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): $(2008)^2$
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:
\(\sum_{r=1}^{2008} \det(M_r) = (2008)^2\)
If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]
The numbers or functions that are kept in a matrix are termed the elements or the entries of the 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.”