A) 2007
B) 2008
C) \[{{(2008)}^{2}}\]
D) \[{{(2007)}^{2}}\]
E) \[2009\]
Correct Answer: C
Solution :
Given, \[{{M}_{r}}=\left[ \begin{matrix} r & r-1 \\ r-1 & r \\ \end{matrix} \right]\] \[\therefore \]\[\det ({{M}_{r}})={{r}^{2}}-{{(r-1)}^{2}}=2r-1\] \[\therefore \]\[\det ({{M}_{1}})+\det ({{M}_{2}})+....+\det ({{M}_{2008}})\] \[=1+3+5+...+4015\] \[=\frac{2008}{2}[2+(2008-1)2]\] \[=2008(2008)={{(2008)}^{2}}\]You need to login to perform this action.
You will be redirected in
3 sec