Let A = begin{bmatrix} 1 & -1 & 0 0 & 1 & -1 | vedclass

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(B) Given $A = I + C$,where $I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & -1 & 0 \ 0 & 0 & -1

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$910$

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(B) Given $A = I + C$,where $I = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & -1 & 0 \ 0 & 0 & -1 \ 0 & 0 & 0 \end{bmatrix}$.Since $I$ and $C$ commute,we use the binomial expansion $(I+C)^n = I + nC + \frac{n(n-1)}{2}C^2$.Note that $C^2 = \begin{bmatrix} 0 & 0 & 1 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$ and $C^3 = O$.Thus,$A^n = (I+C)^n = I + nC + \frac{n(n-1)}{2}C^2$.The element $b_{13}$ is the entry in the first row and third column of $B = 7A^{20} - 20A^7 + 2I$.The $(1,3)$ entry of $I$ is $0$,the $(1,3)$ entry of $C$ is $0$,and the $(1,3)$ entry of $C^2$ is $1$.Therefore,the $(1,3)$ entry of $A^n$ is $\frac{n(n-1)}{2}$.$b_{13} = 7 	imes \left( \frac{20 	imes 19}{2} ight) - 20 	imes \left( \frac{7 	imes 6}{2} ight) + 2(0)$.$b_{13} = 7 	imes 190 - 20 	imes 21 = 1330 - 420 = 910$.

Let $A = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}$ and $B = 7A^{20} - 20A^{7} + 2I$,where $I$ is an identity matrix of order $3 \times 3$. If $B = [b_{ij}]$,then $b_{13}$ is equal to $....$

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