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$\begin{array}{c}|A鈭�位I|=0\\ |\left[\begin{array}{cc}7& 10\\ 鈭�4& 鈭�5\end{array}\right]鈭�位\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0\end{array}$Consider the equation$\frac{d\stackrel{鈫�}{x}}{dt}=\left[\begin{array}{cc}7& 10\\ 鈭�4& 鈭�5\end{array}\right]\stackrel{鈫�}{x}$ with the initial value$\stackrel{鈫�}{x}\left(0\right)=\left[\begin{array}{c}1\\ 0\end{array}\right]$ .

Compare the equations$\frac{d\stackrel{鈫�}{x}}{dt}=\left[\begin{array}{cc}7& 10\\ 鈭�4& 鈭�5\end{array}\right]\stackrel{鈫�}{x}$and $\frac{d\stackrel{鈫�}{x}}{dt}=A\stackrel{鈫�}{x}$as follows.

$A=\left[\begin{array}{cc}7& 10\\ 鈭�4& 鈭�5\end{array}\right]$

Assume$位$ is an Eigen value of the matrix $\left[\begin{array}{cc}7& 10\\ 鈭�4& 鈭�5\end{array}\right]$implies $|A鈭�位I|=0$.

Substitute the values$\left[\begin{array}{cc}7& 10\\ 鈭�4& 鈭�5\end{array}\right]$for $A$and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$for$I$ in the equation$|A鈭�位I|=0$ as follows.

$\begin{array}{c}|A鈭�位I|=0\\ |\left[\begin{array}{cc}7& 10\\ 鈭�4& 鈭�5\end{array}\right]鈭�位\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0\end{array}$

Simplify the equation$|\left[\begin{array}{cc}7& 10\\ 鈭�4& 鈭�5\end{array}\right]鈭�位\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0$as follows.

$\begin{array}{c}|\left[\begin{array}{cc}7& 10\\ 鈭�4& 鈭�5\end{array}\right]鈭�位\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0\\ |\left[\begin{array}{cc}7& 10\\ 鈭�4& 鈭�5\end{array}\right]鈭�\left[\begin{array}{cc}位& 0\\ 0& 位\end{array}\right]|=0\\ \left|\left[\begin{array}{cc}7鈭�位& 10\\ 鈭�4& 鈭�5鈭�位\end{array}\right]\right|=0\\ (7鈭�位)(鈭�5鈭�位)+40=0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{c}(7鈭�位)(鈭�5鈭�位)+40=0\\ 鈭�35鈭�7位+5位+{位}^2+40=0\\ {位}^2鈭�2位+5=0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{c}位=\frac{鈭�b卤\sqrt{b^2鈭�4ac}}{2a}\\ 位=\frac{鈭�(鈭�2)卤\sqrt{{(鈭�2)}^2鈭�4\left(1\right)\left(5\right)}}{2\left(1\right)}\\ 位=\frac{2卤\sqrt{4鈭�20}}{2}\\ 位=\frac{2卤4i}{2}\end{array}$

Therefore, the Eigen values of $A$are $位=1卤i2$.

Substitute the values$1+i2$ for$位$ in the equation$\left|\left[\begin{array}{cc}7鈭�位& 10\\ 鈭�4& 鈭�5鈭�位\end{array}\right]\right|=0$as follows.

$\begin{array}{c}\left|\left[\begin{array}{cc}7鈭�位& 10\\ 鈭�4& 鈭�5鈭�位\end{array}\right]\right|=0\\ \left|\left[\begin{array}{cc}7鈭�(1+i2)& 10\\ 鈭�4& 鈭�5鈭�(1+i2)\end{array}\right]\right|=0\\ \left|\left[\begin{array}{cc}7鈭�1鈭�2i& 10\\ 鈭�4& 鈭�5鈭�1鈭�2i\end{array}\right]\right|=0\\ \left|\left[\begin{array}{cc}6鈭�2i& 10\\ 鈭�4& 鈭�6鈭�2i\end{array}\right]\right|=0\end{array}$

As$E_{1+2i}=\ker \left[\begin{array}{cc}6鈭�2i& 10\\ 鈭�4& 鈭�6鈭�2i\end{array}\right]=span\left[\begin{array}{c}5\\ 3鈭�i\end{array}\right]$ , the values$\stackrel{鈫�}{v}+i\stackrel{鈫�}{w}$ is defined as follows.

$\stackrel{鈫�}{v}+i\stackrel{鈫�}{w}=\left[\begin{array}{c}5\\ 3\end{array}\right]+i\left[\begin{array}{c}0\\ 鈭�1\end{array}\right]$

Therefore, the value of$S$ is $S=\left[\begin{array}{cc}0& 5\\ 鈭�1& 3\end{array}\right]$.

The inverse of the matrix $S=\left[\begin{array}{cc}0& 5\\ 鈭�1& 3\end{array}\right]$is defined as follows.

$\begin{array}{c}{S}^{鈭�1}=\frac{1}{5}\left[\begin{array}{cc}3& 鈭�5\\ 1& 0\end{array}\right]\\ S^{鈭�1}=\left[\begin{array}{cc}\frac{3}{5}& 鈭�1\\ \frac{1}{5}& 0\end{array}\right]\end{array}$

As$\stackrel{鈫�}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}\cos \left(qt\right)& 鈭�\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right]S^{鈭�1}{\stackrel{鈫�}{x}}_0$ , substitute the value$\left[\begin{array}{cc}0& 5\\ 鈭�1& 3\end{array}\right]$for $S$,$\left[\begin{array}{cc}\frac{3}{5}& 鈭�1\\ \frac{1}{5}& 0\end{array}\right]$for$S^{鈭�1}$ ,$\left[\begin{array}{c}1\\ 0\end{array}\right]$ for ${\stackrel{鈫�}{x}}_0$,$1$ for $p$and$2$ for $q$in the equation $\stackrel{鈫�}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}\cos \left(qt\right)& 鈭�\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right]S^{鈭�1}{\stackrel{鈫�}{x}}_0$as follows.

$\begin{array}{c}\stackrel{鈫�}{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}\cos \left(qt\right)& 鈭�\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right]S^{鈭�1}{\stackrel{鈫�}{x}}_0\\ \stackrel{鈫�}{x}\left(t\right)=e^{\left(1\right)t}\left[\begin{array}{cc}0& 5\\ 鈭�1& 3\end{array}\right]\left[\begin{array}{cc}\cos \left(2t\right)& 鈭�\sin \left(2t\right)\\ \sin \left(2t\right)& \cos \left(2t\right)\end{array}\right]\left[\begin{array}{cc}\frac{3}{5}& 鈭�1\\ \frac{1}{5}& 0\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{鈫�}{x}\left(t\right)=e^t\left[\begin{array}{cc}5\sin \left(2t\right)& 5\cos \left(2t\right)\\ 鈭�\cos \left(2t\right)+3\sin \left(2t\right)& \sin \left(2t\right)+3\cos \left(2t\right)\end{array}\right]\left[\begin{array}{cc}\frac{3}{5}& 鈭�1\\ \frac{1}{5}& 0\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \stackrel{鈫�}{x}\left(t\right)=e^t\left[\begin{array}{cc}5\sin \left(2t\right)& 5\cos \left(2t\right)\\ 鈭�\cos \left(2t\right)+3\sin \left(2t\right)& \sin \left(2t\right)+3\cos \left(2t\right)\end{array}\right]\left[\begin{array}{c}\frac{3}{5}\\ \frac{1}{5}\end{array}\right]\end{array}$

Further, simplify the equation as follows.

$\begin{array}{l}\stackrel{鈫�}{x}\left(t\right)=e^t\left[\begin{array}{cc}5\sin \left(2t\right)& 5\cos \left(2t\right)\\ 鈭�\cos \left(2t\right)+3\sin \left(2t\right)& \sin \left(2t\right)+3\cos \left(2t\right)\end{array}\right]\left[\begin{array}{c}\frac{3}{5}\\ \frac{1}{5}\end{array}\right]\\ \stackrel{鈫�}{x}\left(t\right)=e^t\left[\begin{array}{c}3\sin \left(2t\right)+\cos \left(2t\right)\\ 鈭�\frac{3}{5}\cos \left(2t\right)+\frac{9}{5}\sin \left(2t\right)+\frac{1}{5}\sin \left(2t\right)+\frac{3}{5}\cos \left(2t\right)\end{array}\right]\\ \stackrel{鈫�}{x}\left(t\right)=e^t\left[\begin{array}{c}3\sin \left(2t\right)+\cos \left(2t\right)\\ 2\sin \left(2t\right)\end{array}\right]\end{array}$

Therefore, the solution of the system is$\stackrel{鈫�}{x}\left(t\right)=e^t\left[\begin{array}{c}3\sin \left(2t\right)+\cos \left(2t\right)\\ 2\sin \left(2t\right)\end{array}\right]$ .

As${位}_{1,2}=1卤2i$and$p>0$, draw the graph of the solution$\stackrel{鈫�}{x}\left(t\right)=e^t\left[\begin{array}{c}3\sin \left(2t\right)+\cos \left(2t\right)\\ 2\sin \left(2t\right)\end{array}\right]$as follows.

Hence, the solution of the system$\frac{d\stackrel{鈫�}{x}}{dt}=\left[\begin{array}{cc}7& 10\\ 鈭�4& 鈭�5\end{array}\right]\stackrel{鈫�}{x}$with the initial value $\stackrel{鈫�}{x}\left(0\right)=\left[\begin{array}{c}1\\ 0\end{array}\right]$is$\stackrel{鈫�}{x}\left(t\right)=e^t\left[\begin{array}{c}3\sin \left(2t\right)+\cos \left(2t\right)\\ 2\sin \left(2t\right)\end{array}\right]$and the graph of the solution is a spiral in counterclockwise direction outward the origin.