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let-A-1-2-1-1-1-calculste-A-n-2-determine-e-A-and-e-2A-3-find-cos-A-and-sinA-is-cos-2-A-sin-2-A-I-4-determine-sh-A-and-ch-A-is-ch-2-A-sh-2-A-I-

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Question Number 91268 by mathmax by abdo last updated on 29/Apr/20
let A = (((1 2)),((1 −1)) ) 1) calculste A^n 2) determine e^A and e^(−2A) 3)find cos(A)and sinA is cos^2 A +sin^2 A =I? 4) determine sh(A) and ch(A) is ch^2 A−sh^2 A =I ?
$${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:−\mathrm{1}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:{calculste}\:{A}^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:\:{e}^{{A}} \:\:{and}\:{e}^{−\mathrm{2}{A}} \\ $$$$\left.\mathrm{3}\right){find}\:{cos}\left({A}\right){and}\:{sinA}\:\:\:{is}\:{cos}^{\mathrm{2}} {A}\:+{sin}^{\mathrm{2}} {A}\:={I}? \\ $$$$\left.\mathrm{4}\right)\:{determine}\:{sh}\left({A}\right)\:{and}\:{ch}\left({A}\right)\:\:{is}\:{ch}^{\mathrm{2}} {A}−{sh}^{\mathrm{2}} {A}\:={I}\:\:? \\ $$
Commented by mathmax by abdo last updated on 29/Apr/20
1) the caracteristic polynom is p(x)=det (A−xI) = determinant (((1−x 2)),((1 −1−x)))=−(1+x)(1−x)−2 =−(1−x^2 )−2 =−1+x^2 −2 =x^2 −3 =(x−(√3))(x+(√3)) p(x)=0 ⇔x =+^− (√3) x^n =Qp(x)+u_n x +v_n ⇒((√3))^n =u_n (√3)+v_n (−(√3))^n =−(√3)u_n +v_n ⇒((√3))^n −(−(√3))^n =2(√3)u_n ⇒ u_n =((((√3))^n −(−(√3))^n )/(2(√3))) and ((√3))^n +(−(√3))^n =2v_n ⇒ v_n =((((√3))^n +(−(√3))^n )/2) A^n =Q(A)p(A)+u_n A +v_n I =u_n A +v_n I (P(A)=0 by cayley hamiton) ⇒A^n =((((√3))^n −(−(√3))^n )/(2(√3))) (((1 2)),((1 −1)) ) +((((√3))^n +(−(√3))^n )/2) (((1 0)),((0 1)) ) A^n = (((((((√3))^n −(−(√3))^n )/(2(√3))) ((((√3))^n −(−(√3))^n )/( (√3))))),((((((√3))^n −(−(√3))^n )/(2(√3))) (((−(√3))^n −((√3))^n )/(2(√3))))) ) + (((((((√3))^n +(−(√3))^n )/2) 0)),((0 ((((√3))^n +(−(√3))^n )/2))) ) = (((((((√3))^n −(−(√3))^n +((√3))((√3))^n +(√3)(−(√3))^n )/(2(√3))) ((((√3))^n −(−(√3))^n )/( (√3))))),((((((√3))^n −(−(√3))^n )/(2(√3))) (((−(√3))^n −((√3))^n +(√3)((√3))^n +(√3)(−(√3))^n )/(2(√3))))) ) A^n = ((((((1+(√3))((√3))^n +((√3)−1)(−(√3))^n )/(2(√3))) ((((√3))^n −(−(√3))^n )/( (√3))))),((((((√3))^n −(−(√3))^n )/(2(√3))) ((((√3)−1)((√3))^n +(1+(√3))(−(√3))^n )/(2(√3))))) )
$$\left.\mathrm{1}\right)\:{the}\:{caracteristic}\:{polynom}\:{is}\:{p}\left({x}\right)={det}\:\left({A}−{xI}\right) \\ $$$$=\begin{vmatrix}{\mathrm{1}−{x}\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}−{x}}\end{vmatrix}=−\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}\right)−\mathrm{2}\:=−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)−\mathrm{2} \\ $$$$=−\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}\:={x}^{\mathrm{2}} −\mathrm{3}\:=\left({x}−\sqrt{\mathrm{3}}\right)\left({x}+\sqrt{\mathrm{3}}\right) \\ $$$${p}\left({x}\right)=\mathrm{0}\:\Leftrightarrow{x}\:=\overset{−} {+}\sqrt{\mathrm{3}} \\ $$$${x}^{{n}} \:={Qp}\left({x}\right)+{u}_{{n}} {x}\:+{v}_{{n}} \:\Rightarrow\left(\sqrt{\mathrm{3}}\right)^{{n}} \:={u}_{{n}} \sqrt{\mathrm{3}}+{v}_{{n}} \\ $$$$\left(−\sqrt{\mathrm{3}}\right)^{{n}} \:=−\sqrt{\mathrm{3}}{u}_{{n}} \:+{v}_{{n}} \:\Rightarrow\left(\sqrt{\mathrm{3}}\right)^{{n}} −\left(−\sqrt{\mathrm{3}}\right)^{{n}} \:=\mathrm{2}\sqrt{\mathrm{3}}{u}_{{n}} \:\Rightarrow \\ $$$${u}_{{n}} =\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} −\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:{and}\:\:\left(\sqrt{\mathrm{3}}\right)^{{n}} \:+\left(−\sqrt{\mathrm{3}}\right)^{{n}} \:=\mathrm{2}{v}_{{n}} \:\Rightarrow \\ $$$${v}_{{n}} =\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} \:+\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}} \\ $$$${A}^{{n}} \:={Q}\left({A}\right){p}\left({A}\right)+{u}_{{n}} \:{A}\:+{v}_{{n}} {I}\:\:\:\:={u}_{{n}} {A}\:+{v}_{{n}} {I}\:\:\:\left({P}\left({A}\right)=\mathrm{0}\:{by}\:{cayley}\:{hamiton}\right) \\ $$$$\Rightarrow{A}^{{n}} \:=\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} −\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{1}\:\:\:\:\:\:\:\:−\mathrm{1}}\end{pmatrix}\:+\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} +\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${A}^{{n}} \:=\begin{pmatrix}{\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} −\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} −\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\:\sqrt{\mathrm{3}}}}\\{\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} −\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\left(−\sqrt{\mathrm{3}}\right)^{{n}} −\left(\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}}}\end{pmatrix} \\ $$$$+\begin{pmatrix}{\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} \:+\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} +\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} −\left(−\sqrt{\mathrm{3}}\right)^{{n}} \:+\left(\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{3}}\right)^{{n}} +\sqrt{\mathrm{3}}\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} −\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\:\sqrt{\mathrm{3}}}}\\{\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} −\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\left(−\sqrt{\mathrm{3}}\right)^{{n}} −\left(\sqrt{\mathrm{3}}\right)^{{n}} \:+\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}\right)^{{n}} +\sqrt{\mathrm{3}}\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}}}\end{pmatrix} \\ $$$${A}^{{n}} \:=\begin{pmatrix}{\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{3}}\right)^{{n}} +\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} −\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\:\sqrt{\mathrm{3}}}}\\{\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} −\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{3}}\right)^{{n}} +\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\left(−\sqrt{\mathrm{3}}\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}}}\end{pmatrix} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 29/Apr/20
A^(2p) = ((((((1+(√3))3^p +((√3)−1)3^p )/(2(√3))) 0)),((0 ((((√3)−1)3^p +(1+(√3))3^p )/(2(√3))))) ) = (((3^p 0)),((0 3^p )) ) A^(2p+1) = ((((((1+(√3))(√3)3^p +((√3)−1)(−(√3))3^p )/(2(√3))) (((√3)3^p −(−(√3))3^p )/( (√3))))),(((((√3)3^p −(−(√3))3^p )/(2(√3))) ((((√3)−1)(√3)3^p +(1+(√3))(−(√3))3^p )/(2(√3))))) ) = (((3^p 2.3^p )),((2.3^p −3^p )) )
$${A}^{\mathrm{2}{p}} \:=\:\:\begin{pmatrix}{\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\mathrm{3}^{{p}} +\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\mathrm{3}^{{p}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\mathrm{3}^{{p}} \:+\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\mathrm{3}^{{p}} }{\mathrm{2}\sqrt{\mathrm{3}}}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\mathrm{3}^{{p}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}^{{p}} }\end{pmatrix} \\ $$$${A}^{\mathrm{2}{p}+\mathrm{1}} \:\:=\begin{pmatrix}{\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{3}}\mathrm{3}^{{p}} \:+\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(−\sqrt{\mathrm{3}}\right)\mathrm{3}^{{p}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{3}}\mathrm{3}^{{p}} \:−\left(−\sqrt{\mathrm{3}}\right)\mathrm{3}^{{p}} }{\:\sqrt{\mathrm{3}}}}\\{\frac{\sqrt{\mathrm{3}}\mathrm{3}^{{p}} −\left(−\sqrt{\mathrm{3}}\right)\mathrm{3}^{{p}} }{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\sqrt{\mathrm{3}}\mathrm{3}^{{p}} +\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\left(−\sqrt{\mathrm{3}}\right)\mathrm{3}^{{p}} }{\mathrm{2}\sqrt{\mathrm{3}}}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\mathrm{3}^{{p}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}.\mathrm{3}^{{p}} }\\{\mathrm{2}.\mathrm{3}^{{p}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{3}^{{p}} }\end{pmatrix} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 29/Apr/20
sorry A^(2p+1) = (((3^p 2 .3^p )),((3^p −3^p )) )
$${sorry}\:\:{A}^{\mathrm{2}{p}+\mathrm{1}} =\begin{pmatrix}{\mathrm{3}^{{p}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:.\mathrm{3}^{{p}} }\\{\mathrm{3}^{{p}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{3}^{{p}} }\end{pmatrix} \\ $$
Commented by mathmax by abdo last updated on 29/Apr/20
2) e^A =Σ_(n=0) ^∞ (A^n /(n!)) =Σ_(n=0) ^∞ (A^(2n) /((2n)!)) +Σ_(n=0) ^∞ (A^(2n+1) /((2n+1)!)) =Σ_(n=0) ^∞ (1/((2n)!)) (((3^n 0)),((0 3^n )) ) +Σ_(n=0) ^∞ (1/((2n+1)!)) (((3^n 23^n )),((3^n −3^n )) ) = (((Σ_(n=0) ^∞ (3^n /((2n)!)) 0)),((0 Σ_(n=0) ^∞ (3^n /((2n)!)))) ) + (((Σ_(n=0) ^∞ (3^n /((2n+1)!)) 2Σ_(n=0) ^∞ (3^n /((2n+1)!)))),((Σ_(n=0) ^∞ (3^n /((2n+1)!)) −Σ_(n=0) ^∞ (3^n /((2n+1)!)))) )
$$\left.\mathrm{2}\right)\:{e}^{{A}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{A}^{{n}} }{{n}!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{A}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{A}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)!}\begin{pmatrix}{\mathrm{3}^{{n}} \:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}^{{n}} }\end{pmatrix}\:\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\begin{pmatrix}{\mathrm{3}^{{n}} \:\:\:\:\:\:\:\:\:\:\mathrm{23}^{{n}} }\\{\mathrm{3}^{{n}} \:\:\:\:\:\:\:\:\:\:−\mathrm{3}^{{n}} }\end{pmatrix} \\ $$$$=\begin{pmatrix}{\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{{n}} }{\left(\mathrm{2}{n}\right)!}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{{n}} }{\left(\mathrm{2}{n}\right)!}}\end{pmatrix}\:+\begin{pmatrix}{\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\:\:\:\:\:\:\:\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}}\\{\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:\:\:\:\:\:\:\:\:\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}}\end{pmatrix} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 29/Apr/20
we have Σ_(n=0) ^∞ (3^n /((2n)!)) =Σ_(n=0) ^∞ ((((√3))^(2n) )/((2n)!)) =cos((√3)) Σ_(n=0) ^∞ (3^n /((2n+1)!)) =(1/( (√3)))Σ_(n=0) ^∞ ((((√3))^(2n+1) )/((2n+1)!)) =(1/( (√3)))sin((√3)) ⇒ e^A = (((cos((√3)) 0)),((0 cos((√3)))) ) + ((((1/( (√3)))sin((√3)) (2/( (√3)))sin((√3)))),(((1/( (√3)))sin((√3)) −(1/( (√3)))sin((√3)))) ) = (((cos((√3))+(1/( (√3)))sin((√3)) (2/( (√3)))sin((√3)))),(((1/( (√3)))sin((√3)) cos((√3))−(1/( (√3) ))sin((√3)))) )
$${we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{{n}} }{\left(\mathrm{2}{n}\right)!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\:={cos}\left(\sqrt{\mathrm{3}}\right) \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{sin}\left(\sqrt{\mathrm{3}}\right)\:\Rightarrow \\ $$$${e}^{{A}} \:=\begin{pmatrix}{{cos}\left(\sqrt{\mathrm{3}}\right)\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cos}\left(\sqrt{\mathrm{3}}\right)}\end{pmatrix}\:\:+\begin{pmatrix}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{sin}\left(\sqrt{\mathrm{3}}\right)\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{sin}\left(\sqrt{\mathrm{3}}\right)}\\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{sin}\left(\sqrt{\mathrm{3}}\right)\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{sin}\left(\sqrt{\mathrm{3}}\right)}\end{pmatrix} \\ $$$$=\begin{pmatrix}{{cos}\left(\sqrt{\mathrm{3}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{sin}\left(\sqrt{\mathrm{3}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{sin}\left(\sqrt{\mathrm{3}}\right)}\\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}{sin}\left(\sqrt{\mathrm{3}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cos}\left(\sqrt{\mathrm{3}}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:}{sin}\left(\sqrt{\mathrm{3}}\right)}\end{pmatrix} \\ $$

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