We first need the eigenvalues and eigenvectors for the matrix.

\[\begin{align*}\det \left( {A - \lambda I} \right) & = \left| {\begin{array}{*{20}{c}}{3 - \lambda }&9\\{ - 4}&{ - 3 - \lambda }\end{array}} \right|\\ & = {\lambda ^2} + 27\hspace{0.25in}{\lambda _{1,2}} = \pm 3\sqrt 3 \,i\end{align*}\]

So, now that we have the eigenvalues recall that we only need to get the eigenvector for one of the eigenvalues since we can get the second eigenvector for free from the first eigenvector.

\({\lambda _1} = 3\sqrt 3 \,i\):

We need to solve the following system.

\[\left( {\begin{array}{*{20}{c}}{3 - 3\sqrt 3 \,i}&9\\{ - 4}&{ - 3 - 3\sqrt 3 \,i}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{{\eta _2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\]

Using the first equation we get,

\[\begin{align*}\left( {3 - 3\sqrt 3 \,i} \right){\eta _1} + 9{\eta _2} & = 0\\ {\eta _2} & = - \frac{1}{3}\left( {1 - \sqrt 3 \,i} \right){\eta _1}\end{align*}\]

So, the first eigenvector is,

\[\begin{align*}\vec \eta & = \left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{ - \frac{1}{3}\left( {1 - \sqrt 3 \,i} \right){\eta _1}}\end{array}} \right)\\ {{\vec \eta }^{\left( 1 \right)}} & = \left( {\begin{array}{*{20}{c}}3\\{ - 1 + \sqrt 3 \,i}\end{array}} \right)\hspace{0.25in}{\eta _1} = 3\end{align*}\]

When finding the eigenvectors in these cases make sure that the complex number appears in the numerator of any fractions since we’ll need it in the numerator later on. Also try to clear out any fractions by appropriately picking the constant. This will make our life easier down the road.

Now, the second eigenvector is,

\[{\vec \eta ^{\left( 2 \right)}} = \left( {\begin{array}{*{20}{c}}3\\{ - 1 - \sqrt 3 \,i}\end{array}} \right)\]

However, as we will see we won’t need this eigenvector.

The solution that we get from the first eigenvalue and eigenvector is,

\[{\vec x_1}\left( t \right) = {{\bf{e}}^{3\sqrt 3 \,i\,t}}\left( {\begin{array}{*{20}{c}}3\\{-1 + \sqrt 3 \,i}\end{array}} \right)\]

So, as we can see there are complex numbers in both the exponential and vector that we will need to get rid of in order to use this as a solution. Recall from the complex roots section of the second order differential equation chapter that we can use Euler’s formula to get the complex number out of the exponential. Doing this gives us,

\[{\vec x_1}\left( t \right) = \left( {\cos \left( {3\sqrt 3 t} \right) + i\sin \left( {3\sqrt 3 t} \right)} \right)\left( {\begin{array}{*{20}{c}}3\\{ - 1 + \sqrt 3 \,i}\end{array}} \right)\]

The next step is to multiply the cosines and sines into the vector.

\[{\vec x_1}\left( t \right) = \left( {\begin{array}{*{20}{c}}{3\cos \left( {3\sqrt 3 t} \right) + 3i\sin \left( {3\sqrt 3 t} \right)}\\{ - \cos \left( {3\sqrt 3 t} \right) - i\sin \left( {3\sqrt 3 t} \right) + \sqrt 3 \,i\cos \left( {3\sqrt 3 t} \right) - \sqrt 3 \sin \left( {3\sqrt 3 t} \right)}\end{array}} \right)\]

Now combine the terms with an “\(i\)” in them and split these terms off from those terms that don’t contain an “\(i\)”. Also factor the “\(i\)” out of this vector.

\[\begin{align*}{{\vec x}_1}\left( t \right) & = \left( {\begin{array}{*{20}{c}}{3\cos \left( {3\sqrt 3 t} \right)}\\{ - \cos \left( {3\sqrt 3 t} \right) - \sqrt 3 \sin \left( {3\sqrt 3 t} \right)}\end{array}} \right) + i\left( {\begin{array}{*{20}{c}}{3\sin \left( {3\sqrt 3 t} \right)}\\{ - \sin \left( {3\sqrt 3 t} \right) + \sqrt 3 \,\cos \left( {3\sqrt 3 t} \right)}\end{array}} \right)\\ & = \vec u\left( t \right) + i\,\vec v\left( t \right)\end{align*}\]

Now, it can be shown (we’ll leave the details to you) that \(\vec u\left( t \right)\) and \(\vec v\left( t \right)\) are two linearly independent solutions to the system of differential equations. This means that we can use them to form a general solution and they are both real solutions.

So, the general solution to a system with complex roots is

\[\vec x\left( t \right) = {c_1}\vec u\left( t \right) + {c_2}\vec v\left( t \right)\]

where \(\vec u\left( t \right)\) and \(\vec v\left( t \right)\) are found by writing the first solution as

\[\vec x\left( t \right) = \vec u\left( t \right) + i\,\vec v\left( t \right)\]

For our system then, the general solution is,

\[\vec x\left( t \right) = {c_1}\left( {\begin{array}{*{20}{c}}{3\cos \left( {3\sqrt 3 t} \right)}\\{ - \cos \left( {3\sqrt 3 t} \right) - \sqrt 3 \sin \left( {3\sqrt 3 t} \right)}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{3\sin \left( {3\sqrt 3 t} \right)}\\{ - \sin \left( {3\sqrt 3 t} \right) + \sqrt 3 \,\cos \left( {3\sqrt 3 t} \right)}\end{array}} \right)\]

We now need to apply the initial condition to this to find the constants.

\[\left( {\begin{array}{*{20}{c}}2\\{ - 4}\end{array}} \right) = \vec x\left( 0 \right) = {c_1}\left( {\begin{array}{*{20}{c}}3\\{ - 1}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}0\\{\sqrt 3 }\end{array}} \right)\]

This leads to the following system of equations to be solved,

\[\left. {\begin{array}{*{20}{r}}{3{c_1} = 2}\\{ - {c_1} + \sqrt 3 {c_2} = - 4}\end{array}} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = \frac{2}{3},\,\,\,{c_2} = \frac{{ - 10}}{{3\sqrt 3 }}\]

The actual solution is then,

\[\vec x\left( t \right) = \frac{2}{3}\left( {\begin{array}{*{20}{c}}{3\cos \left( {3\sqrt 3 t} \right)}\\{ - \cos \left( {3\sqrt 3 t} \right) - \sqrt 3 \sin \left( {3\sqrt 3 t} \right)}\end{array}} \right) - \frac{{10}}{{3\sqrt 3 }}\left( {\begin{array}{*{20}{c}}{3\sin \left( {3\sqrt 3 t} \right)}\\{ - \sin \left( {3\sqrt 3 t} \right) + \sqrt 3 \,\cos \left( {3\sqrt 3 t} \right)}\end{array}} \right)\]