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Question Number 81430 by abdomathmax last updated on 13/Feb/20

let the matrix A= (((1 2)),((−1 3)) ) 1) calculste A^n 2) find e^A and e^(−A) 3)find cosA and sinA

$${let}\:{the}\:{matrix}\:{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\mathrm{3}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:{calculste}\:{A}^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:{e}^{{A}} \:{and}\:{e}^{−{A}} \\ $$$$\left.\mathrm{3}\right){find}\:{cosA}\:{and}\:{sinA} \\ $$

Commented by abdomathmax last updated on 13/Feb/20

1) P_c (A) =det (A−xI)= determinant (((1−x 2)),((−1 3−x))) =(1−x)(3−x)+2 =3−x−3x+x^2 +2 =x^2 −4x +5 →Δ^′ =4−5 =−1 ⇒ λ_1 =2+i and λ_2 =2−i x^n =Q (x)P_c (x) +u_n x +v_n ⇒ A^n =u_n A +v_(n I) λ_1 ^n = u_n λ_1 +v_n and λ_2 ^n =u_n λ_2 +v_n ⇒ λ_1 ^n −λ_2 ^n =(λ_1 −λ_2 )u_n ⇒u_n =((λ_1 ^n −λ_2 ^n )/(2i)) =((2i Im(λ_1 ^n ))/(2i)) =Im(λ_1 ^n ) we have λ_1 =(√5)e^(iarctan((1/2))) ⇒ λ_1 ^n =((√5))^n e^(inarctan((1/2))) ⇒u_n =((√5))^n sin(narctan((1/2))) v_n =λ_1 ^n −λ_1 u_n =λ_1 ^n −λ_1 ×((λ_1 ^n −λ_2 ^n )/(λ_1 −λ_2 )) =((λ_1 ^(n+1) −λ_2 λ_1 ^n −λ_1 ^(n+1) +λ_1 λ_2 ^n )/(λ_1 −λ_2 )) =((λ_1 λ_2 ^n −λ_2 λ_1 ^n )/(2i)) A^n =((√5))^n sin(narctan((1/2)))A+v_n I v_n =(((2+i)(2−i)^n −(2−i)(2+i)^n )/(2i)) =Im(2+i)(2−i)^n we have 2+i=(√5)e^(iarctan((1/2))) and (2−i)^n =((√5)e^(−iarctan((1/2))) )^n =((√5))^n e^(−in arctan((1/2))) ⇒ (2+i)(2−i)^n =((√5))^(n+1) e^(−i(n−1)arctan((1/2))) ⇒v_n =−((√5))^(n+1) sin(n−1)arctan((1/2))

$$\left.\mathrm{1}\right)\:{P}_{{c}} \left({A}\right)\:={det}\:\left({A}−{xI}\right)=\begin{vmatrix}{\mathrm{1}−{x}\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\:\:\:\:\mathrm{3}−{x}}\end{vmatrix} \\ $$$$=\left(\mathrm{1}−{x}\right)\left(\mathrm{3}−{x}\right)+\mathrm{2}\:=\mathrm{3}−{x}−\mathrm{3}{x}+{x}^{\mathrm{2}} \:+\mathrm{2} \\ $$$$={x}^{\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{5}\:\rightarrow\Delta^{'} =\mathrm{4}−\mathrm{5}\:=−\mathrm{1}\:\Rightarrow \\ $$$$\lambda_{\mathrm{1}} =\mathrm{2}+{i}\:\:{and}\:\lambda_{\mathrm{2}} =\mathrm{2}−{i} \\ $$$${x}^{{n}} ={Q}\:\left({x}\right){P}_{{c}} \left({x}\right)\:+{u}_{{n}} {x}\:+{v}_{{n}} \:\Rightarrow \\ $$$${A}^{{n}} \:={u}_{{n}} \:{A}\:+{v}_{{n}\:{I}} \\ $$$$\lambda_{\mathrm{1}} ^{{n}} =\:{u}_{{n}} \lambda_{\mathrm{1}} \:+{v}_{{n}} \:{and}\:\lambda_{\mathrm{2}} ^{{n}} \:={u}_{{n}} \lambda_{\mathrm{2}} \:+{v}_{{n}} \:\Rightarrow \\ $$$$\lambda_{\mathrm{1}} ^{{n}} \:−\lambda_{\mathrm{2}} ^{{n}} \:=\left(\lambda_{\mathrm{1}} −\lambda_{\mathrm{2}} \right){u}_{{n}} \:\Rightarrow{u}_{{n}} =\frac{\lambda_{\mathrm{1}} ^{{n}} \:−\lambda_{\mathrm{2}} ^{{n}} }{\mathrm{2}{i}} \\ $$$$=\frac{\mathrm{2}{i}\:{Im}\left(\lambda_{\mathrm{1}} ^{{n}} \right)}{\mathrm{2}{i}}\:={Im}\left(\lambda_{\mathrm{1}} ^{{n}} \right)\:\:{we}\:{have}\:\lambda_{\mathrm{1}} =\sqrt{\mathrm{5}}{e}^{{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\Rightarrow \\ $$$$\lambda_{\mathrm{1}} ^{{n}} \:=\left(\sqrt{\mathrm{5}}\right)^{{n}} \:{e}^{{inarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\Rightarrow{u}_{{n}} =\left(\sqrt{\mathrm{5}}\right)^{{n}} {sin}\left({narctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${v}_{{n}} =\lambda_{\mathrm{1}} ^{{n}} −\lambda_{\mathrm{1}} {u}_{{n}} =\lambda_{\mathrm{1}} ^{{n}} −\lambda_{\mathrm{1}} ×\frac{\lambda_{\mathrm{1}} ^{{n}} \:−\lambda_{\mathrm{2}} ^{{n}} }{\lambda_{\mathrm{1}} \:−\lambda_{\mathrm{2}} } \\ $$$$=\frac{\lambda_{\mathrm{1}} ^{{n}+\mathrm{1}} −\lambda_{\mathrm{2}} \:\lambda_{\mathrm{1}} ^{{n}} \:−\lambda_{\mathrm{1}} ^{{n}+\mathrm{1}} +\lambda_{\mathrm{1}} \:\lambda_{\mathrm{2}} ^{{n}} }{\lambda_{\mathrm{1}} \:−\lambda_{\mathrm{2}} }\:=\frac{\lambda_{\mathrm{1}} \lambda_{\mathrm{2}} ^{{n}} −\lambda_{\mathrm{2}} \lambda_{\mathrm{1}} ^{{n}} }{\mathrm{2}{i}} \\ $$$${A}^{{n}} \:=\left(\sqrt{\mathrm{5}}\right)^{{n}} \:{sin}\left({narctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right){A}+{v}_{{n}} \:{I} \\ $$$${v}_{{n}} \:=\frac{\left(\mathrm{2}+{i}\right)\left(\mathrm{2}−{i}\right)^{{n}} −\left(\mathrm{2}−{i}\right)\left(\mathrm{2}+{i}\right)^{{n}} }{\mathrm{2}{i}} \\ $$$$={Im}\left(\mathrm{2}+{i}\right)\left(\mathrm{2}−{i}\right)^{{n}} \:{we}\:{have} \\ $$$$\mathrm{2}+{i}=\sqrt{\mathrm{5}}{e}^{{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} {and}\:\:\:\left(\mathrm{2}−{i}\right)^{{n}} \:=\left(\sqrt{\mathrm{5}}{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)^{{n}} \\ $$$$=\left(\sqrt{\mathrm{5}}\right)^{{n}} \:{e}^{−{in}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\Rightarrow \\ $$$$\left(\mathrm{2}+{i}\right)\left(\mathrm{2}−{i}\right)^{{n}} \:=\left(\sqrt{\mathrm{5}}\right)^{{n}+\mathrm{1}} \:{e}^{−{i}\left({n}−\mathrm{1}\right){arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\Rightarrow{v}_{{n}} =−\left(\sqrt{\mathrm{5}}\right)^{{n}+\mathrm{1}} \:{sin}\left({n}−\mathrm{1}\right){arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

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