A) \[2007\]
B) \[2008\]
C) \[{{(2008)}^{2}}\]
D) \[{{(2007)}^{2}}\]
Correct Answer: B
Solution :
\[\det ({{M}_{r}})=\left| \begin{matrix} r & r-1 \\ r-1 & r \\ \end{matrix} \right|=2r-1\] \[\sum\limits_{r=1}^{2007}{\det }({{M}_{r}})2\sum\limits_{r=1}^{2007}{(r-2007)}\] \[=2\times \frac{2007\times 2008}{2}-2007\] \[={{(2007)}^{2}}\]You need to login to perform this action.
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