let-be-A-1-1-0-1-B-1-0-1-1-find-e-A-e-B-1-e-A-exponential-matrix-

Question Number 154651 by mathdanisur last updated on 20/Sep/21

let be A = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}); B = (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})

find Ω = e^{A} \cdot {(e^{B})}^{- 1}

(e^{A} - exponential matrix)

Answered by TheHoneyCat last updated on 20/Sep/21

let n \in N

(\begin{matrix} 1 & n \\ 0 & 1 \end{matrix}) \times (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) = (\begin{matrix} 1 & 1 + n \\ 0 & 1 \end{matrix})

so by a trivual recurence

\forall n \in N A^{n} = (\begin{matrix} 1 & n \\ 0 & 1 \end{matrix})

by definition e^{A} = \sum_{n \in N} \frac{1}{n!} A^{n}

so e^{A} = (\begin{matrix} e & e \\ 0 & e \end{matrix}) = e . A

and since B =^{t} A, e^{B} = e .^{t} A

hence Ω = \frac{e}{e} A {(^{t} A)}^{- 1}

= (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}) \times (\begin{matrix} 1 & 0 \\ - 1 & 1 \end{matrix})

= {(\begin{matrix} 0 & 1 \\ - 1 & 1 \end{matrix})}_{}

Commented by mathdanisur last updated on 20/Sep/21

Very nice Ser, thankyou