Question Number 164227 by SANOGO last updated on 15/Jan/22
Answered by TheSupreme last updated on 15/Jan/22
![determinant ((4,6,0),((−3),(−5),0),((−3),(−6),(−5)))−λ [(1,0,0),(0,1,0),(0,0,1) ]{ determinant ((x),(y),(z))}={ determinant ((0),(0),(0))} det(A−λI)=0 [(4−λ)(−5−λ)+18](−5−λ)=0 [−20−4λ+5λ+λ^2 +18](−5−λ)=0 (λ^2 +λ−2)(−5−λ)=0 −(λ+5)(λ−1)(λ+2)=0 {x}=c_1 {w_1 }e^(4t) +c_2 {w_2 }e^t +c_3 {w_3 }e^(−2t) λ=1 [ determinant ((3,6,0),((−3),(−6),0),((−3),(−6),(−6)))]{w}={0} 3w_x +6w_y =0 −6w_z =0 w_2 ={1,−2,0} λ=−2 [ determinant ((6,6,0),((−3),(−3),0),((−3),(−6),(−3)))]{w}={0} w_x +w_y =0 w_z =0 w_3 ={1,−1,0}](https://www.tinkutara.com/question/Q164230.png)
$$\begin{vmatrix}{\mathrm{4}}&{\mathrm{6}}&{\mathrm{0}}\\{−\mathrm{3}}&{−\mathrm{5}}&{\mathrm{0}}\\{−\mathrm{3}}&{−\mathrm{6}}&{−\mathrm{5}}\end{vmatrix}−\lambda\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\left\{\begin{matrix}{{x}}\\{{y}}\\{{z}}\end{matrix}\right\}=\left\{\begin{matrix}{\mathrm{0}}\\{\mathrm{0}}\\{\mathrm{0}}\end{matrix}\right\} \\ $$$${det}\left({A}−\lambda{I}\right)=\mathrm{0} \\ $$$$\left[\left(\mathrm{4}−\lambda\right)\left(−\mathrm{5}−\lambda\right)+\mathrm{18}\right]\left(−\mathrm{5}−\lambda\right)=\mathrm{0} \\ $$$$\left[−\mathrm{20}−\mathrm{4}\lambda+\mathrm{5}\lambda+\lambda^{\mathrm{2}} +\mathrm{18}\right]\left(−\mathrm{5}−\lambda\right)=\mathrm{0} \\ $$$$\left(\lambda^{\mathrm{2}} +\lambda−\mathrm{2}\right)\left(−\mathrm{5}−\lambda\right)=\mathrm{0} \\ $$$$−\left(\lambda+\mathrm{5}\right)\left(\lambda−\mathrm{1}\right)\left(\lambda+\mathrm{2}\right)=\mathrm{0} \\ $$$$\left\{{x}\right\}={c}_{\mathrm{1}} \left\{{w}_{\mathrm{1}} \right\}{e}^{\mathrm{4}{t}} +{c}_{\mathrm{2}} \left\{{w}_{\mathrm{2}} \right\}{e}^{{t}} +{c}_{\mathrm{3}} \left\{{w}_{\mathrm{3}} \right\}{e}^{−\mathrm{2}{t}} \\ $$$$ \\ $$$$\lambda=\mathrm{1} \\ $$$$\left[\begin{matrix}{\mathrm{3}}&{\mathrm{6}}&{\mathrm{0}}\\{−\mathrm{3}}&{−\mathrm{6}}&{\mathrm{0}}\\{−\mathrm{3}}&{−\mathrm{6}}&{−\mathrm{6}}\end{matrix}\right]\left\{{w}\right\}=\left\{\mathrm{0}\right\} \\ $$$$\mathrm{3}{w}_{{x}} +\mathrm{6}{w}_{{y}} =\mathrm{0} \\ $$$$−\mathrm{6}{w}_{{z}} =\mathrm{0}\: \\ $$$${w}_{\mathrm{2}} =\left\{\mathrm{1},−\mathrm{2},\mathrm{0}\right\} \\ $$$$ \\ $$$$\lambda=−\mathrm{2} \\ $$$$\left[\begin{array}{|c|c|c|}{\mathrm{6}}&\hline{\mathrm{6}}&\hline{\mathrm{0}}\\{−\mathrm{3}}&\hline{−\mathrm{3}}&\hline{\mathrm{0}}\\{−\mathrm{3}}&\hline{−\mathrm{6}}&\hline{−\mathrm{3}}\\\hline\end{array}\right]\left\{{w}\right\}=\left\{\mathrm{0}\right\} \\ $$$${w}_{{x}} +{w}_{{y}} =\mathrm{0} \\ $$$${w}_{{z}} =\mathrm{0} \\ $$$${w}_{\mathrm{3}} =\left\{\mathrm{1},−\mathrm{1},\mathrm{0}\right\} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by pluton last updated on 15/Jan/22
![posons X(t)= (( determinant (((x(t))),((y(t))))),((z(t))) )((dX(t))/dt)=AX ou A= [((4 6 0)),((−3 −5 0)) ]il vient que ((dX(t))/X)=Adt donc X(t)=X_o e^(tA) ou X_0 = (( determinant (((x(o))),((y(0))))),((z(0))) ) [−3 −6 −5]](https://www.tinkutara.com/question/Q164273.png)
$${posons}\:{X}\left({t}\right)=\begin{pmatrix}{\begin{matrix}{{x}\left({t}\right)}\\{{y}\left({t}\right)}\end{matrix}}\\{{z}\left({t}\right)}\end{pmatrix}\frac{{dX}\left({t}\right)}{{dt}}={AX} \\ $$$${ou}\:{A}=\begin{bmatrix}{\mathrm{4}\:\:\:\:\:\:\:\:\:\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{−\mathrm{3}\:\:−\mathrm{5}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{bmatrix}{il}\:{vient}\:\:{que}\:\:\frac{{dX}\left({t}\right)}{{X}}={Adt}\:\:{donc}\:{X}\left({t}\right)={X}_{{o}} {e}^{{tA}} \:{ou}\:\:\:{X}_{\mathrm{0}} =\begin{pmatrix}{\begin{matrix}{{x}\left({o}\right)}\\{{y}\left(\mathrm{0}\right)}\end{matrix}}\\{{z}\left(\mathrm{0}\right)}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[−\mathrm{3}\:\:\:\:\:\:−\mathrm{6}\:\:\:\:\:−\mathrm{5}\right] \\ $$