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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. $4$
in Linear Algebra by Boss (17.5k points)
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1 Answer

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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.
by (143 points)

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