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(B) Given $A_{\lambda} = \begin{bmatrix} \lambda & \lambda - 1 \ \lambda - 1 & \lambda \end{bmatrix}$.Calculating the determinant $|A_{\lambda}| = \l

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$(300)^2$

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(B) Given $A_{\lambda} = \begin{bmatrix} \lambda & \lambda - 1 \ \lambda - 1 & \lambda \end{bmatrix}$.Calculating the determinant $|A_{\lambda}| = \lambda(\lambda) - (\lambda - 1)(\lambda - 1) = \lambda^2 - (\lambda - 1)^2$.Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $|A_{\lambda}| = (\lambda - (\lambda - 1))(\lambda + \lambda - 1) = (1)(2\lambda - 1) = 2\lambda - 1$.We need to find the sum $S = \sum_{\lambda=1}^{300} |A_{\lambda}| = \sum_{\lambda=1}^{300} (2\lambda - 1)$.This is the sum of the first $300$ odd numbers.The sum of the first $n$ odd numbers is given by $n^2$.Therefore,$S = (300)^2 = 90000$.

If $A_{\lambda} = \begin{bmatrix} \lambda & \lambda - 1 \\ \lambda - 1 & \lambda \end{bmatrix}; \lambda \in N$,then $|A_1| + |A_2| + \dots + |A_{300}|$ is equal to

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