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 = ((1,1),(0,1) ) ; B = ((1,0),(1,1) ) find 𝛀 = e^A ∙ (e^B )^(−1) (e^A - exponential matrix)

$$\mathrm{let}\:\mathrm{be}\:\:\boldsymbol{\mathrm{A}}\:=\:\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}\:\:;\:\:\boldsymbol{\mathrm{B}}\:=\:\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{find}\:\:\boldsymbol{\Omega}\:=\:\mathrm{e}^{\boldsymbol{\mathrm{A}}} \:\centerdot\:\left(\mathrm{e}^{\boldsymbol{\mathrm{B}}} \right)^{−\mathrm{1}} \\ $$$$\left(\mathrm{e}^{\boldsymbol{\mathrm{A}}} \:-\:\mathrm{exponential}\:\mathrm{matrix}\right) \\ $$

Answered by TheHoneyCat last updated on 20/Sep/21

let n∈N ((1,n),(0,1) )× ((1,1),(0,1) )= ((1,(1+n)),(0,1) ) so by a trivual recurence ∀n∈N A^n = ((1,n),(0,1) ) by definition e^A =Σ_(n∈N) (1/(n!))A^n so e^A = ((e,e),(0,e) )=e.A and since B=^t A, e^B =e.^t A hence 𝛀=(e/e)A(^t A)^(−1) = ((1,1),(0,1) )× ((1,0),((−1),1) ) = ((0,1),((−1),1) )_■

$$\mathrm{let}\:{n}\in\mathbb{N} \\ $$$$\begin{pmatrix}{\mathrm{1}}&{{n}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}×\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}+{n}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{so}\:\mathrm{by}\:\mathrm{a}\:\mathrm{trivual}\:\mathrm{recurence} \\ $$$$\forall{n}\in\mathbb{N}\:\boldsymbol{\mathrm{A}}^{{n}} =\begin{pmatrix}{\mathrm{1}}&{{n}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\mathrm{by}\:\mathrm{definition}\:{e}^{\boldsymbol{\mathrm{A}}} =\underset{{n}\in\mathbb{N}} {\sum}\frac{\mathrm{1}}{{n}!}\boldsymbol{\mathrm{A}}^{{n}} \\ $$$$\mathrm{so}\:{e}^{\boldsymbol{\mathrm{A}}} =\begin{pmatrix}{{e}}&{{e}}\\{\mathrm{0}}&{{e}}\end{pmatrix}={e}.\boldsymbol{\mathrm{A}} \\ $$$$\mathrm{and}\:\mathrm{since}\:\boldsymbol{\mathrm{B}}=^{{t}} \boldsymbol{\mathrm{A}},\:{e}^{\boldsymbol{\mathrm{B}}} ={e}.^{{t}} \boldsymbol{\mathrm{A}} \\ $$$$\mathrm{hence}\:\boldsymbol{\Omega}=\frac{{e}}{{e}}\boldsymbol{\mathrm{A}}\left(^{{t}} \boldsymbol{\mathrm{A}}\right)^{−\mathrm{1}} \\ $$$$=\begin{pmatrix}{\mathrm{1}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}×\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{−\mathrm{1}}&{\mathrm{1}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\mathrm{0}}&{\mathrm{1}}\\{−\mathrm{1}}&{\mathrm{1}}\end{pmatrix}_{\blacksquare} \\ $$

Commented by mathdanisur last updated on 20/Sep/21

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{Ser},\:\mathrm{thankyou} \\ $$

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