Internal problem ID [12387]
Internal file name [OUTPUT/11040_Wednesday_October_04_2023_01_27_17_AM_85648862/index.tex]
Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A.
Dobrushkin. CRC Press 2015
Section: Chapter 8.3 Systems of Linear Differential Equations (Variation of Parameters).
Problems page 514
Problem number: Problem 6(a).
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "system of linear ODEs"
Solve \begin {align*} x^{\prime }\left (t \right )&=-3 x \left (t \right )-3 y+z \left (t \right )\\ y^{\prime }&=2 y+2 z \left (t \right )+29 \,{\mathrm e}^{-t}\\ z^{\prime }\left (t \right )&=5 x \left (t \right )+y+z \left (t \right )+39 \,{\mathrm e}^{t} \end {align*}
With initial conditions \[ [x \left (0\right ) = 1, y \left (0\right ) = 2, z \left (0\right ) = 3] \]
In this method, we will assume we have found the matrix exponential \(e^{A t}\) allready. There are different methods to determine this but will not be shown here. This is a system of linear ODE’s given as \begin {align*} \vec {x}'(t) &= A\, \vec {x}(t) + \vec {G}(t) \end {align*}
Or \begin {align*} \left [\begin {array}{c} x^{\prime }\left (t \right ) \\ y^{\prime } \\ z^{\prime }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{ccc} -3 & -3 & 1 \\ 0 & 2 & 2 \\ 5 & 1 & 1 \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \\ z \left (t \right ) \end {array}\right ] + \left [\begin {array}{c} 0 \\ 29 \,{\mathrm e}^{-t} \\ 39 \,{\mathrm e}^{t} \end {array}\right ] \end {align*}
Since the system is nonhomogeneous, then the solution is given by \begin {align*} \vec {x}(t) &= \vec {x}_h(t) + \vec {x}_p(t) \end {align*}
Where \(\vec {x}_h(t)\) is the homogeneous solution to \(\vec {x}'(t) = A\, \vec {x}(t)\) and \(\vec {x}_p(t)\) is a particular solution to \(\vec {x}'(t) = A\, \vec {x}(t) + \vec {G}(t)\). The particular solution will be found using variation of parameters method applied to the fundamental matrix. For the above matrix \(A\), the matrix exponential can be found to be \begin {align*} e^{A t} &= \text {Expression too large to display}\\ &= \text {Expression too large to display} \end {align*}
Therefore the homogeneous solution is \begin {align*} \vec {x}_h(t) &= e^{A t} \vec {x}_0 \\ &= \text {Expression too large to display} \left [\begin {array}{c} 1 \\ 2 \\ 3 \end {array}\right ] \\ &= \text {Expression too large to display}\\ &= \text {Expression too large to display} \end {align*}
The particular solution given by \begin {align*} \vec {x}_p (t) &= e^{A t} \int { e^{-A t} \vec {G}(t) \,dt} \end {align*}
But \begin {align*} e^{-A t} &= (e^{A t})^{-1} \\ &= \text {Expression too large to display} \end {align*}
Hence \begin {align*} \vec {x}_p (t) &= \text {Expression too large to display} \int { \text {Expression too large to display} \left [\begin {array}{c} 0 \\ 29 \,{\mathrm e}^{-t} \\ 39 \,{\mathrm e}^{t} \end {array}\right ]\,dt}\\ &= \text {Expression too large to display} \text {Expression too large to display}\\ &= \left [\begin {array}{c} -\frac {7 \left (689 \,{\mathrm e}^{\frac {t \left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42\right )}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}}-261 \,{\mathrm e}^{\frac {t \left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42\right )}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}}\right ) {\mathrm e}^{-\frac {\left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+42\right ) t}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}}}{477} \\ \frac {\left (5512 \,{\mathrm e}^{\frac {t \left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42\right )}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}}-2349 \,{\mathrm e}^{\frac {t \left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42\right )}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}}\right ) {\mathrm e}^{-\frac {\left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+42\right ) t}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}}}{477} \\ -\frac {13 \left (212 \,{\mathrm e}^{\frac {t \left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42\right )}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}}+261 \,{\mathrm e}^{\frac {t \left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42\right )}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}}\right ) {\mathrm e}^{-\frac {\left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+42\right ) t}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}}}{477} \end {array}\right ] \end {align*}
Hence the complete solution is \begin {align*} \vec {x}(t) &= \vec {x}_h(t) + \vec {x}_p (t) \\ &= \text {Expression too large to display} \end {align*}
This is a system of linear ODE’s given as \begin {align*} \vec {x}'(t) &= A\, \vec {x}(t) + \vec {G}(t) \end {align*}
Or \begin {align*} \left [\begin {array}{c} x^{\prime }\left (t \right ) \\ y^{\prime } \\ z^{\prime }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{ccc} -3 & -3 & 1 \\ 0 & 2 & 2 \\ 5 & 1 & 1 \end {array}\right ]\, \left [\begin {array}{c} x \left (t \right ) \\ y \\ z \left (t \right ) \end {array}\right ] + \left [\begin {array}{c} 0 \\ 29 \,{\mathrm e}^{-t} \\ 39 \,{\mathrm e}^{t} \end {array}\right ] \end {align*}
Since the system is nonhomogeneous, then the solution is given by \begin {align*} \vec {x}(t) &= \vec {x}_h(t) + \vec {x}_p(t) \end {align*}
Where \(\vec {x}_h(t)\) is the homogeneous solution to \(\vec {x}'(t) = A\, \vec {x}(t)\) and \(\vec {x}_p(t)\) is a particular solution to \(\vec {x}'(t) = A\, \vec {x}(t) + \vec {G}(t)\). The particular solution will be found using variation of parameters method applied to the fundamental matrix.
The first step is find the homogeneous solution. We start by finding the eigenvalues of \(A\). This is done by solving the following equation for the eigenvalues \(\lambda \) \begin {align*} \operatorname {det} \left ( A- \lambda I \right ) &= 0 \end {align*}
Expanding gives \begin {align*} \operatorname {det} \left (\left [\begin {array}{ccc} -3 & -3 & 1 \\ 0 & 2 & 2 \\ 5 & 1 & 1 \end {array}\right ]-\lambda \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right ) &= 0 \end {align*}
Therefore \begin {align*} \operatorname {det} \left (\left [\begin {array}{ccc} -3-\lambda & -3 & 1 \\ 0 & 2-\lambda & 2 \\ 5 & 1 & 1-\lambda \end {array}\right ]\right ) &= 0 \end {align*}
Which gives the characteristic equation \begin {align*} \lambda ^{3}-14 \lambda +40&=0 \end {align*}
The roots of the above are the eigenvalues. \begin {align*} \lambda _1 &= -\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\\ \lambda _2 &= \frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\\ \lambda _3 &= \frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2} \end {align*}
This table summarises the above result
| eigenvalue | algebraic multiplicity | type of eigenvalue |
| \(-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\) | \(1\) | real eigenvalue |
| \(\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\) | \(1\) | complex eigenvalue |
| \(\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\) | \(1\) | complex eigenvalue |
Now the eigenvector for each eigenvalue are found.
Considering the eigenvalue \(\lambda _{1} = -\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\)
We need to solve \(A \vec {v} = \lambda \vec {v}\) or \((A-\lambda I) \vec {v} = \vec {0}\) which becomes \begin {align*} \left (\left [\begin {array}{ccc} -3 & -3 & 1 \\ 0 & 2 & 2 \\ 5 & 1 & 1 \end {array}\right ] - \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right ) \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{ccc} \frac {\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-9 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} & -3 & 1 \\ 0 & \frac {\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} & 2 \\ 5 & 1 & \frac {\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -3+\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}&-3&1&0\\ 0&2+\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}&2&0\\ 5&1&1+\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}&0 \end {array} \right ] \] \begin {align*} R_{3} = R_{3}-\frac {5 R_{1}}{-3+\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-9 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}&-3&1&0\\ 0&\frac {\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}&2&0\\ 0&\frac {\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+36 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}{\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-9 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}&-\frac {2 \left (\left (\sqrt {6042}+48\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-6 \sqrt {6042}+2 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-246\right )}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \left (-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+9 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42\right )}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {3 \left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+36 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} R_{2}}{\left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-9 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42\right ) \left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-9 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}&-3&1&0\\ 0&\frac {\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}&2&0\\ 0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{ccc} \frac {\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-9 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} & -3 & 1 \\ 0 & \frac {\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}{3 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} & 2 \\ 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{3}\}\) and the leading variables are \(\{v_{1}, v_{2}\}\). Let \(v_{3} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\left \{v_{1} = -\frac {3 t \left (4 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{5 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}+24}, v_{2} = -\frac {6 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42}\right \}\)
Hence the solution is \[ \left [\begin {array}{c} -\frac {3 t \left (4 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{5 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}+24} \\ -\frac {6 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {3 t \left (4 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{5 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}+24} \\ -\frac {6 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42} \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} -\frac {3 t \left (4 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{5 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}+24} \\ -\frac {6 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42} \\ t \end {array}\right ] = t \left [\begin {array}{c} -\frac {3 \left (4 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{5 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}+24} \\ -\frac {6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42} \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} -\frac {3 t \left (4 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{5 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}+24} \\ -\frac {6 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {3 \left (4 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{5 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}+24} \\ -\frac {6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42} \\ 1 \end {array}\right ] \] Which is normalized to \[ \left [\begin {array}{c} -\frac {3 t \left (4 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{5 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}+24} \\ -\frac {6 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {3 \left (4 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{5 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}+24} \\ -\frac {6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+6 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+42} \\ 1 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{2} = \frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\)
We need to solve \(A \vec {v} = \lambda \vec {v}\) or \((A-\lambda I) \vec {v} = \vec {0}\) which becomes \begin {align*} \left (\left [\begin {array}{ccc} -3 & -3 & 1 \\ 0 & 2 & 2 \\ 5 & 1 & 1 \end {array}\right ] - \left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{ccc} -\frac {42+18 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}} & -3 & 1 \\ 0 & \frac {-42+12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}-i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}} & 2 \\ 5 & 1 & \frac {-42+6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}-i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}} \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -3-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}-\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}&-3&1&0\\ 0&2-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}-\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}&2&0\\ 5&1&1-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}-\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}&0 \end {array} \right ] \] \begin {align*} R_{3} = R_{3}-\frac {5 R_{1}}{-3-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}-\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -\frac {42+18 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}&-3&1&0\\ 0&\frac {-42+12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}-i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}&2&0\\ 0&\frac {i \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-72 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42}{i \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}+18 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42}&\frac {2 \left (48-48 i \sqrt {3}-3 i \sqrt {2014}+\sqrt {6042}\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-492-492 i \sqrt {3}-36 i \sqrt {2014}-12 \sqrt {6042}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \left (42+18 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}+\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}\right )}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {6 \left (i \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-72 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42\right ) \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}} R_{2}}{\left (i \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}+18 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42\right ) \left (-42+12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}-i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -\frac {42+18 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}&-3&1&0\\ 0&\frac {-42+12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}-i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}&2&0\\ 0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{ccc} -\frac {42+18 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}} & -3 & 1 \\ 0 & \frac {-42+12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}-i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}} & 2 \\ 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{3}\}\) and the leading variables are \(\{v_{1}, v_{2}\}\). Let \(v_{3} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\left \{v_{1} = -\frac {3 t \left (-7 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+90 i \sqrt {3}+3 i \sqrt {2014}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24-3 i \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-69 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+24 i \sqrt {3}-9 i \sqrt {2014}-3 \sqrt {6042}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}, v_{2} = \frac {12 t \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}{i \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42}\right \}\)
Hence the solution is \[ \left [\begin {array}{c} -\frac {3 t \left (-7 \,\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+90 \,\operatorname {I} \sqrt {3}+3 \,\operatorname {I} \sqrt {2014}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24-3 \,\operatorname {I} \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-69 \,\operatorname {I} \sqrt {3}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+24 \,\operatorname {I} \sqrt {3}-9 \,\operatorname {I} \sqrt {2014}-3 \sqrt {6042}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ \frac {12 t \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}{\operatorname {I} \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 \,\operatorname {I} \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {3 t \left (-7 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+90 i \sqrt {3}+3 i \sqrt {2014}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24-3 i \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-69 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+24 i \sqrt {3}-9 i \sqrt {2014}-3 \sqrt {6042}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ \frac {12 t \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}{i \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42} \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} -\frac {3 t \left (-7 \,\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+90 \,\operatorname {I} \sqrt {3}+3 \,\operatorname {I} \sqrt {2014}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24-3 \,\operatorname {I} \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-69 \,\operatorname {I} \sqrt {3}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+24 \,\operatorname {I} \sqrt {3}-9 \,\operatorname {I} \sqrt {2014}-3 \sqrt {6042}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ \frac {12 t \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}{\operatorname {I} \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 \,\operatorname {I} \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42} \\ t \end {array}\right ] = t \left [\begin {array}{c} -\frac {3 \left (-7 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+90 i \sqrt {3}+3 i \sqrt {2014}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24-3 i \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-69 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+24 i \sqrt {3}-9 i \sqrt {2014}-3 \sqrt {6042}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ \frac {12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}{i \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42} \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} -\frac {3 t \left (-7 \,\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+90 \,\operatorname {I} \sqrt {3}+3 \,\operatorname {I} \sqrt {2014}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24-3 \,\operatorname {I} \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-69 \,\operatorname {I} \sqrt {3}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+24 \,\operatorname {I} \sqrt {3}-9 \,\operatorname {I} \sqrt {2014}-3 \sqrt {6042}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ \frac {12 t \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}{\operatorname {I} \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 \,\operatorname {I} \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {3 \left (-7 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+90 i \sqrt {3}+3 i \sqrt {2014}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24-3 i \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-69 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+24 i \sqrt {3}-9 i \sqrt {2014}-3 \sqrt {6042}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ \frac {12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}{i \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42} \\ 1 \end {array}\right ] \] Which is normalized to \[ \left [\begin {array}{c} -\frac {3 t \left (-7 \,\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+90 \,\operatorname {I} \sqrt {3}+3 \,\operatorname {I} \sqrt {2014}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24-3 \,\operatorname {I} \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-69 \,\operatorname {I} \sqrt {3}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+24 \,\operatorname {I} \sqrt {3}-9 \,\operatorname {I} \sqrt {2014}-3 \sqrt {6042}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ \frac {12 t \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}{\operatorname {I} \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 \,\operatorname {I} \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {3 \left (-7 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+90 i \sqrt {3}+3 i \sqrt {2014}+\sqrt {6042}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24-3 i \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-69 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+24 i \sqrt {3}-9 i \sqrt {2014}-3 \sqrt {6042}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ \frac {12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}{i \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}+\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+42} \\ 1 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{3} = \frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\)
We need to solve \(A \vec {v} = \lambda \vec {v}\) or \((A-\lambda I) \vec {v} = \vec {0}\) which becomes \begin {align*} \left (\left [\begin {array}{ccc} -3 & -3 & 1 \\ 0 & 2 & 2 \\ 5 & 1 & 1 \end {array}\right ] - \left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) \left [\begin {array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{ccc} \frac {-42-18 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}} & -3 & 1 \\ 0 & \frac {-42+12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}} & 2 \\ 5 & 1 & \frac {-42+6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}} \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -3-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}-\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}&-3&1&0\\ 0&2-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}-\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}&2&0\\ 5&1&1-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}-\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}&0 \end {array} \right ] \] \begin {align*} R_{3} = R_{3}-\frac {5 R_{1}}{-3-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}-\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {-42-18 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}&-3&1&0\\ 0&\frac {-42+12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}&2&0\\ 0&\frac {i \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-42 i \sqrt {3}+72 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42}{i \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-18 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42}&\frac {-2 \left (48 i \sqrt {3}+3 i \sqrt {2014}+\sqrt {6042}+48\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+492-492 i \sqrt {3}-36 i \sqrt {2014}+12 \sqrt {6042}+8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}}{\left (-42-18 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {6 \left (i \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-42 i \sqrt {3}+72 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42\right ) \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}} R_{2}}{\left (i \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-18 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42\right ) \left (-42+12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} \frac {-42-18 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}&-3&1&0\\ 0&\frac {-42+12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}}&2&0\\ 0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{ccc} \frac {-42-18 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}} & -3 & 1 \\ 0 & \frac {-42+12 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}+i \left (\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}-42\right ) \sqrt {3}-\left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {2}{3}}}{6 \left (540+6 \sqrt {3}\, \sqrt {2014}\right )^{\frac {1}{3}}} & 2 \\ 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{3}\}\) and the leading variables are \(\{v_{1}, v_{2}\}\). Let \(v_{3} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\left \{v_{1} = -\frac {3 t \left (-3 i \sqrt {2014}+7 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}-90 i \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24+3 i \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-3 \sqrt {6042}-24 i \sqrt {3}+9 i \sqrt {2014}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}, v_{2} = -\frac {12 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{i \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42}\right \}\)
Hence the solution is \[ \left [\begin {array}{c} -\frac {3 t \left (-3 \,\operatorname {I} \sqrt {2014}+7 \,\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}-90 \,\operatorname {I} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24+3 \,\operatorname {I} \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \,\operatorname {I} \sqrt {3}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}-24 \,\operatorname {I} \sqrt {3}+9 \,\operatorname {I} \sqrt {2014}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ -\frac {12 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-42 \,\operatorname {I} \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {3 t \left (-3 i \sqrt {2014}+7 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}-90 i \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24+3 i \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-3 \sqrt {6042}-24 i \sqrt {3}+9 i \sqrt {2014}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ -\frac {12 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{i \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42} \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} -\frac {3 t \left (-3 \,\operatorname {I} \sqrt {2014}+7 \,\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}-90 \,\operatorname {I} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24+3 \,\operatorname {I} \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \,\operatorname {I} \sqrt {3}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}-24 \,\operatorname {I} \sqrt {3}+9 \,\operatorname {I} \sqrt {2014}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ -\frac {12 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-42 \,\operatorname {I} \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42} \\ t \end {array}\right ] = t \left [\begin {array}{c} -\frac {3 \left (-3 i \sqrt {2014}+7 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}-90 i \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24+3 i \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-3 \sqrt {6042}-24 i \sqrt {3}+9 i \sqrt {2014}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ -\frac {12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{i \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42} \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} -\frac {3 t \left (-3 \,\operatorname {I} \sqrt {2014}+7 \,\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}-90 \,\operatorname {I} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24+3 \,\operatorname {I} \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \,\operatorname {I} \sqrt {3}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}-24 \,\operatorname {I} \sqrt {3}+9 \,\operatorname {I} \sqrt {2014}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ -\frac {12 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-42 \,\operatorname {I} \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {3 \left (-3 i \sqrt {2014}+7 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}-90 i \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24+3 i \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-3 \sqrt {6042}-24 i \sqrt {3}+9 i \sqrt {2014}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ -\frac {12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{i \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42} \\ 1 \end {array}\right ] \] Which is normalized to \[ \left [\begin {array}{c} -\frac {3 t \left (-3 \,\operatorname {I} \sqrt {2014}+7 \,\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}-90 \,\operatorname {I} \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24+3 \,\operatorname {I} \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 \,\operatorname {I} \sqrt {3}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-3 \sqrt {6042}-24 \,\operatorname {I} \sqrt {3}+9 \,\operatorname {I} \sqrt {2014}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ -\frac {12 t \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{\operatorname {I} \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-42 \,\operatorname {I} \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42} \\ t \end {array}\right ] = \left [\begin {array}{c} -\frac {3 \left (-3 i \sqrt {2014}+7 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}+\sqrt {6042}-90 i \sqrt {3}-8 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+7 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+90\right )}{24+3 i \sqrt {2014}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\sqrt {6042}\, \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+69 i \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}} \sqrt {3}-3 \sqrt {6042}-24 i \sqrt {3}+9 i \sqrt {2014}-10 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+69 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}} \\ -\frac {12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{i \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {3}-42 i \sqrt {3}-\left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+12 \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}-42} \\ 1 \end {array}\right ] \] The following table gives a summary of this result. It shows for each eigenvalue the algebraic multiplicity \(m\), and its geometric multiplicity \(k\) and the eigenvectors associated with the eigenvalue. If \(m>k\) then the eigenvalue is defective which means the number of normal linearly independent eigenvectors associated with this eigenvalue (called the geometric multiplicity \(k\)) does not equal the algebraic multiplicity \(m\), and we need to determine an additional \(m-k\) generalized eigenvectors for this eigenvalue.
| multiplicity
| ||||
| eigenvalue | algebraic \(m\) | geometric \(k\) | defective? | eigenvectors |
| \(-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} \frac {-8-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}}{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-2\right ) \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+3\right )} \\ \frac {2}{-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-2} \\ 1 \end {array}\right ]\) |
| \(\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} \frac {-8+\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}}{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2\right ) \left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}+3\right )} \\ \frac {2}{\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2} \\ 1 \end {array}\right ]\) |
| \(\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} \frac {-8+\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}}{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2\right ) \left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}+3\right )} \\ \frac {2}{\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2} \\ 1 \end {array}\right ]\) |
Now that we found the eigenvalues and associated eigenvectors, we will go over each eigenvalue and generate the solution basis. The only problem we need to take care of is if the eigenvalue is defective. Since eigenvalue \(-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\) is real and distinct then the corresponding eigenvector solution is \begin {align*} \vec {x}_{1}(t) &= \vec {v}_{1} e^{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right ) t}\\ &= \left [\begin {array}{c} \frac {-8-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}}{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-2\right ) \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+3\right )} \\ \frac {2}{-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-2} \\ 1 \end {array}\right ] e^{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right ) t} \end {align*}
Therefore the homogeneous solution is \begin {align*} \vec {x}_h(t) &= c_{1} \vec {x}_{1}(t) + c_{2} \vec {x}_{2}(t) + c_{3} \vec {x}_{3}(t) \end {align*}
Which is written as \begin {align*} \left [\begin {array}{c} x \left (t \right ) \\ y \\ z \left (t \right ) \end {array}\right ] &= c_{1} \left [\begin {array}{c} \frac {{\mathrm e}^{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right ) t} \left (-8-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-2\right ) \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+3\right )} \\ \frac {2 \,{\mathrm e}^{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right ) t}}{-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-2} \\ {\mathrm e}^{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right ) t} \end {array}\right ] + c_{2} \left [\begin {array}{c} \frac {{\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} \left (-8+\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right )}{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2\right ) \left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}+3\right )} \\ \frac {2 \,{\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t}}{\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2} \\ {\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} \end {array}\right ] + c_{3} \left [\begin {array}{c} \frac {{\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} \left (-8+\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right )}{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2\right ) \left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}+3\right )} \\ \frac {2 \,{\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t}}{\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2} \\ {\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} \end {array}\right ] \end {align*}
Now that we found homogeneous solution above, we need to find a particular solution \(\vec {x}_p(t)\). We will use Variation of parameters. The fundamental matrix is \[ \Phi =\begin {bmatrix} \vec {x}_{1} & \vec {x}_{2} & \cdots \end {bmatrix} \] Where \(\vec {x}_i\) are the solution basis found above. Therefore the fundamental matrix is \begin {align*} \Phi (t)&= \text {Expression too large to display} \end {align*}
The particular solution is then given by \begin {align*} \vec {x}_p(t) &= \Phi \int { \Phi ^{-1} \vec {G}(t) \, dt}\\ \end {align*}
But \begin {align*} \Phi ^{-1} &= \text {Expression too large to display} \end {align*}
Hence \begin {align*} \vec {x}_p(t) &= \text {Expression too large to display} \int { \text {Expression too large to display} \left [\begin {array}{c} 0 \\ 29 \,{\mathrm e}^{-t} \\ 39 \,{\mathrm e}^{t} \end {array}\right ] \, dt}\\ &= \text {Expression too large to display} \int { \text {Expression too large to display} \, dt}\\ &= \text {Expression too large to display} \text {Expression too large to display} \\ &= \left [\begin {array}{c} \frac {7 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \left (689 \,{\mathrm e}^{t}-261 \,{\mathrm e}^{-t}\right ) \left (21 \left (20199392243-20199392243 i \sqrt {3}-779599731 i \sqrt {2014}+259866577 \sqrt {6042}\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+877678813 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {6042}-13875408631008 i \sqrt {3}-535521330117 i \sqrt {2014}+68222525850 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-178507110039 \sqrt {6042}-13875408631008\right )}{1431 \left (-142298484724368+444 \left (-51544877093-663124410 \sqrt {6042}\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\left (-1830670135266+59502370013 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}\right ) \sqrt {6042}+3 i \left (1830670135266+59502370013 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}\right ) \sqrt {2014}+48 i \left (2964551765091+96357004382 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}\right ) \sqrt {3}+4625136210336 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}\right )} \\ \frac {43165007182128 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \left (-\frac {2349 \,{\mathrm e}^{-t}}{5512}+{\mathrm e}^{t}\right ) \left (\left (i \sqrt {3}+\frac {50374407 i \sqrt {2014}}{1305182849}+\frac {16791469 \sqrt {6042}}{1305182849}+1\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\left (\frac {230355932 i \sqrt {3}}{1305182849}+\frac {17781409 i \sqrt {2014}}{2610365698}-\frac {17781409 \sqrt {6042}}{7831097094}-\frac {230355932}{1305182849}\right ) \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-\frac {47442098 \sqrt {6042}}{1305182849}-\frac {3687704100}{1305182849}\right )}{\left (663490237622796 i \sqrt {3}+25607422183572 i \sqrt {2014}-8535807394524 \sqrt {6042}-663490237622796\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\left (-135779284476 \sqrt {6042}-10554209134200\right ) \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+3759303033918576 i \sqrt {3}+145090438179444 i \sqrt {2014}+48363479393148 \sqrt {6042}+3759303033918576} \\ -\frac {377 \,{\mathrm e}^{-t}}{53}-\frac {52 \,{\mathrm e}^{t}}{9} \end {array}\right ] \end {align*}
Now that we found particular solution, the final solution is \begin {align*} \vec {x}(t) &= \vec {x}_h(t) + \vec {x}_p(t)\\ \left [\begin {array}{c} x \left (t \right ) \\ y \\ z \left (t \right ) \end {array}\right ] &= \left [\begin {array}{c} \frac {c_{1} {\mathrm e}^{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right ) t} \left (-8-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-2\right ) \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+3\right )} \\ \frac {2 c_{1} {\mathrm e}^{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right ) t}}{-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-2} \\ c_{1} {\mathrm e}^{\left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}-\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right ) t} \end {array}\right ] + \left [\begin {array}{c} \frac {c_{2} {\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} \left (-8+\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right )}{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2\right ) \left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}+3\right )} \\ \frac {2 c_{2} {\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t}}{\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2} \\ c_{2} {\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} \end {array}\right ] + \left [\begin {array}{c} \frac {c_{3} {\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} \left (-8+\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right )}{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2\right ) \left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}+3\right )} \\ \frac {2 c_{3} {\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t}}{\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}-2} \\ c_{3} {\mathrm e}^{\left (\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{6}+\frac {7}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (-\frac {\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}{3}+\frac {14}{\left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}}\right )}{2}\right ) t} \end {array}\right ] + \left [\begin {array}{c} \frac {7 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \left (689 \,{\mathrm e}^{t}-261 \,{\mathrm e}^{-t}\right ) \left (21 \left (20199392243-20199392243 i \sqrt {3}-779599731 i \sqrt {2014}+259866577 \sqrt {6042}\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+877678813 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \sqrt {6042}-13875408631008 i \sqrt {3}-535521330117 i \sqrt {2014}+68222525850 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-178507110039 \sqrt {6042}-13875408631008\right )}{1431 \left (-142298484724368+444 \left (-51544877093-663124410 \sqrt {6042}\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\left (-1830670135266+59502370013 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}\right ) \sqrt {6042}+3 i \left (1830670135266+59502370013 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}\right ) \sqrt {2014}+48 i \left (2964551765091+96357004382 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}\right ) \sqrt {3}+4625136210336 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}\right )} \\ \frac {43165007182128 \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}} \left (-\frac {2349 \,{\mathrm e}^{-t}}{5512}+{\mathrm e}^{t}\right ) \left (\left (i \sqrt {3}+\frac {50374407 i \sqrt {2014}}{1305182849}+\frac {16791469 \sqrt {6042}}{1305182849}+1\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\left (\frac {230355932 i \sqrt {3}}{1305182849}+\frac {17781409 i \sqrt {2014}}{2610365698}-\frac {17781409 \sqrt {6042}}{7831097094}-\frac {230355932}{1305182849}\right ) \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}-\frac {47442098 \sqrt {6042}}{1305182849}-\frac {3687704100}{1305182849}\right )}{\left (663490237622796 i \sqrt {3}+25607422183572 i \sqrt {2014}-8535807394524 \sqrt {6042}-663490237622796\right ) \left (540+6 \sqrt {6042}\right )^{\frac {1}{3}}+\left (-135779284476 \sqrt {6042}-10554209134200\right ) \left (540+6 \sqrt {6042}\right )^{\frac {2}{3}}+3759303033918576 i \sqrt {3}+145090438179444 i \sqrt {2014}+48363479393148 \sqrt {6042}+3759303033918576} \\ -\frac {377 \,{\mathrm e}^{-t}}{53}-\frac {52 \,{\mathrm e}^{t}}{9} \end {array}\right ] \end {align*}
Which becomes \begin {align*} \text {Expression too large to display} \end {align*}
Since initial conditions are given, the solution above needs to be updated by solving for the constants of integrations using the given initial conditions \begin {align*} \left [\begin {array}{c} x \left (0\right )=1 \\ y \left (0\right )=2 \\ z \left (0\right )=3 \end {array}\right ]\tag {1} \end {align*}
Substituting initial conditions into the above solution at \(t=0\) gives \begin {align*} \text {Expression too large to display} \end {align*}
Solving for the constants of integrations gives \begin {align*} \text {Expression too large to display} \end {align*}
Substituting these constants back in original solution in Eq. (1) gives
\begin {align*} \text {Expression too large to display} \end {align*}
The following are plots of each solution against another.
The following are plots of each solution.
✓ Solution by Maple
Time used: 5.609 (sec). Leaf size: 949416
dsolve([diff(x(t),t) = -3*x(t)-3*y(t)+z(t), diff(y(t),t) = 2*y(t)+2*z(t)+29*exp(-t), diff(z(t),t) = 5*x(t)+y(t)+z(t)+39*exp(t), x(0) = 1, y(0) = 2, z(0) = 3], singsol=all)
\begin{align*} \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.228 (sec). Leaf size: 3462
DSolve[{x'[t]==-3*x[t]-3*y[t]+z[t],y'[t]==2*y[t]+2*z[t]+29*Exp[-t],z'[t]==5*x[t]+y[t]+z[t]+39*Exp[t]},{x[0]==1,y[0]==2,z[0]==3},{x[t],y[t],z[t]},t,IncludeSingularSolutions -> True]
Too large to display