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

Question

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

Dr. Abubacker · Accepted Answer

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

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

Explanation

Reviewed by Dr. Abubacker

Want more practice?