Question

If the matrix $M_r$ is given by $ {{M}_{r}}=\left[ \begin{matrix} r & r-1 \\ r-1 & r \\ \end{matrix} \right],r=1,2,3,...., $ then the value of $ \det ({{M}_{1}})+\det ({{M}_{2}})+....+\det ({{M}_{2008}}) $ is

• A. $2007$
• B. $2008$
• C. $ {{(2008)}^{2}} $
• D. $ {{(2007)}^{2}} $
• E. $ 2009 $

Answer 1

Answer: (C)

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}} $