# ISI2017-DCG-7

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If  $\begin{vmatrix} 10! & 11! & 12! \\ 11! & 12! & 13! \\ 12! & 13! & 14! \end{vmatrix} = k(10!)(11!)(12!)$, then the value of $k$ is

1. $1$
2. $2$
3. $3$
4. $4$

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take 10! common from first row, 11! common from second row and 12! common from the third row.

now apply elementary operation R3 --> R3-R2 after that R2-->R2-R1.now get the value from the newly formed determinant.  after solving the determinant we will get determinant= 2. by comparing we can say that k=2.

$\begin{vmatrix} 10! &11! &12! \\ 11! & 12! & 13!\\ 12! &13! & 14! \end{vmatrix}$

$\begin{vmatrix} 10! &11.10! &12.11.10! \\ 11! & 12.11! & 13.12.11!\\ 12! &13.12! & 14.13.12! \end{vmatrix}$

Taking 10! common from R1, 11! from R2, 12! from R3.

(10!)(11!)(12!) $\begin{vmatrix} 1 &11 &12.11 \\ 1 & 12 & 13.12\\ 1 &13 & 14.13 \end{vmatrix}$

R2→ R2 – R1

R3→ R3 – R2

(10!)(11!)(12!) $\begin{vmatrix} 1 &11 &12.11 \\ 0 & 1 & 24\\ 0 &1& 26 \end{vmatrix}$

R3→ R3 – R2

(10!)(11!)(12!)$\begin{vmatrix} 1 &11 &12.11 \\ 0 & 1 & 24\\ 0 &0& 2 \end{vmatrix}$

2(10!)(11!)(12!). Therefore, K=2.

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