If A_n = begin{bmatrix} 1-n & n n & 1-n end{ | vedclass

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(B) Given,$A_n = \begin{bmatrix} 1-n & n \ n & 1-n \end{bmatrix}$.The determinant $|A_n|$ is calculated as:$|A_n| = (1-n)(1-n) - (n)(n)$$|A_n| = 1 -

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$-(2021)^2$

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(B) Given,$A_n = \begin{bmatrix} 1-n & n \ n & 1-n \end{bmatrix}$.The determinant $|A_n|$ is calculated as:$|A_n| = (1-n)(1-n) - (n)(n)$$|A_n| = 1 - 2n + n^2 - n^2 = 1 - 2n$.Now,we need to find the sum $S = \sum_{n=1}^{2021} |A_n| = \sum_{n=1}^{2021} (1 - 2n)$.This can be expanded as:$S = (1-2) + (1-4) + (1-6) + \dots + (1 - 2 	imes 2021)$$S = (1 + 1 + \dots + 1) - 2(1 + 2 + 3 + \dots + 2021)$There are $2021$ terms of $1$,so the first part is $2021$.The second part is an arithmetic series sum: $2 	imes \frac{2021(2021+1)}{2} = 2021 	imes 2022$.Thus,$S = 2021 - 2021 	imes 2022$.$S = 2021(1 - 2022) = 2021(-2021) = -(2021)^2$.

If $A_n = \begin{bmatrix} 1-n & n \\ n & 1-n \end{bmatrix}$,then $|A_1| + |A_2| + \dots + |A_{2021}| = $

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