If the matrix A_{lambda} = begin{bmatrix} lam | vedclass

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(B) The determinant of matrix $A_{\lambda}$ is given by:$|A_{\lambda}| = \begin{vmatrix} \lambda & \lambda - 1 \ \lambda - 1 & \lambda \end{vmatrix}

Vedclass · Accepted Answer

$(300)^2$

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(B) The determinant of matrix $A_{\lambda}$ is given by:$|A_{\lambda}| = \begin{vmatrix} \lambda & \lambda - 1 \ \lambda - 1 & \lambda \end{vmatrix} = \lambda^2 - (\lambda - 1)^2$$= \lambda^2 - (\lambda^2 - 2\lambda + 1) = 2\lambda - 1$We need to calculate the sum $S = \sum_{\lambda=1}^{300} |A_{\lambda}| = \sum_{\lambda=1}^{300} (2\lambda - 1)$.This is the sum of the first $300$ odd natural numbers.$S = 1 + 3 + 5 + \dots + (2(300) - 1)$$S = 1 + 3 + 5 + \dots + 599$Using the formula for the sum of an arithmetic progression $S_n = \frac{n}{2}(a + l)$:$S = \frac{300}{2}(1 + 599) = 150 	imes 600 = 90000$Since $(300)^2 = 90000$,the correct value is $(300)^2$.

If the matrix $A_{\lambda} = \begin{bmatrix} \lambda & \lambda - 1 \\ \lambda - 1 & \lambda \end{bmatrix}$,where $\lambda \in N$,then the value of $|A_1| + |A_2| + |A_3| + \dots + |A_{300}|$ is:

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