Internal problem ID [353]
Internal file name [OUTPUT/353_Sunday_June_05_2022_01_39_28_AM_35276350/index.tex]
Book: Differential equations and linear algebra, 4th ed., Edwards and Penney
Section: Section 7.3, The eigenvalue method for linear systems. Page 395
Problem number: problem 50.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "system of linear ODEs"
Solve \begin {align*} x_{1}^{\prime }\left (t \right )&=9 x_{1} \left (t \right )+13 x_{2} \left (t \right )-13 x_{6} \left (t \right )\\ x_{2}^{\prime }\left (t \right )&=-14 x_{1} \left (t \right )+19 x_{2} \left (t \right )-10 x_{3} \left (t \right )-20 x_{4} \left (t \right )+10 x_{5} \left (t \right )+4 x_{6} \left (t \right )\\ x_{3}^{\prime }\left (t \right )&=-30 x_{1} \left (t \right )+12 x_{2} \left (t \right )-7 x_{3} \left (t \right )-30 x_{4} \left (t \right )+12 x_{5} \left (t \right )+18 x_{6} \left (t \right )\\ x_{4}^{\prime }\left (t \right )&=-12 x_{1} \left (t \right )+10 x_{2} \left (t \right )-10 x_{3} \left (t \right )-9 x_{4} \left (t \right )+10 x_{5} \left (t \right )+2 x_{6} \left (t \right )\\ x_{5}^{\prime }\left (t \right )&=6 x_{1} \left (t \right )+9 x_{2} \left (t \right )+6 x_{4} \left (t \right )+5 x_{5} \left (t \right )-15 x_{6} \left (t \right )\\ x_{6}^{\prime }\left (t \right )&=-14 x_{1} \left (t \right )+23 x_{2} \left (t \right )-10 x_{3} \left (t \right )-20 x_{4} \left (t \right )+10 x_{5} \left (t \right ) \end {align*}
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) \end {align*}
Or \begin {align*} \left [\begin {array}{c} x_{1}^{\prime }\left (t \right ) \\ x_{2}^{\prime }\left (t \right ) \\ x_{3}^{\prime }\left (t \right ) \\ x_{4}^{\prime }\left (t \right ) \\ x_{5}^{\prime }\left (t \right ) \\ x_{6}^{\prime }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{cccccc} 9 & 13 & 0 & 0 & 0 & -13 \\ -14 & 19 & -10 & -20 & 10 & 4 \\ -30 & 12 & -7 & -30 & 12 & 18 \\ -12 & 10 & -10 & -9 & 10 & 2 \\ 6 & 9 & 0 & 6 & 5 & -15 \\ -14 & 23 & -10 & -20 & 10 & 0 \end {array}\right ]\, \left [\begin {array}{c} x_{1} \left (t \right ) \\ x_{2} \left (t \right ) \\ x_{3} \left (t \right ) \\ x_{4} \left (t \right ) \\ x_{5} \left (t \right ) \\ x_{6} \left (t \right ) \end {array}\right ] \end {align*}
For the above matrix \(A\), the matrix exponential can be found to be \begin {align*} e^{A t} &= \left [\begin {array}{cccccc} {\mathrm e}^{9 t} & \left ({\mathrm e}^{13 t}-1\right ) {\mathrm e}^{-4 t} & 0 & 0 & 0 & -\left ({\mathrm e}^{13 t}-1\right ) {\mathrm e}^{-4 t} \\ \left ({\mathrm e}^{16 t}-3 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{16 t}+{\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & -\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & -2 \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left (-{\mathrm e}^{16 t}+2 \,{\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} \\ -\left ({\mathrm e}^{18 t}+{\mathrm e}^{12 t}-2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{12 t}-1\right ) {\mathrm e}^{-7 t} & {\mathrm e}^{-7 t} & -\left ({\mathrm e}^{18 t}+{\mathrm e}^{12 t}-2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{12 t}-1\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{18 t}-1\right ) {\mathrm e}^{-7 t} \\ \left ({\mathrm e}^{18 t}-3 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & -\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{18 t}-2 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left (-{\mathrm e}^{18 t}+2 \,{\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} \\ {\mathrm e}^{11 t}-{\mathrm e}^{5 t} & \left ({\mathrm e}^{9 t}-1\right ) {\mathrm e}^{-4 t} & 0 & {\mathrm e}^{11 t}-{\mathrm e}^{5 t} & {\mathrm e}^{5 t} & -\left ({\mathrm e}^{15 t}-1\right ) {\mathrm e}^{-4 t} \\ \left ({\mathrm e}^{16 t}-3 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{16 t}+{\mathrm e}^{10 t}-{\mathrm e}^{3 t}-1\right ) {\mathrm e}^{-7 t} & -\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & -2 \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left (-{\mathrm e}^{16 t}+2 \,{\mathrm e}^{10 t}+{\mathrm e}^{3 t}-1\right ) {\mathrm e}^{-7 t} \end {array}\right ] \end {align*}
Therefore the homogeneous solution is \begin {align*} \vec {x}_h(t) &= e^{A t} \vec {c} \\ &= \left [\begin {array}{cccccc} {\mathrm e}^{9 t} & \left ({\mathrm e}^{13 t}-1\right ) {\mathrm e}^{-4 t} & 0 & 0 & 0 & -\left ({\mathrm e}^{13 t}-1\right ) {\mathrm e}^{-4 t} \\ \left ({\mathrm e}^{16 t}-3 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{16 t}+{\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & -\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & -2 \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left (-{\mathrm e}^{16 t}+2 \,{\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} \\ -\left ({\mathrm e}^{18 t}+{\mathrm e}^{12 t}-2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{12 t}-1\right ) {\mathrm e}^{-7 t} & {\mathrm e}^{-7 t} & -\left ({\mathrm e}^{18 t}+{\mathrm e}^{12 t}-2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{12 t}-1\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{18 t}-1\right ) {\mathrm e}^{-7 t} \\ \left ({\mathrm e}^{18 t}-3 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & -\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{18 t}-2 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left (-{\mathrm e}^{18 t}+2 \,{\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} \\ {\mathrm e}^{11 t}-{\mathrm e}^{5 t} & \left ({\mathrm e}^{9 t}-1\right ) {\mathrm e}^{-4 t} & 0 & {\mathrm e}^{11 t}-{\mathrm e}^{5 t} & {\mathrm e}^{5 t} & -\left ({\mathrm e}^{15 t}-1\right ) {\mathrm e}^{-4 t} \\ \left ({\mathrm e}^{16 t}-3 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{16 t}+{\mathrm e}^{10 t}-{\mathrm e}^{3 t}-1\right ) {\mathrm e}^{-7 t} & -\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & -2 \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} & \left (-{\mathrm e}^{16 t}+2 \,{\mathrm e}^{10 t}+{\mathrm e}^{3 t}-1\right ) {\mathrm e}^{-7 t} \end {array}\right ] \left [\begin {array}{c} c_{1} \\ c_{2} \\ c_{3} \\ c_{4} \\ c_{5} \\ c_{6} \end {array}\right ] \\ &= \left [\begin {array}{c} {\mathrm e}^{9 t} c_{1}+\left ({\mathrm e}^{13 t}-1\right ) {\mathrm e}^{-4 t} c_{2}-\left ({\mathrm e}^{13 t}-1\right ) {\mathrm e}^{-4 t} c_{6} \\ \left ({\mathrm e}^{16 t}-3 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} c_{1}+\left ({\mathrm e}^{16 t}+{\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{2}-\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{3}-2 \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{4}+\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{5}+\left (-{\mathrm e}^{16 t}+2 \,{\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{6} \\ -\left ({\mathrm e}^{18 t}+{\mathrm e}^{12 t}-2\right ) {\mathrm e}^{-7 t} c_{1}+\left ({\mathrm e}^{12 t}-1\right ) {\mathrm e}^{-7 t} c_{2}+{\mathrm e}^{-7 t} c_{3}-\left ({\mathrm e}^{18 t}+{\mathrm e}^{12 t}-2\right ) {\mathrm e}^{-7 t} c_{4}+\left ({\mathrm e}^{12 t}-1\right ) {\mathrm e}^{-7 t} c_{5}+\left ({\mathrm e}^{18 t}-1\right ) {\mathrm e}^{-7 t} c_{6} \\ \left ({\mathrm e}^{18 t}-3 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} c_{1}+\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{2}-\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{3}+\left ({\mathrm e}^{18 t}-2 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} c_{4}+\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{5}+\left (-{\mathrm e}^{18 t}+2 \,{\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{6} \\ \left ({\mathrm e}^{11 t}-{\mathrm e}^{5 t}\right ) c_{1}+\left ({\mathrm e}^{9 t}-1\right ) {\mathrm e}^{-4 t} c_{2}+\left ({\mathrm e}^{11 t}-{\mathrm e}^{5 t}\right ) c_{4}+{\mathrm e}^{5 t} c_{5}-\left ({\mathrm e}^{15 t}-1\right ) {\mathrm e}^{-4 t} c_{6} \\ \left ({\mathrm e}^{16 t}-3 \,{\mathrm e}^{10 t}+2\right ) {\mathrm e}^{-7 t} c_{1}+\left ({\mathrm e}^{16 t}+{\mathrm e}^{10 t}-{\mathrm e}^{3 t}-1\right ) {\mathrm e}^{-7 t} c_{2}-\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{3}-2 \left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{4}+\left ({\mathrm e}^{10 t}-1\right ) {\mathrm e}^{-7 t} c_{5}+\left (-{\mathrm e}^{16 t}+2 \,{\mathrm e}^{10 t}+{\mathrm e}^{3 t}-1\right ) {\mathrm e}^{-7 t} c_{6} \end {array}\right ]\\ &= \left [\begin {array}{c} {\mathrm e}^{-4 t} \left (\left (c_{1}+c_{2}-c_{6}\right ) {\mathrm e}^{13 t}-c_{2}+c_{6}\right ) \\ \left (\left (-3 c_{1}+c_{2}-c_{3}-2 c_{4}+c_{5}+2 c_{6}\right ) {\mathrm e}^{10 t}+\left (c_{1}+c_{2}-c_{6}\right ) {\mathrm e}^{16 t}+2 c_{1}-c_{2}+c_{3}+2 c_{4}-c_{5}-c_{6}\right ) {\mathrm e}^{-7 t} \\ -\left (\left (c_{1}-c_{2}+c_{4}-c_{5}\right ) {\mathrm e}^{12 t}+\left (c_{1}+c_{4}-c_{6}\right ) {\mathrm e}^{18 t}-2 c_{1}+c_{2}-c_{3}-2 c_{4}+c_{5}+c_{6}\right ) {\mathrm e}^{-7 t} \\ {\mathrm e}^{-7 t} \left (\left (-3 c_{1}+c_{2}-c_{3}-2 c_{4}+c_{5}+2 c_{6}\right ) {\mathrm e}^{10 t}+\left (c_{1}+c_{4}-c_{6}\right ) {\mathrm e}^{18 t}+2 c_{1}-c_{2}+c_{3}+2 c_{4}-c_{5}-c_{6}\right ) \\ -\left (\left (c_{1}-c_{2}+c_{4}-c_{5}\right ) {\mathrm e}^{9 t}+\left (-c_{1}-c_{4}+c_{6}\right ) {\mathrm e}^{15 t}+c_{2}-c_{6}\right ) {\mathrm e}^{-4 t} \\ -3 \,{\mathrm e}^{-7 t} \left (\left (c_{1}-\frac {c_{2}}{3}+\frac {c_{3}}{3}+\frac {2 c_{4}}{3}-\frac {c_{5}}{3}-\frac {2 c_{6}}{3}\right ) {\mathrm e}^{10 t}+\frac {\left (-c_{1}-c_{2}+c_{6}\right ) {\mathrm e}^{16 t}}{3}+\frac {\left (c_{2}-c_{6}\right ) {\mathrm e}^{3 t}}{3}-\frac {2 c_{1}}{3}+\frac {c_{2}}{3}-\frac {c_{3}}{3}-\frac {2 c_{4}}{3}+\frac {c_{5}}{3}+\frac {c_{6}}{3}\right ) \end {array}\right ] \end {align*}
Since no forcing function is given, then the final solution is \(\vec {x}_h(t)\) above.
This is a system of linear ODE’s given as \begin {align*} \vec {x}'(t) &= A\, \vec {x}(t) \end {align*}
Or \begin {align*} \left [\begin {array}{c} x_{1}^{\prime }\left (t \right ) \\ x_{2}^{\prime }\left (t \right ) \\ x_{3}^{\prime }\left (t \right ) \\ x_{4}^{\prime }\left (t \right ) \\ x_{5}^{\prime }\left (t \right ) \\ x_{6}^{\prime }\left (t \right ) \end {array}\right ] &= \left [\begin {array}{cccccc} 9 & 13 & 0 & 0 & 0 & -13 \\ -14 & 19 & -10 & -20 & 10 & 4 \\ -30 & 12 & -7 & -30 & 12 & 18 \\ -12 & 10 & -10 & -9 & 10 & 2 \\ 6 & 9 & 0 & 6 & 5 & -15 \\ -14 & 23 & -10 & -20 & 10 & 0 \end {array}\right ]\, \left [\begin {array}{c} x_{1} \left (t \right ) \\ x_{2} \left (t \right ) \\ x_{3} \left (t \right ) \\ x_{4} \left (t \right ) \\ x_{5} \left (t \right ) \\ x_{6} \left (t \right ) \end {array}\right ] \end {align*}
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}{cccccc} 9 & 13 & 0 & 0 & 0 & -13 \\ -14 & 19 & -10 & -20 & 10 & 4 \\ -30 & 12 & -7 & -30 & 12 & 18 \\ -12 & 10 & -10 & -9 & 10 & 2 \\ 6 & 9 & 0 & 6 & 5 & -15 \\ -14 & 23 & -10 & -20 & 10 & 0 \end {array}\right ]-\lambda \left [\begin {array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end {array}\right ]\right ) &= 0 \end {align*}
Therefore \begin {align*} \operatorname {det} \left (\left [\begin {array}{cccccc} 9-\lambda & 13 & 0 & 0 & 0 & -13 \\ -14 & 19-\lambda & -10 & -20 & 10 & 4 \\ -30 & 12 & -7-\lambda & -30 & 12 & 18 \\ -12 & 10 & -10 & -9-\lambda & 10 & 2 \\ 6 & 9 & 0 & 6 & 5-\lambda & -15 \\ -14 & 23 & -10 & -20 & 10 & -\lambda \end {array}\right ]\right ) &= 0 \end {align*}
Which gives the characteristic equation \begin {align*} \lambda ^{6}-17 \lambda ^{5}-6 \lambda ^{4}+1138 \lambda ^{3}-2855 \lambda ^{2}-14241 \lambda +41580&=0 \end {align*}
The roots of the above are the eigenvalues. \begin {align*} \lambda _1 &= 5\\ \lambda _2 &= 9\\ \lambda _3 &= 3\\ \lambda _4 &= 11\\ \lambda _5 &= -7\\ \lambda _6 &= -4 \end {align*}
This table summarises the above result
| eigenvalue | algebraic multiplicity | type of eigenvalue |
| \(-4\) | \(1\) | real eigenvalue |
| \(3\) | \(1\) | real eigenvalue |
| \(5\) | \(1\) | real eigenvalue |
| \(-7\) | \(1\) | real eigenvalue |
| \(9\) | \(1\) | real eigenvalue |
| \(11\) | \(1\) | real eigenvalue |
Now the eigenvector for each eigenvalue are found.
Considering the eigenvalue \(\lambda _{1} = -7\)
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}{cccccc} 9 & 13 & 0 & 0 & 0 & -13 \\ -14 & 19 & -10 & -20 & 10 & 4 \\ -30 & 12 & -7 & -30 & 12 & 18 \\ -12 & 10 & -10 & -9 & 10 & 2 \\ 6 & 9 & 0 & 6 & 5 & -15 \\ -14 & 23 & -10 & -20 & 10 & 0 \end {array}\right ] - \left (-7\right ) \left [\begin {array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{cccccc} 16 & 13 & 0 & 0 & 0 & -13 \\ -14 & 26 & -10 & -20 & 10 & 4 \\ -30 & 12 & 0 & -30 & 12 & 18 \\ -12 & 10 & -10 & -2 & 10 & 2 \\ 6 & 9 & 0 & 6 & 12 & -15 \\ -14 & 23 & -10 & -20 & 10 & 7 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ -14&26&-10&-20&10&4&0\\ -30&12&0&-30&12&18&0\\ -12&10&-10&-2&10&2&0\\ 6&9&0&6&12&-15&0\\ -14&23&-10&-20&10&7&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}+\frac {7 R_{1}}{8} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ -30&12&0&-30&12&18&0\\ -12&10&-10&-2&10&2&0\\ 6&9&0&6&12&-15&0\\ -14&23&-10&-20&10&7&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}+\frac {15 R_{1}}{8} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&{\frac {291}{8}}&0&-30&12&-{\frac {51}{8}}&0\\ -12&10&-10&-2&10&2&0\\ 6&9&0&6&12&-15&0\\ -14&23&-10&-20&10&7&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}+\frac {3 R_{1}}{4} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&{\frac {291}{8}}&0&-30&12&-{\frac {51}{8}}&0\\ 0&{\frac {79}{4}}&-10&-2&10&-{\frac {31}{4}}&0\\ 6&9&0&6&12&-15&0\\ -14&23&-10&-20&10&7&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {3 R_{1}}{8} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&{\frac {291}{8}}&0&-30&12&-{\frac {51}{8}}&0\\ 0&{\frac {79}{4}}&-10&-2&10&-{\frac {31}{4}}&0\\ 0&{\frac {33}{8}}&0&6&12&-{\frac {81}{8}}&0\\ -14&23&-10&-20&10&7&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}+\frac {7 R_{1}}{8} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&{\frac {291}{8}}&0&-30&12&-{\frac {51}{8}}&0\\ 0&{\frac {79}{4}}&-10&-2&10&-{\frac {31}{4}}&0\\ 0&{\frac {33}{8}}&0&6&12&-{\frac {81}{8}}&0\\ 0&{\frac {275}{8}}&-10&-20&10&-{\frac {35}{8}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {291 R_{2}}{299} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&0&{\frac {2910}{299}}&-{\frac {3150}{299}}&{\frac {678}{299}}&{\frac {240}{299}}&0\\ 0&{\frac {79}{4}}&-10&-2&10&-{\frac {31}{4}}&0\\ 0&{\frac {33}{8}}&0&6&12&-{\frac {81}{8}}&0\\ 0&{\frac {275}{8}}&-10&-20&10&-{\frac {35}{8}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}-\frac {158 R_{2}}{299} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&0&{\frac {2910}{299}}&-{\frac {3150}{299}}&{\frac {678}{299}}&{\frac {240}{299}}&0\\ 0&0&-{\frac {1410}{299}}&{\frac {2562}{299}}&{\frac {1410}{299}}&-{\frac {1152}{299}}&0\\ 0&{\frac {33}{8}}&0&6&12&-{\frac {81}{8}}&0\\ 0&{\frac {275}{8}}&-10&-20&10&-{\frac {35}{8}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {33 R_{2}}{299} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&0&{\frac {2910}{299}}&-{\frac {3150}{299}}&{\frac {678}{299}}&{\frac {240}{299}}&0\\ 0&0&-{\frac {1410}{299}}&{\frac {2562}{299}}&{\frac {1410}{299}}&-{\frac {1152}{299}}&0\\ 0&0&{\frac {330}{299}}&{\frac {2454}{299}}&{\frac {3258}{299}}&-{\frac {2784}{299}}&0\\ 0&{\frac {275}{8}}&-10&-20&10&-{\frac {35}{8}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-\frac {275 R_{2}}{299} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&0&{\frac {2910}{299}}&-{\frac {3150}{299}}&{\frac {678}{299}}&{\frac {240}{299}}&0\\ 0&0&-{\frac {1410}{299}}&{\frac {2562}{299}}&{\frac {1410}{299}}&-{\frac {1152}{299}}&0\\ 0&0&{\frac {330}{299}}&{\frac {2454}{299}}&{\frac {3258}{299}}&-{\frac {2784}{299}}&0\\ 0&0&-{\frac {240}{299}}&-{\frac {480}{299}}&{\frac {240}{299}}&{\frac {720}{299}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}+\frac {47 R_{3}}{97} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&0&{\frac {2910}{299}}&-{\frac {3150}{299}}&{\frac {678}{299}}&{\frac {240}{299}}&0\\ 0&0&0&{\frac {336}{97}}&{\frac {564}{97}}&-{\frac {336}{97}}&0\\ 0&0&{\frac {330}{299}}&{\frac {2454}{299}}&{\frac {3258}{299}}&-{\frac {2784}{299}}&0\\ 0&0&-{\frac {240}{299}}&-{\frac {480}{299}}&{\frac {240}{299}}&{\frac {720}{299}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {11 R_{3}}{97} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&0&{\frac {2910}{299}}&-{\frac {3150}{299}}&{\frac {678}{299}}&{\frac {240}{299}}&0\\ 0&0&0&{\frac {336}{97}}&{\frac {564}{97}}&-{\frac {336}{97}}&0\\ 0&0&0&{\frac {912}{97}}&{\frac {1032}{97}}&-{\frac {912}{97}}&0\\ 0&0&-{\frac {240}{299}}&-{\frac {480}{299}}&{\frac {240}{299}}&{\frac {720}{299}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}+\frac {8 R_{3}}{97} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&0&{\frac {2910}{299}}&-{\frac {3150}{299}}&{\frac {678}{299}}&{\frac {240}{299}}&0\\ 0&0&0&{\frac {336}{97}}&{\frac {564}{97}}&-{\frac {336}{97}}&0\\ 0&0&0&{\frac {912}{97}}&{\frac {1032}{97}}&-{\frac {912}{97}}&0\\ 0&0&0&-{\frac {240}{97}}&{\frac {96}{97}}&{\frac {240}{97}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {19 R_{4}}{7} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&0&{\frac {2910}{299}}&-{\frac {3150}{299}}&{\frac {678}{299}}&{\frac {240}{299}}&0\\ 0&0&0&{\frac {336}{97}}&{\frac {564}{97}}&-{\frac {336}{97}}&0\\ 0&0&0&0&-{\frac {36}{7}}&0&0\\ 0&0&0&-{\frac {240}{97}}&{\frac {96}{97}}&{\frac {240}{97}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}+\frac {5 R_{4}}{7} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&0&{\frac {2910}{299}}&-{\frac {3150}{299}}&{\frac {678}{299}}&{\frac {240}{299}}&0\\ 0&0&0&{\frac {336}{97}}&{\frac {564}{97}}&-{\frac {336}{97}}&0\\ 0&0&0&0&-{\frac {36}{7}}&0&0\\ 0&0&0&0&{\frac {36}{7}}&0&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}+R_{5} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 16&13&0&0&0&-13&0\\ 0&{\frac {299}{8}}&-10&-20&10&-{\frac {59}{8}}&0\\ 0&0&{\frac {2910}{299}}&-{\frac {3150}{299}}&{\frac {678}{299}}&{\frac {240}{299}}&0\\ 0&0&0&{\frac {336}{97}}&{\frac {564}{97}}&-{\frac {336}{97}}&0\\ 0&0&0&0&-{\frac {36}{7}}&0&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{cccccc} 16 & 13 & 0 & 0 & 0 & -13 \\ 0 & \frac {299}{8} & -10 & -20 & 10 & -\frac {59}{8} \\ 0 & 0 & \frac {2910}{299} & -\frac {3150}{299} & \frac {678}{299} & \frac {240}{299} \\ 0 & 0 & 0 & \frac {336}{97} & \frac {564}{97} & -\frac {336}{97} \\ 0 & 0 & 0 & 0 & -\frac {36}{7} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{6}\}\) and the leading variables are \(\{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\}\). Let \(v_{6} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\{v_{1} = 0, v_{2} = t, v_{3} = t, v_{4} = t, v_{5} = 0\}\)
Hence the solution is \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = \left [\begin {array}{c} 0 \\ t \\ t \\ t \\ 0 \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = t \left [\begin {array}{c} 0 \\ 1 \\ 1 \\ 1 \\ 0 \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = \left [\begin {array}{c} 0 \\ 1 \\ 1 \\ 1 \\ 0 \\ 1 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{2} = -4\)
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}{cccccc} 9 & 13 & 0 & 0 & 0 & -13 \\ -14 & 19 & -10 & -20 & 10 & 4 \\ -30 & 12 & -7 & -30 & 12 & 18 \\ -12 & 10 & -10 & -9 & 10 & 2 \\ 6 & 9 & 0 & 6 & 5 & -15 \\ -14 & 23 & -10 & -20 & 10 & 0 \end {array}\right ] - \left (-4\right ) \left [\begin {array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{cccccc} 13 & 13 & 0 & 0 & 0 & -13 \\ -14 & 23 & -10 & -20 & 10 & 4 \\ -30 & 12 & -3 & -30 & 12 & 18 \\ -12 & 10 & -10 & -5 & 10 & 2 \\ 6 & 9 & 0 & 6 & 9 & -15 \\ -14 & 23 & -10 & -20 & 10 & 4 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ -14&23&-10&-20&10&4&0\\ -30&12&-3&-30&12&18&0\\ -12&10&-10&-5&10&2&0\\ 6&9&0&6&9&-15&0\\ -14&23&-10&-20&10&4&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}+\frac {14 R_{1}}{13} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ -30&12&-3&-30&12&18&0\\ -12&10&-10&-5&10&2&0\\ 6&9&0&6&9&-15&0\\ -14&23&-10&-20&10&4&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}+\frac {30 R_{1}}{13} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ 0&42&-3&-30&12&-12&0\\ -12&10&-10&-5&10&2&0\\ 6&9&0&6&9&-15&0\\ -14&23&-10&-20&10&4&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}+\frac {12 R_{1}}{13} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ 0&42&-3&-30&12&-12&0\\ 0&22&-10&-5&10&-10&0\\ 6&9&0&6&9&-15&0\\ -14&23&-10&-20&10&4&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {6 R_{1}}{13} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ 0&42&-3&-30&12&-12&0\\ 0&22&-10&-5&10&-10&0\\ 0&3&0&6&9&-9&0\\ -14&23&-10&-20&10&4&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}+\frac {14 R_{1}}{13} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ 0&42&-3&-30&12&-12&0\\ 0&22&-10&-5&10&-10&0\\ 0&3&0&6&9&-9&0\\ 0&37&-10&-20&10&-10&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {42 R_{2}}{37} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ 0&0&{\frac {309}{37}}&-{\frac {270}{37}}&{\frac {24}{37}}&-{\frac {24}{37}}&0\\ 0&22&-10&-5&10&-10&0\\ 0&3&0&6&9&-9&0\\ 0&37&-10&-20&10&-10&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}-\frac {22 R_{2}}{37} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ 0&0&{\frac {309}{37}}&-{\frac {270}{37}}&{\frac {24}{37}}&-{\frac {24}{37}}&0\\ 0&0&-{\frac {150}{37}}&{\frac {255}{37}}&{\frac {150}{37}}&-{\frac {150}{37}}&0\\ 0&3&0&6&9&-9&0\\ 0&37&-10&-20&10&-10&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {3 R_{2}}{37} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ 0&0&{\frac {309}{37}}&-{\frac {270}{37}}&{\frac {24}{37}}&-{\frac {24}{37}}&0\\ 0&0&-{\frac {150}{37}}&{\frac {255}{37}}&{\frac {150}{37}}&-{\frac {150}{37}}&0\\ 0&0&{\frac {30}{37}}&{\frac {282}{37}}&{\frac {303}{37}}&-{\frac {303}{37}}&0\\ 0&37&-10&-20&10&-10&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-R_{2} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ 0&0&{\frac {309}{37}}&-{\frac {270}{37}}&{\frac {24}{37}}&-{\frac {24}{37}}&0\\ 0&0&-{\frac {150}{37}}&{\frac {255}{37}}&{\frac {150}{37}}&-{\frac {150}{37}}&0\\ 0&0&{\frac {30}{37}}&{\frac {282}{37}}&{\frac {303}{37}}&-{\frac {303}{37}}&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}+\frac {50 R_{3}}{103} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ 0&0&{\frac {309}{37}}&-{\frac {270}{37}}&{\frac {24}{37}}&-{\frac {24}{37}}&0\\ 0&0&0&{\frac {345}{103}}&{\frac {450}{103}}&-{\frac {450}{103}}&0\\ 0&0&{\frac {30}{37}}&{\frac {282}{37}}&{\frac {303}{37}}&-{\frac {303}{37}}&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {10 R_{3}}{103} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ 0&0&{\frac {309}{37}}&-{\frac {270}{37}}&{\frac {24}{37}}&-{\frac {24}{37}}&0\\ 0&0&0&{\frac {345}{103}}&{\frac {450}{103}}&-{\frac {450}{103}}&0\\ 0&0&0&{\frac {858}{103}}&{\frac {837}{103}}&-{\frac {837}{103}}&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {286 R_{4}}{115} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 13&13&0&0&0&-13&0\\ 0&37&-10&-20&10&-10&0\\ 0&0&{\frac {309}{37}}&-{\frac {270}{37}}&{\frac {24}{37}}&-{\frac {24}{37}}&0\\ 0&0&0&{\frac {345}{103}}&{\frac {450}{103}}&-{\frac {450}{103}}&0\\ 0&0&0&0&-{\frac {63}{23}}&{\frac {63}{23}}&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{cccccc} 13 & 13 & 0 & 0 & 0 & -13 \\ 0 & 37 & -10 & -20 & 10 & -10 \\ 0 & 0 & \frac {309}{37} & -\frac {270}{37} & \frac {24}{37} & -\frac {24}{37} \\ 0 & 0 & 0 & \frac {345}{103} & \frac {450}{103} & -\frac {450}{103} \\ 0 & 0 & 0 & 0 & -\frac {63}{23} & \frac {63}{23} \\ 0 & 0 & 0 & 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{6}\}\) and the leading variables are \(\{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\}\). Let \(v_{6} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\{v_{1} = t, v_{2} = 0, v_{3} = 0, v_{4} = 0, v_{5} = t\}\)
Hence the solution is \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = \left [\begin {array}{c} t \\ 0 \\ 0 \\ 0 \\ t \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = t \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{3} = 3\)
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}{cccccc} 9 & 13 & 0 & 0 & 0 & -13 \\ -14 & 19 & -10 & -20 & 10 & 4 \\ -30 & 12 & -7 & -30 & 12 & 18 \\ -12 & 10 & -10 & -9 & 10 & 2 \\ 6 & 9 & 0 & 6 & 5 & -15 \\ -14 & 23 & -10 & -20 & 10 & 0 \end {array}\right ] - \left (3\right ) \left [\begin {array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{cccccc} 6 & 13 & 0 & 0 & 0 & -13 \\ -14 & 16 & -10 & -20 & 10 & 4 \\ -30 & 12 & -10 & -30 & 12 & 18 \\ -12 & 10 & -10 & -12 & 10 & 2 \\ 6 & 9 & 0 & 6 & 2 & -15 \\ -14 & 23 & -10 & -20 & 10 & -3 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ -14&16&-10&-20&10&4&0\\ -30&12&-10&-30&12&18&0\\ -12&10&-10&-12&10&2&0\\ 6&9&0&6&2&-15&0\\ -14&23&-10&-20&10&-3&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}+\frac {7 R_{1}}{3} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ -30&12&-10&-30&12&18&0\\ -12&10&-10&-12&10&2&0\\ 6&9&0&6&2&-15&0\\ -14&23&-10&-20&10&-3&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}+5 R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&77&-10&-30&12&-47&0\\ -12&10&-10&-12&10&2&0\\ 6&9&0&6&2&-15&0\\ -14&23&-10&-20&10&-3&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}+2 R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&77&-10&-30&12&-47&0\\ 0&36&-10&-12&10&-24&0\\ 6&9&0&6&2&-15&0\\ -14&23&-10&-20&10&-3&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&77&-10&-30&12&-47&0\\ 0&36&-10&-12&10&-24&0\\ 0&-4&0&6&2&-2&0\\ -14&23&-10&-20&10&-3&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}+\frac {7 R_{1}}{3} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&77&-10&-30&12&-47&0\\ 0&36&-10&-12&10&-24&0\\ 0&-4&0&6&2&-2&0\\ 0&{\frac {160}{3}}&-10&-20&10&-{\frac {100}{3}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {231 R_{2}}{139} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&0&{\frac {920}{139}}&{\frac {450}{139}}&-{\frac {642}{139}}&-{\frac {450}{139}}&0\\ 0&36&-10&-12&10&-24&0\\ 0&-4&0&6&2&-2&0\\ 0&{\frac {160}{3}}&-10&-20&10&-{\frac {100}{3}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}-\frac {108 R_{2}}{139} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&0&{\frac {920}{139}}&{\frac {450}{139}}&-{\frac {642}{139}}&-{\frac {450}{139}}&0\\ 0&0&-{\frac {310}{139}}&{\frac {492}{139}}&{\frac {310}{139}}&-{\frac {492}{139}}&0\\ 0&-4&0&6&2&-2&0\\ 0&{\frac {160}{3}}&-10&-20&10&-{\frac {100}{3}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}+\frac {12 R_{2}}{139} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&0&{\frac {920}{139}}&{\frac {450}{139}}&-{\frac {642}{139}}&-{\frac {450}{139}}&0\\ 0&0&-{\frac {310}{139}}&{\frac {492}{139}}&{\frac {310}{139}}&-{\frac {492}{139}}&0\\ 0&0&-{\frac {120}{139}}&{\frac {594}{139}}&{\frac {398}{139}}&-{\frac {594}{139}}&0\\ 0&{\frac {160}{3}}&-10&-20&10&-{\frac {100}{3}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-\frac {160 R_{2}}{139} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&0&{\frac {920}{139}}&{\frac {450}{139}}&-{\frac {642}{139}}&-{\frac {450}{139}}&0\\ 0&0&-{\frac {310}{139}}&{\frac {492}{139}}&{\frac {310}{139}}&-{\frac {492}{139}}&0\\ 0&0&-{\frac {120}{139}}&{\frac {594}{139}}&{\frac {398}{139}}&-{\frac {594}{139}}&0\\ 0&0&{\frac {210}{139}}&{\frac {420}{139}}&-{\frac {210}{139}}&-{\frac {420}{139}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}+\frac {31 R_{3}}{92} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&0&{\frac {920}{139}}&{\frac {450}{139}}&-{\frac {642}{139}}&-{\frac {450}{139}}&0\\ 0&0&0&{\frac {213}{46}}&{\frac {31}{46}}&-{\frac {213}{46}}&0\\ 0&0&-{\frac {120}{139}}&{\frac {594}{139}}&{\frac {398}{139}}&-{\frac {594}{139}}&0\\ 0&0&{\frac {210}{139}}&{\frac {420}{139}}&-{\frac {210}{139}}&-{\frac {420}{139}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}+\frac {3 R_{3}}{23} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&0&{\frac {920}{139}}&{\frac {450}{139}}&-{\frac {642}{139}}&-{\frac {450}{139}}&0\\ 0&0&0&{\frac {213}{46}}&{\frac {31}{46}}&-{\frac {213}{46}}&0\\ 0&0&0&{\frac {108}{23}}&{\frac {52}{23}}&-{\frac {108}{23}}&0\\ 0&0&{\frac {210}{139}}&{\frac {420}{139}}&-{\frac {210}{139}}&-{\frac {420}{139}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-\frac {21 R_{3}}{92} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&0&{\frac {920}{139}}&{\frac {450}{139}}&-{\frac {642}{139}}&-{\frac {450}{139}}&0\\ 0&0&0&{\frac {213}{46}}&{\frac {31}{46}}&-{\frac {213}{46}}&0\\ 0&0&0&{\frac {108}{23}}&{\frac {52}{23}}&-{\frac {108}{23}}&0\\ 0&0&0&{\frac {105}{46}}&-{\frac {21}{46}}&-{\frac {105}{46}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {72 R_{4}}{71} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&0&{\frac {920}{139}}&{\frac {450}{139}}&-{\frac {642}{139}}&-{\frac {450}{139}}&0\\ 0&0&0&{\frac {213}{46}}&{\frac {31}{46}}&-{\frac {213}{46}}&0\\ 0&0&0&0&{\frac {112}{71}}&0&0\\ 0&0&0&{\frac {105}{46}}&-{\frac {21}{46}}&-{\frac {105}{46}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-\frac {35 R_{4}}{71} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&0&{\frac {920}{139}}&{\frac {450}{139}}&-{\frac {642}{139}}&-{\frac {450}{139}}&0\\ 0&0&0&{\frac {213}{46}}&{\frac {31}{46}}&-{\frac {213}{46}}&0\\ 0&0&0&0&{\frac {112}{71}}&0&0\\ 0&0&0&0&-{\frac {56}{71}}&0&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}+\frac {R_{5}}{2} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 6&13&0&0&0&-13&0\\ 0&{\frac {139}{3}}&-10&-20&10&-{\frac {79}{3}}&0\\ 0&0&{\frac {920}{139}}&{\frac {450}{139}}&-{\frac {642}{139}}&-{\frac {450}{139}}&0\\ 0&0&0&{\frac {213}{46}}&{\frac {31}{46}}&-{\frac {213}{46}}&0\\ 0&0&0&0&{\frac {112}{71}}&0&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{cccccc} 6 & 13 & 0 & 0 & 0 & -13 \\ 0 & \frac {139}{3} & -10 & -20 & 10 & -\frac {79}{3} \\ 0 & 0 & \frac {920}{139} & \frac {450}{139} & -\frac {642}{139} & -\frac {450}{139} \\ 0 & 0 & 0 & \frac {213}{46} & \frac {31}{46} & -\frac {213}{46} \\ 0 & 0 & 0 & 0 & \frac {112}{71} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{6}\}\) and the leading variables are \(\{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\}\). Let \(v_{6} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\{v_{1} = 0, v_{2} = t, v_{3} = 0, v_{4} = t, v_{5} = 0\}\)
Hence the solution is \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = \left [\begin {array}{c} 0 \\ t \\ 0 \\ t \\ 0 \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = t \left [\begin {array}{c} 0 \\ 1 \\ 0 \\ 1 \\ 0 \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = \left [\begin {array}{c} 0 \\ 1 \\ 0 \\ 1 \\ 0 \\ 1 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{4} = 5\)
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}{cccccc} 9 & 13 & 0 & 0 & 0 & -13 \\ -14 & 19 & -10 & -20 & 10 & 4 \\ -30 & 12 & -7 & -30 & 12 & 18 \\ -12 & 10 & -10 & -9 & 10 & 2 \\ 6 & 9 & 0 & 6 & 5 & -15 \\ -14 & 23 & -10 & -20 & 10 & 0 \end {array}\right ] - \left (5\right ) \left [\begin {array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{cccccc} 4 & 13 & 0 & 0 & 0 & -13 \\ -14 & 14 & -10 & -20 & 10 & 4 \\ -30 & 12 & -12 & -30 & 12 & 18 \\ -12 & 10 & -10 & -14 & 10 & 2 \\ 6 & 9 & 0 & 6 & 0 & -15 \\ -14 & 23 & -10 & -20 & 10 & -5 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ -14&14&-10&-20&10&4&0\\ -30&12&-12&-30&12&18&0\\ -12&10&-10&-14&10&2&0\\ 6&9&0&6&0&-15&0\\ -14&23&-10&-20&10&-5&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}+\frac {7 R_{1}}{2} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ -30&12&-12&-30&12&18&0\\ -12&10&-10&-14&10&2&0\\ 6&9&0&6&0&-15&0\\ -14&23&-10&-20&10&-5&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}+\frac {15 R_{1}}{2} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&{\frac {219}{2}}&-12&-30&12&-{\frac {159}{2}}&0\\ -12&10&-10&-14&10&2&0\\ 6&9&0&6&0&-15&0\\ -14&23&-10&-20&10&-5&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}+3 R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&{\frac {219}{2}}&-12&-30&12&-{\frac {159}{2}}&0\\ 0&49&-10&-14&10&-37&0\\ 6&9&0&6&0&-15&0\\ -14&23&-10&-20&10&-5&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {3 R_{1}}{2} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&{\frac {219}{2}}&-12&-30&12&-{\frac {159}{2}}&0\\ 0&49&-10&-14&10&-37&0\\ 0&-{\frac {21}{2}}&0&6&0&{\frac {9}{2}}&0\\ -14&23&-10&-20&10&-5&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}+\frac {7 R_{1}}{2} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&{\frac {219}{2}}&-12&-30&12&-{\frac {159}{2}}&0\\ 0&49&-10&-14&10&-37&0\\ 0&-{\frac {21}{2}}&0&6&0&{\frac {9}{2}}&0\\ 0&{\frac {137}{2}}&-10&-20&10&-{\frac {101}{2}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {219 R_{2}}{119} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&0&{\frac {762}{119}}&{\frac {810}{119}}&-{\frac {762}{119}}&-{\frac {372}{119}}&0\\ 0&49&-10&-14&10&-37&0\\ 0&-{\frac {21}{2}}&0&6&0&{\frac {9}{2}}&0\\ 0&{\frac {137}{2}}&-10&-20&10&-{\frac {101}{2}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}-\frac {14 R_{2}}{17} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&0&{\frac {762}{119}}&{\frac {810}{119}}&-{\frac {762}{119}}&-{\frac {372}{119}}&0\\ 0&0&-{\frac {30}{17}}&{\frac {42}{17}}&{\frac {30}{17}}&-{\frac {48}{17}}&0\\ 0&-{\frac {21}{2}}&0&6&0&{\frac {9}{2}}&0\\ 0&{\frac {137}{2}}&-10&-20&10&-{\frac {101}{2}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}+\frac {3 R_{2}}{17} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&0&{\frac {762}{119}}&{\frac {810}{119}}&-{\frac {762}{119}}&-{\frac {372}{119}}&0\\ 0&0&-{\frac {30}{17}}&{\frac {42}{17}}&{\frac {30}{17}}&-{\frac {48}{17}}&0\\ 0&0&-{\frac {30}{17}}&{\frac {42}{17}}&{\frac {30}{17}}&-{\frac {48}{17}}&0\\ 0&{\frac {137}{2}}&-10&-20&10&-{\frac {101}{2}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-\frac {137 R_{2}}{119} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&0&{\frac {762}{119}}&{\frac {810}{119}}&-{\frac {762}{119}}&-{\frac {372}{119}}&0\\ 0&0&-{\frac {30}{17}}&{\frac {42}{17}}&{\frac {30}{17}}&-{\frac {48}{17}}&0\\ 0&0&-{\frac {30}{17}}&{\frac {42}{17}}&{\frac {30}{17}}&-{\frac {48}{17}}&0\\ 0&0&{\frac {180}{119}}&{\frac {360}{119}}&-{\frac {180}{119}}&-{\frac {324}{119}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}+\frac {35 R_{3}}{127} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&0&{\frac {762}{119}}&{\frac {810}{119}}&-{\frac {762}{119}}&-{\frac {372}{119}}&0\\ 0&0&0&{\frac {552}{127}}&0&-{\frac {468}{127}}&0\\ 0&0&-{\frac {30}{17}}&{\frac {42}{17}}&{\frac {30}{17}}&-{\frac {48}{17}}&0\\ 0&0&{\frac {180}{119}}&{\frac {360}{119}}&-{\frac {180}{119}}&-{\frac {324}{119}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}+\frac {35 R_{3}}{127} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&0&{\frac {762}{119}}&{\frac {810}{119}}&-{\frac {762}{119}}&-{\frac {372}{119}}&0\\ 0&0&0&{\frac {552}{127}}&0&-{\frac {468}{127}}&0\\ 0&0&0&{\frac {552}{127}}&0&-{\frac {468}{127}}&0\\ 0&0&{\frac {180}{119}}&{\frac {360}{119}}&-{\frac {180}{119}}&-{\frac {324}{119}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-\frac {30 R_{3}}{127} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&0&{\frac {762}{119}}&{\frac {810}{119}}&-{\frac {762}{119}}&-{\frac {372}{119}}&0\\ 0&0&0&{\frac {552}{127}}&0&-{\frac {468}{127}}&0\\ 0&0&0&{\frac {552}{127}}&0&-{\frac {468}{127}}&0\\ 0&0&0&{\frac {180}{127}}&0&-{\frac {252}{127}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-R_{4} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&0&{\frac {762}{119}}&{\frac {810}{119}}&-{\frac {762}{119}}&-{\frac {372}{119}}&0\\ 0&0&0&{\frac {552}{127}}&0&-{\frac {468}{127}}&0\\ 0&0&0&0&0&0&0\\ 0&0&0&{\frac {180}{127}}&0&-{\frac {252}{127}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-\frac {15 R_{4}}{46} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&0&{\frac {762}{119}}&{\frac {810}{119}}&-{\frac {762}{119}}&-{\frac {372}{119}}&0\\ 0&0&0&{\frac {552}{127}}&0&-{\frac {468}{127}}&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&-{\frac {18}{23}}&0 \end {array} \right ] \end {align*}
Since the current pivot \(A(5,6)\) is zero, then the current pivot row is replaced with a row with a non-zero pivot. Swapping row \(5\) and row \(6\) gives \[ \left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 4&13&0&0&0&-13&0\\ 0&{\frac {119}{2}}&-10&-20&10&-{\frac {83}{2}}&0\\ 0&0&{\frac {762}{119}}&{\frac {810}{119}}&-{\frac {762}{119}}&-{\frac {372}{119}}&0\\ 0&0&0&{\frac {552}{127}}&0&-{\frac {468}{127}}&0\\ 0&0&0&0&0&-{\frac {18}{23}}&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \] Therefore the system in Echelon form is \[ \left [\begin {array}{cccccc} 4 & 13 & 0 & 0 & 0 & -13 \\ 0 & \frac {119}{2} & -10 & -20 & 10 & -\frac {83}{2} \\ 0 & 0 & \frac {762}{119} & \frac {810}{119} & -\frac {762}{119} & -\frac {372}{119} \\ 0 & 0 & 0 & \frac {552}{127} & 0 & -\frac {468}{127} \\ 0 & 0 & 0 & 0 & 0 & -\frac {18}{23} \\ 0 & 0 & 0 & 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{5}\}\) and the leading variables are \(\{v_{1}, v_{2}, v_{3}, v_{4}, v_{6}\}\). Let \(v_{5} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\{v_{1} = 0, v_{2} = 0, v_{3} = t, v_{4} = 0, v_{6} = 0\}\)
Hence the solution is \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ t \\ v_{6} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ t \\ 0 \\ t \\ 0 \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ t \\ v_{6} \end {array}\right ] = t \left [\begin {array}{c} 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ t \\ v_{6} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{5} = 9\)
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}{cccccc} 9 & 13 & 0 & 0 & 0 & -13 \\ -14 & 19 & -10 & -20 & 10 & 4 \\ -30 & 12 & -7 & -30 & 12 & 18 \\ -12 & 10 & -10 & -9 & 10 & 2 \\ 6 & 9 & 0 & 6 & 5 & -15 \\ -14 & 23 & -10 & -20 & 10 & 0 \end {array}\right ] - \left (9\right ) \left [\begin {array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{cccccc} 0 & 13 & 0 & 0 & 0 & -13 \\ -14 & 10 & -10 & -20 & 10 & 4 \\ -30 & 12 & -16 & -30 & 12 & 18 \\ -12 & 10 & -10 & -18 & 10 & 2 \\ 6 & 9 & 0 & 6 & -4 & -15 \\ -14 & 23 & -10 & -20 & 10 & -9 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 0&13&0&0&0&-13&0\\ -14&10&-10&-20&10&4&0\\ -30&12&-16&-30&12&18&0\\ -12&10&-10&-18&10&2&0\\ 6&9&0&6&-4&-15&0\\ -14&23&-10&-20&10&-9&0 \end {array} \right ] \] Since the current pivot \(A(1,1)\) is zero, then the current pivot row is replaced with a row with a non-zero pivot. Swapping row \(1\) and row \(2\) gives \[ \left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ -30&12&-16&-30&12&18&0\\ -12&10&-10&-18&10&2&0\\ 6&9&0&6&-4&-15&0\\ -14&23&-10&-20&10&-9&0 \end {array} \right ] \] \begin {align*} R_{3} = R_{3}-\frac {15 R_{1}}{7} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ 0&-{\frac {66}{7}}&{\frac {38}{7}}&{\frac {90}{7}}&-{\frac {66}{7}}&{\frac {66}{7}}&0\\ -12&10&-10&-18&10&2&0\\ 6&9&0&6&-4&-15&0\\ -14&23&-10&-20&10&-9&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}-\frac {6 R_{1}}{7} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ 0&-{\frac {66}{7}}&{\frac {38}{7}}&{\frac {90}{7}}&-{\frac {66}{7}}&{\frac {66}{7}}&0\\ 0&{\frac {10}{7}}&-{\frac {10}{7}}&-{\frac {6}{7}}&{\frac {10}{7}}&-{\frac {10}{7}}&0\\ 6&9&0&6&-4&-15&0\\ -14&23&-10&-20&10&-9&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}+\frac {3 R_{1}}{7} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ 0&-{\frac {66}{7}}&{\frac {38}{7}}&{\frac {90}{7}}&-{\frac {66}{7}}&{\frac {66}{7}}&0\\ 0&{\frac {10}{7}}&-{\frac {10}{7}}&-{\frac {6}{7}}&{\frac {10}{7}}&-{\frac {10}{7}}&0\\ 0&{\frac {93}{7}}&-{\frac {30}{7}}&-{\frac {18}{7}}&{\frac {2}{7}}&-{\frac {93}{7}}&0\\ -14&23&-10&-20&10&-9&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ 0&-{\frac {66}{7}}&{\frac {38}{7}}&{\frac {90}{7}}&-{\frac {66}{7}}&{\frac {66}{7}}&0\\ 0&{\frac {10}{7}}&-{\frac {10}{7}}&-{\frac {6}{7}}&{\frac {10}{7}}&-{\frac {10}{7}}&0\\ 0&{\frac {93}{7}}&-{\frac {30}{7}}&-{\frac {18}{7}}&{\frac {2}{7}}&-{\frac {93}{7}}&0\\ 0&13&0&0&0&-13&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}+\frac {66 R_{2}}{91} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ 0&0&{\frac {38}{7}}&{\frac {90}{7}}&-{\frac {66}{7}}&0&0\\ 0&{\frac {10}{7}}&-{\frac {10}{7}}&-{\frac {6}{7}}&{\frac {10}{7}}&-{\frac {10}{7}}&0\\ 0&{\frac {93}{7}}&-{\frac {30}{7}}&-{\frac {18}{7}}&{\frac {2}{7}}&-{\frac {93}{7}}&0\\ 0&13&0&0&0&-13&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}-\frac {10 R_{2}}{91} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ 0&0&{\frac {38}{7}}&{\frac {90}{7}}&-{\frac {66}{7}}&0&0\\ 0&0&-{\frac {10}{7}}&-{\frac {6}{7}}&{\frac {10}{7}}&0&0\\ 0&{\frac {93}{7}}&-{\frac {30}{7}}&-{\frac {18}{7}}&{\frac {2}{7}}&-{\frac {93}{7}}&0\\ 0&13&0&0&0&-13&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {93 R_{2}}{91} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ 0&0&{\frac {38}{7}}&{\frac {90}{7}}&-{\frac {66}{7}}&0&0\\ 0&0&-{\frac {10}{7}}&-{\frac {6}{7}}&{\frac {10}{7}}&0&0\\ 0&0&-{\frac {30}{7}}&-{\frac {18}{7}}&{\frac {2}{7}}&0&0\\ 0&13&0&0&0&-13&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-R_{2} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ 0&0&{\frac {38}{7}}&{\frac {90}{7}}&-{\frac {66}{7}}&0&0\\ 0&0&-{\frac {10}{7}}&-{\frac {6}{7}}&{\frac {10}{7}}&0&0\\ 0&0&-{\frac {30}{7}}&-{\frac {18}{7}}&{\frac {2}{7}}&0&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}+\frac {5 R_{3}}{19} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ 0&0&{\frac {38}{7}}&{\frac {90}{7}}&-{\frac {66}{7}}&0&0\\ 0&0&0&{\frac {48}{19}}&-{\frac {20}{19}}&0&0\\ 0&0&-{\frac {30}{7}}&-{\frac {18}{7}}&{\frac {2}{7}}&0&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}+\frac {15 R_{3}}{19} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ 0&0&{\frac {38}{7}}&{\frac {90}{7}}&-{\frac {66}{7}}&0&0\\ 0&0&0&{\frac {48}{19}}&-{\frac {20}{19}}&0&0\\ 0&0&0&{\frac {144}{19}}&-{\frac {136}{19}}&0&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-3 R_{4} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&10&-10&-20&10&4&0\\ 0&13&0&0&0&-13&0\\ 0&0&{\frac {38}{7}}&{\frac {90}{7}}&-{\frac {66}{7}}&0&0\\ 0&0&0&{\frac {48}{19}}&-{\frac {20}{19}}&0&0\\ 0&0&0&0&-4&0&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{cccccc} -14 & 10 & -10 & -20 & 10 & 4 \\ 0 & 13 & 0 & 0 & 0 & -13 \\ 0 & 0 & \frac {38}{7} & \frac {90}{7} & -\frac {66}{7} & 0 \\ 0 & 0 & 0 & \frac {48}{19} & -\frac {20}{19} & 0 \\ 0 & 0 & 0 & 0 & -4 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{6}\}\) and the leading variables are \(\{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\}\). Let \(v_{6} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\{v_{1} = t, v_{2} = t, v_{3} = 0, v_{4} = 0, v_{5} = 0\}\)
Hence the solution is \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = \left [\begin {array}{c} t \\ t \\ 0 \\ 0 \\ 0 \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = t \left [\begin {array}{c} 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ t \end {array}\right ] = \left [\begin {array}{c} 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \end {array}\right ] \] Considering the eigenvalue \(\lambda _{6} = 11\)
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}{cccccc} 9 & 13 & 0 & 0 & 0 & -13 \\ -14 & 19 & -10 & -20 & 10 & 4 \\ -30 & 12 & -7 & -30 & 12 & 18 \\ -12 & 10 & -10 & -9 & 10 & 2 \\ 6 & 9 & 0 & 6 & 5 & -15 \\ -14 & 23 & -10 & -20 & 10 & 0 \end {array}\right ] - \left (11\right ) \left [\begin {array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end {array}\right ]\right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\\ \left [\begin {array}{cccccc} -2 & 13 & 0 & 0 & 0 & -13 \\ -14 & 8 & -10 & -20 & 10 & 4 \\ -30 & 12 & -18 & -30 & 12 & 18 \\ -12 & 10 & -10 & -20 & 10 & 2 \\ 6 & 9 & 0 & 6 & -6 & -15 \\ -14 & 23 & -10 & -20 & 10 & -11 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ]&=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}
Now forward elimination is applied to solve for the eigenvector \(\vec {v}\). The augmented matrix is \[ \left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ -14&8&-10&-20&10&4&0\\ -30&12&-18&-30&12&18&0\\ -12&10&-10&-20&10&2&0\\ 6&9&0&6&-6&-15&0\\ -14&23&-10&-20&10&-11&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}-7 R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ -30&12&-18&-30&12&18&0\\ -12&10&-10&-20&10&2&0\\ 6&9&0&6&-6&-15&0\\ -14&23&-10&-20&10&-11&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-15 R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&-183&-18&-30&12&213&0\\ -12&10&-10&-20&10&2&0\\ 6&9&0&6&-6&-15&0\\ -14&23&-10&-20&10&-11&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}-6 R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&-183&-18&-30&12&213&0\\ 0&-68&-10&-20&10&80&0\\ 6&9&0&6&-6&-15&0\\ -14&23&-10&-20&10&-11&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}+3 R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&-183&-18&-30&12&213&0\\ 0&-68&-10&-20&10&80&0\\ 0&48&0&6&-6&-54&0\\ -14&23&-10&-20&10&-11&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-7 R_{1} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&-183&-18&-30&12&213&0\\ 0&-68&-10&-20&10&80&0\\ 0&48&0&6&-6&-54&0\\ 0&-68&-10&-20&10&80&0 \end {array} \right ] \end {align*}
\begin {align*} R_{3} = R_{3}-\frac {183 R_{2}}{83} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&0&{\frac {336}{83}}&{\frac {1170}{83}}&-{\frac {834}{83}}&{\frac {294}{83}}&0\\ 0&-68&-10&-20&10&80&0\\ 0&48&0&6&-6&-54&0\\ 0&-68&-10&-20&10&80&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}-\frac {68 R_{2}}{83} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&0&{\frac {336}{83}}&{\frac {1170}{83}}&-{\frac {834}{83}}&{\frac {294}{83}}&0\\ 0&0&-{\frac {150}{83}}&-{\frac {300}{83}}&{\frac {150}{83}}&{\frac {180}{83}}&0\\ 0&48&0&6&-6&-54&0\\ 0&-68&-10&-20&10&80&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}+\frac {48 R_{2}}{83} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&0&{\frac {336}{83}}&{\frac {1170}{83}}&-{\frac {834}{83}}&{\frac {294}{83}}&0\\ 0&0&-{\frac {150}{83}}&-{\frac {300}{83}}&{\frac {150}{83}}&{\frac {180}{83}}&0\\ 0&0&-{\frac {480}{83}}&-{\frac {462}{83}}&-{\frac {18}{83}}&{\frac {78}{83}}&0\\ 0&-68&-10&-20&10&80&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-\frac {68 R_{2}}{83} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&0&{\frac {336}{83}}&{\frac {1170}{83}}&-{\frac {834}{83}}&{\frac {294}{83}}&0\\ 0&0&-{\frac {150}{83}}&-{\frac {300}{83}}&{\frac {150}{83}}&{\frac {180}{83}}&0\\ 0&0&-{\frac {480}{83}}&-{\frac {462}{83}}&-{\frac {18}{83}}&{\frac {78}{83}}&0\\ 0&0&-{\frac {150}{83}}&-{\frac {300}{83}}&{\frac {150}{83}}&{\frac {180}{83}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{4} = R_{4}+\frac {25 R_{3}}{56} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&0&{\frac {336}{83}}&{\frac {1170}{83}}&-{\frac {834}{83}}&{\frac {294}{83}}&0\\ 0&0&0&{\frac {75}{28}}&-{\frac {75}{28}}&{\frac {15}{4}}&0\\ 0&0&-{\frac {480}{83}}&-{\frac {462}{83}}&-{\frac {18}{83}}&{\frac {78}{83}}&0\\ 0&0&-{\frac {150}{83}}&-{\frac {300}{83}}&{\frac {150}{83}}&{\frac {180}{83}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}+\frac {10 R_{3}}{7} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&0&{\frac {336}{83}}&{\frac {1170}{83}}&-{\frac {834}{83}}&{\frac {294}{83}}&0\\ 0&0&0&{\frac {75}{28}}&-{\frac {75}{28}}&{\frac {15}{4}}&0\\ 0&0&0&{\frac {102}{7}}&-{\frac {102}{7}}&6&0\\ 0&0&-{\frac {150}{83}}&-{\frac {300}{83}}&{\frac {150}{83}}&{\frac {180}{83}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}+\frac {25 R_{3}}{56} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&0&{\frac {336}{83}}&{\frac {1170}{83}}&-{\frac {834}{83}}&{\frac {294}{83}}&0\\ 0&0&0&{\frac {75}{28}}&-{\frac {75}{28}}&{\frac {15}{4}}&0\\ 0&0&0&{\frac {102}{7}}&-{\frac {102}{7}}&6&0\\ 0&0&0&{\frac {75}{28}}&-{\frac {75}{28}}&{\frac {15}{4}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{5} = R_{5}-\frac {136 R_{4}}{25} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&0&{\frac {336}{83}}&{\frac {1170}{83}}&-{\frac {834}{83}}&{\frac {294}{83}}&0\\ 0&0&0&{\frac {75}{28}}&-{\frac {75}{28}}&{\frac {15}{4}}&0\\ 0&0&0&0&0&-{\frac {72}{5}}&0\\ 0&0&0&{\frac {75}{28}}&-{\frac {75}{28}}&{\frac {15}{4}}&0 \end {array} \right ] \end {align*}
\begin {align*} R_{6} = R_{6}-R_{4} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}cccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -2&13&0&0&0&-13&0\\ 0&-83&-10&-20&10&95&0\\ 0&0&{\frac {336}{83}}&{\frac {1170}{83}}&-{\frac {834}{83}}&{\frac {294}{83}}&0\\ 0&0&0&{\frac {75}{28}}&-{\frac {75}{28}}&{\frac {15}{4}}&0\\ 0&0&0&0&0&-{\frac {72}{5}}&0\\ 0&0&0&0&0&0&0 \end {array} \right ] \end {align*}
Therefore the system in Echelon form is \[ \left [\begin {array}{cccccc} -2 & 13 & 0 & 0 & 0 & -13 \\ 0 & -83 & -10 & -20 & 10 & 95 \\ 0 & 0 & \frac {336}{83} & \frac {1170}{83} & -\frac {834}{83} & \frac {294}{83} \\ 0 & 0 & 0 & \frac {75}{28} & -\frac {75}{28} & \frac {15}{4} \\ 0 & 0 & 0 & 0 & 0 & -\frac {72}{5} \\ 0 & 0 & 0 & 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \\ v_{6} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{5}\}\) and the leading variables are \(\{v_{1}, v_{2}, v_{3}, v_{4}, v_{6}\}\). Let \(v_{5} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\{v_{1} = 0, v_{2} = 0, v_{3} = -t, v_{4} = t, v_{6} = 0\}\)
Hence the solution is \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ t \\ v_{6} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ -t \\ t \\ t \\ 0 \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ t \\ v_{6} \end {array}\right ] = t \left [\begin {array}{c} 0 \\ 0 \\ -1 \\ 1 \\ 1 \\ 0 \end {array}\right ] \] Let \(t = 1\) the eigenvector becomes \[ \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ t \\ v_{6} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ -1 \\ 1 \\ 1 \\ 0 \end {array}\right ] \] 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 |
| \(5\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \end {array}\right ]\) |
| \(9\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \end {array}\right ]\) |
| \(3\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} 0 \\ 1 \\ 0 \\ 1 \\ 0 \\ 1 \end {array}\right ]\) |
| \(11\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} 0 \\ 0 \\ -1 \\ 1 \\ 1 \\ 0 \end {array}\right ]\) |
| \(-7\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} 0 \\ 1 \\ 1 \\ 1 \\ 0 \\ 1 \end {array}\right ]\) |
| \(-4\) | \(1\) | \(1\) | No | \(\left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \end {array}\right ]\) |
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 \(5\) is real and distinct then the corresponding eigenvector solution is \begin {align*} \vec {x}_{1}(t) &= \vec {v}_{1} e^{5 t}\\ &= \left [\begin {array}{c} 0 \\ 0 \\ 1 \\ 0 \\ 1 \\ 0 \end {array}\right ] e^{5 t} \end {align*}
Since eigenvalue \(9\) is real and distinct then the corresponding eigenvector solution is \begin {align*} \vec {x}_{2}(t) &= \vec {v}_{2} e^{9 t}\\ &= \left [\begin {array}{c} 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 1 \end {array}\right ] e^{9 t} \end {align*}
Since eigenvalue \(3\) is real and distinct then the corresponding eigenvector solution is \begin {align*} \vec {x}_{3}(t) &= \vec {v}_{3} e^{3 t}\\ &= \left [\begin {array}{c} 0 \\ 1 \\ 0 \\ 1 \\ 0 \\ 1 \end {array}\right ] e^{3 t} \end {align*}
Since eigenvalue \(11\) is real and distinct then the corresponding eigenvector solution is \begin {align*} \vec {x}_{4}(t) &= \vec {v}_{4} e^{11 t}\\ &= \left [\begin {array}{c} 0 \\ 0 \\ -1 \\ 1 \\ 1 \\ 0 \end {array}\right ] e^{11 t} \end {align*}
Since eigenvalue \(-7\) is real and distinct then the corresponding eigenvector solution is \begin {align*} \vec {x}_{5}(t) &= \vec {v}_{5} e^{-7 t}\\ &= \left [\begin {array}{c} 0 \\ 1 \\ 1 \\ 1 \\ 0 \\ 1 \end {array}\right ] e^{-7 t} \end {align*}
Since eigenvalue \(-4\) is real and distinct then the corresponding eigenvector solution is \begin {align*} \vec {x}_{6}(t) &= \vec {v}_{6} e^{-4 t}\\ &= \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \end {array}\right ] e^{-4 t} \end {align*}
Therefore the final 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) + c_{4} \vec {x}_{4}(t) + c_{5} \vec {x}_{5}(t) + c_{6} \vec {x}_{6}(t) \end {align*}
Which is written as \begin {align*} \left [\begin {array}{c} x_{1} \left (t \right ) \\ x_{2} \left (t \right ) \\ x_{3} \left (t \right ) \\ x_{4} \left (t \right ) \\ x_{5} \left (t \right ) \\ x_{6} \left (t \right ) \end {array}\right ] &= c_{1} \left [\begin {array}{c} 0 \\ 0 \\ {\mathrm e}^{5 t} \\ 0 \\ {\mathrm e}^{5 t} \\ 0 \end {array}\right ] + c_{2} \left [\begin {array}{c} {\mathrm e}^{9 t} \\ {\mathrm e}^{9 t} \\ 0 \\ 0 \\ 0 \\ {\mathrm e}^{9 t} \end {array}\right ] + c_{3} \left [\begin {array}{c} 0 \\ {\mathrm e}^{3 t} \\ 0 \\ {\mathrm e}^{3 t} \\ 0 \\ {\mathrm e}^{3 t} \end {array}\right ] + c_{4} \left [\begin {array}{c} 0 \\ 0 \\ -{\mathrm e}^{11 t} \\ {\mathrm e}^{11 t} \\ {\mathrm e}^{11 t} \\ 0 \end {array}\right ] + c_{5} \left [\begin {array}{c} 0 \\ {\mathrm e}^{-7 t} \\ {\mathrm e}^{-7 t} \\ {\mathrm e}^{-7 t} \\ 0 \\ {\mathrm e}^{-7 t} \end {array}\right ] + c_{6} \left [\begin {array}{c} {\mathrm e}^{-4 t} \\ 0 \\ 0 \\ 0 \\ {\mathrm e}^{-4 t} \\ {\mathrm e}^{-4 t} \end {array}\right ] \end {align*}
Which becomes \begin {align*} \left [\begin {array}{c} x_{1} \left (t \right ) \\ x_{2} \left (t \right ) \\ x_{3} \left (t \right ) \\ x_{4} \left (t \right ) \\ x_{5} \left (t \right ) \\ x_{6} \left (t \right ) \end {array}\right ] = \left [\begin {array}{c} \left (c_{2} {\mathrm e}^{13 t}+c_{6}\right ) {\mathrm e}^{-4 t} \\ \left (c_{2} {\mathrm e}^{16 t}+c_{3} {\mathrm e}^{10 t}+c_{5}\right ) {\mathrm e}^{-7 t} \\ \left (-c_{4} {\mathrm e}^{18 t}+c_{1} {\mathrm e}^{12 t}+c_{5}\right ) {\mathrm e}^{-7 t} \\ \left (c_{4} {\mathrm e}^{18 t}+c_{3} {\mathrm e}^{10 t}+c_{5}\right ) {\mathrm e}^{-7 t} \\ \left (c_{4} {\mathrm e}^{15 t}+c_{1} {\mathrm e}^{9 t}+c_{6}\right ) {\mathrm e}^{-4 t} \\ \left (c_{2} {\mathrm e}^{16 t}+c_{3} {\mathrm e}^{10 t}+c_{6} {\mathrm e}^{3 t}+c_{5}\right ) {\mathrm e}^{-7 t} \end {array}\right ] \end {align*}
✓ Solution by Maple
Time used: 0.141 (sec). Leaf size: 135
dsolve([diff(x__1(t),t)=9*x__1(t)+13*x__2(t)+0*x__3(t)+0*x__4(t)+0*x__5(t)-13*x__6(t),diff(x__2(t),t)=-14*x__1(t)+19*x__2(t)-10*x__3(t)-20*x__4(t)+10*x__5(t)+4*x__6(t),diff(x__3(t),t)=-30*x__1(t)+12*x__2(t)-7*x__3(t)-30*x__4(t)+12*x__5(t)+18*x__6(t),diff(x__4(t),t)=-12*x__1(t)+10*x__2(t)-10*x__3(t)-9*x__4(t)+10*x__5(t)+2*x__6(t),diff(x__5(t),t)=6*x__1(t)+9*x__2(t)+0*x__3(t)+6*x__4(t)+5*x__5(t)-15*x__6(t),diff(x__6(t),t)=-14*x__1(t)+23*x__2(t)-10*x__3(t)-20*x__4(t)+10*x__5(t)+0*x__6(t)],singsol=all)
\begin{align*} x_{1} \left (t \right ) &= c_{5} {\mathrm e}^{-4 t}+c_{6} {\mathrm e}^{9 t} \\ x_{2} \left (t \right ) &= c_{6} {\mathrm e}^{9 t}+c_{4} {\mathrm e}^{3 t}+{\mathrm e}^{-7 t} c_{3} \\ x_{3} \left (t \right ) &= {\mathrm e}^{-7 t} c_{3} -{\mathrm e}^{11 t} c_{2} +{\mathrm e}^{5 t} c_{1} \\ x_{4} \left (t \right ) &= {\mathrm e}^{11 t} c_{2} +c_{4} {\mathrm e}^{3 t}+{\mathrm e}^{-7 t} c_{3} \\ x_{5} \left (t \right ) &= {\mathrm e}^{11 t} c_{2} +{\mathrm e}^{5 t} c_{1} +c_{5} {\mathrm e}^{-4 t} \\ x_{6} \left (t \right ) &= c_{6} {\mathrm e}^{9 t}+c_{5} {\mathrm e}^{-4 t}+c_{4} {\mathrm e}^{3 t}+{\mathrm e}^{-7 t} c_{3} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.107 (sec). Leaf size: 1882
DSolve[{x1'[t]==9*x1[t]+13*x2[t]-13*x6[t],x2'[t]==-14*x1[t]+19*x2[t]-10*x3[t]-20*x4[t]+10*x5[t]+4*x6[t],x3'[t]==-30*x1[t]+12*x2[t]-7*x3[t]-30*x4[t]+12*x5[t]+18*x6[t],x4'[t]==-12*x1[t]+10*x2[t]-10*x3[t]-9*x4[t]+10*x5[t]+2*x6[t],x5'[t]==6*x1[t]+9*x2[t]+6*x4[t]+5*x5[t]-15*x6[t],x6'[t]==-14*x1[t]+23*x2[t]-10*x3[t]-20*x4[t]-10*x5[t]},{x1[t],x2[t],x3[t],x4[t],x5[t],x6[t]},t,IncludeSingularSolutions -> True]
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