Question

If the matrix M_r is given by M_r=\(\begin{pmatrix}r &r-1 \\ r-1&r \end{pmatrix}\) for r=1,2,3....then det(M₁)+det(M₂)+.......+det(M₂₀₀₈)=

• A. 2007
• B. 2008
• C. (2008)²
• D. (2007)²

Answer 1

The Correct Option is C

Given: A matrix $M_r = \begin{bmatrix} r & r - 1 \\ r - 1 & r \end{bmatrix}$ 

Step 1: Compute the determinant of $M_r$

$\det(M_r) = r \cdot r - (r - 1)^2 = r^2 - (r^2 - 2r + 1) = 2r - 1$

Step 2: Sum of determinants from $r = 1$ to $2008$

$\sum_{r=1}^{2008} \det(M_r) = \sum_{r=1}^{2008} (2r - 1)$

This is an arithmetic series with first term $1$ and last term $2 \cdot 2008 - 1 = 4015$.

Number of terms = 2008

$\sum = \frac{2008}{2} \cdot (1 + 4015) = 1004 \cdot 4016$

Now, note that: $1004 \cdot 4016 = (2008/2) \cdot (2 \cdot 2008) = \frac{2008 \cdot 4016}{2}$

$\Rightarrow \sum = \frac{2008 \cdot 4016}{2} = 2008^2$

Final Answer: Option (C): $(2008)^2$

Answer 2

Step 1: Calculate the Determinant of Mr

The determinant of a 2x2 matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is \(ad - bc\). So,

\(\det(M_r) = r \cdot r - (r-1)(r-1) = r^2 - (r^2 - 2r + 1) = r^2 - r^2 + 2r - 1 = 2r - 1\)

Step 2: Evaluate the Sum

We want to find the sum:

\(\sum_{r=1}^{2008} \det(M_r) = \sum_{r=1}^{2008} (2r - 1) = 2 \sum_{r=1}^{2008} r - \sum_{r=1}^{2008} 1\)

We know that \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\) and \(\sum_{r=1}^{n} 1 = n\). Therefore:

\(\sum_{r=1}^{2008} \det(M_r) = 2 \cdot \frac{2008(2008 + 1)}{2} - 2008\)

\(= 2008(2009) - 2008 = 2008(2009 - 1) = 2008 \cdot 2008 = (2008)^2\)

Conclusion:

The sum of the determinants is:

\(\sum_{r=1}^{2008} \det(M_r) = (2008)^2\)