WBJEE · Maths · Determinants
If the matrix \(M_r\) is given by \(M_r=\left(\begin{array}{cc}r & r-1 \\ r-1 & r\end{array}\right)\) for \(r=1,2,3, \ldots\) then \(\operatorname{det}\left(\mathrm{M}_1\right)+\operatorname{det}\left(\mathrm{M}_2\right)+\ldots . .+\operatorname{det}\left(\mathrm{M}_{2008}\right)=\)
- A 2007
- B 2008
- C \((2008)^2\)
- D \((2007)^2\)
Answer & Solution
Correct Answer
(C) \((2008)^2\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {Hint : } \operatorname{det}\left(M_r\right)=r^2-(r-1)^2=r^2-\left(r^2-2 r+1\right)=2 r-1 \\ & \begin{aligned} \therefore \sum_{r=1}^{2008} \operatorname{det}\left(M_r\right)=\sum_{r=1}^{2008}(2 r-1)= & +3+5+\cdots+4015 \\ = & (2008)^2 \end{aligned}…
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