & a \\ d & d \end{vmatrix} \\ & \Delta^{\prime}=\left(2 d^{2}\right)(d)(a+2 d-a)=4 d^{4} \\ & \therefore \quad \Delta=\frac{4 d^{4}}{a(a+d)^{2}(a+2 d)^{3}(a+3 d)^{2}(a+4 d)} \end{aligned} $
