If the matrix $M_{r}$ is given by $M_{r} = \begin{bmatrix} r & r - 1 \ r - 1 & r \end{bmatrix}, = 1,2,3,_{.}\ldots,$then the value of det $\left( M_{1} ight) + \det\left( M_{2} ight) +_{.}\ldots$+det$(M_{2008})$is

Question

If the matrix $M_{r}$ is given by $M_{r} = \begin{bmatrix} r & r - 1 \ r - 1 & r \end{bmatrix}, = 1,2,3,_{.}\ldots,$then the value of det $\left( M_{1} ight) + \det\left( M_{2} ight) +_{.}\ldots$+det$(M_{2008})$is

Dr. Abubacker · Accepted Answer

(c)
$\because\det{\left( M_{r} ight) = r^{2} - (r - 1)^{2} = 2r - 1}$
$	herefore\det{\left( M_{1} ight) + \det{(M_{2})} +}$…$+ det(M^{2008})$
$= 1 + 3 + 5 +$….+4015
$= \frac{2008}{2}$[$2 + (2008 - 1)2brack$
$= 2008(2008) = (2008)^{2}$

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

Explanation

Reviewed by Dr. Abubacker

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