+1 vote
36 views

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. $4$

recategorized | 36 views

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.
by (143 points)