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Question Number 141308 by BHOOPENDRA last updated on 17/May/21

Answered by Dwaipayan Shikari last updated on 17/May/21

$$\mathbf{M}=\mathbf{I}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{t^n}{n!}{\mathbf{A}}^n$$$$\mathbf{A}=\left(\begin{array}{c}-3-1\\ 83\end{array}\right)\Rightarrow {\mathbf{A}}^2=\left(\begin{array}{c}-3-1\\ 83\end{array}\right)\left(\begin{array}{c}-3-1\\ 83\end{array}\right)=\left(\begin{array}{c}10\\ 01\end{array}\right)=\mathbf{I}$$$${\mathbf{A}}^3=\mathbf{IA}$$$$\mathbf{M}=\mathbf{I}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{t^{2n}}{\left(2n\right)!}{\mathbf{A}}^{2n}+\underset{n=0w}{\sum ^{\mathrm{\infty }}}\frac{t^{2n+1}}{(2n+1)!}{\mathbf{A}}^{2n+1}{\mathbf{A}}^{2n}=\mathbf{I}$$$$=\mathbf{I}+\mathbf{I}(\cosh \left(t\right)-1)+\mathbf{A}\left(\sinh \left(t\right)\right)$$$$=\mathbf{I}\cosh \left(t\right)+\mathbf{A}\sinh \left(t\right)$$$$$$

Commented by BHOOPENDRA last updated on 17/May/21

$$txsirbpartplease$$

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