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Value of the determinant $\begin{vmatrix} 10!&11!&12!\\11!&12!&13!\\12!&13!&14!\end{vmatrix}$ is

$\begin{array}{1 1}(A)\;2(10!11!)&(B)\;2(10!13!)\\(C)\;2(10!11!12!)&(D)\;2(11!12!13!)\end{array}$

Can you answer this question?

Given :
$\begin{vmatrix} 10!&11!&12!\\11!&12!&13!\\12!&13!&14!\end{vmatrix}$
$\Rightarrow \Delta=10!11!12!\begin{vmatrix}1&11&12.11\\1&12&13.12\\1&13&14.13\end{vmatrix}$
$\Rightarrow 10!11!12!\begin{vmatrix}1&11&12.11\\0&1&12.2\\0&1&13.2\end{vmatrix}$
Applying $R_2-R_1=R_2$,$R_3=R_3-R_2$
$\Rightarrow 10!11!12![1(13.2-12.2)+0(11.13.2-12.11)+0(11.12.2-12.11)]$
$\Rightarrow 10!11!12![(13.2-12.2)]$
$\Rightarrow 2(10!11!12!)$
Hence (C) is the correct answer.
answered Apr 24, 2014