Matrices and Determinants 2 Question 36
40. For a fixed positive integer $n$, if
$ D=\begin{vmatrix} n ! & (n+1) ! & (n+2) ! \\ (n+1) ! & (n+2) ! & (n+3) ! \\ (n+2) ! & (n+3) ! & (n+4) ! \end{vmatrix} $
then show that $\frac{D}{(n !)^{3}}-4$ is divisible by $n . \quad(1992,4 \mathrm{M})$
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Solution:
- Given, $ D=\begin{vmatrix} n ! & (n+1) ! & (n+2) ! \\ (n+1) ! & (n+2) ! & (n+3) ! \\ (n+2) ! & (n+3) ! & (n+4) ! \end{vmatrix} $
Taking $n$ !, $(n+1)$ ! and $(n+2)$ ! common from $R_{1}, R_{2}$ and $R_{3}$, respectively.
$ \therefore \quad D=n !(n+1) !(n+2) ! \begin{vmatrix} 1 & (n+1) & (n+1)(n+2) \\ 1 & (n+2) & (n+2)(n+3) \\ 1 & (n+3) & (n+3)(n+4) \end{vmatrix} $
Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{2}$, we get
$ D=n !(n+1) !(n+2) ! \begin{vmatrix} 1 & (n+1) & (n+1)(n+2) \\ 0 & 1 & 2 n+4 \\ 0 & 1 & 2 n+6 \end{vmatrix} $
Expanding along $C_{1}$, we get
$ \begin{aligned} & D=(n !)(n+1) !(n+2) ![(2 n+6)-(2 n+4)] \\ & D=(n !)(n+1) !(n+2) ![2] \end{aligned} $
On dividing both side by $(n !)^{3}$
$ \begin{aligned} & \Rightarrow \quad \frac{D}{(n !)^{3}}=\frac{(n !)(n !)(n+1)(n !)(n+1)(n+2) 2}{(n !)^{3}} \\ & \Rightarrow \quad \frac{D}{(n !)^{3}}=2(n+1)(n+1)(n+2) \\ & \Rightarrow \quad \frac{D}{(n !)^{3}}=2\left(n^{3}+4 n^{2}+5 n+2\right)=2 n\left(n^{2}+4 n+5\right)+4 \\ & \Rightarrow \frac{D}{(n !)^{3}}-4=2 n\left(n^{2}+4 n+5\right) \end{aligned} $
which shows that $\frac{D}{(n !)^{3}}-4$ is divisible by $n$.
