Question:

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

Updated On: Apr 11, 2025
  • 2007
  • 2008
  • (2008)2
  • (2007)2
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The Correct Option is C

Approach Solution - 1

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): $(2008)^2$

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Approach Solution -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:

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

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Concepts Used:

Matrix Transformation

The numbers or functions that are kept in a matrix are termed the elements or the entries of the matrix.

Transpose Matrix:

The matrix acquired by interchanging the rows and columns of the parent matrix is termed the Transpose matrix. The definition of a transpose matrix goes as follows - “A Matrix which is devised by turning all the rows of a given matrix into columns and vice-versa.”