18.8 problem 7

Internal problem ID [12837]
Internal file name [OUTPUT/11490_Saturday_November_04_2023_08_47_36_AM_4836427/index.tex]

Book: Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section: Chapter 8. Linear Systems of First-Order Differential Equations. Exercises 8.2 page 362
Problem number: 7.
ODE order: 4.
ODE degree: 1.

The type(s) of ODE detected by this program : "find eigenvalues and eigenvectors"

Find the eigenvalues and associated eigenvectors of the matrix \[ \left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] \] The first step is to determine the characteristic polynomial of the matrix in order to find the eigenvalues of the matrix \(A\). This is given by \begin {align*} \det (A-\lambda I) &= 0 \\ \det \left (\left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] - \lambda \left [\begin {array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end {array}\right ]\right ) &= 0 \\ \det \left [\begin {array}{ccccc} 1-\lambda & 3 & 5 & 2 & 4 \\ 5 & 2-\lambda & 4 & 1 & 3 \\ 4 & 1 & 3-\lambda & 5 & 2 \\ 3 & 5 & 2 & 4-\lambda & 1 \\ 2 & 4 & 1 & 3 & 5-\lambda \end {array}\right ] &= 0 \\ -\lambda ^{5}+15 \lambda ^{4}+125 \lambda -1875 &= 0 \end {align*}

The eigenvalues are the roots of the above characteristic polynomial. Solving for the roots gives \begin {align*} \lambda _1&=15\\ \lambda _2&=5^{\frac {3}{4}}\\ \lambda _3&=i 5^{\frac {3}{4}}\\ \lambda _4&=-5^{\frac {3}{4}}\\ \lambda _5&=-i 5^{\frac {3}{4}} \end {align*}

This table summarises the above result

eigenvalue	algebraic multiplicity	type of eigenvalue
\(15\)	\(1\)	real eigenvalue
\(5^{\frac {3}{4}}\)	\(1\)	real eigenvalue
\(-i 5^{\frac {3}{4}}\)	\(1\)	complex eigenvalue
\(-5^{\frac {3}{4}}\)	\(1\)	real eigenvalue
\(i 5^{\frac {3}{4}}\)	\(1\)	complex eigenvalue

For each eigenvalue \(\lambda \) found above, we now find the corresponding eigenvector. Considering \(\lambda = 15\)

We need now to determine the eigenvector \(\boldsymbol {v}\) where \begin {align*} A \boldsymbol {v} &= \lambda \boldsymbol {v} \\ A \boldsymbol {v} - \lambda \boldsymbol {v} &= \boldsymbol {0} \\ (A - \lambda I ) \boldsymbol {v} &= \boldsymbol {0} \\ \left (\left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] - \left (15\right ) \left [\begin {array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end {array}\right ] \right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \left (\left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] - \left [\begin {array}{ccccc} 15 & 0 & 0 & 0 & 0 \\ 0 & 15 & 0 & 0 & 0 \\ 0 & 0 & 15 & 0 & 0 \\ 0 & 0 & 0 & 15 & 0 \\ 0 & 0 & 0 & 0 & 15 \end {array}\right ] \right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \left [\begin {array}{ccccc} -14 & 3 & 5 & 2 & 4 \\ 5 & -13 & 4 & 1 & 3 \\ 4 & 1 & -12 & 5 & 2 \\ 3 & 5 & 2 & -11 & 1 \\ 2 & 4 & 1 & 3 & -10 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}

We will now do Gaussian elimination in order to solve for the eigenvector. The augmented matrix is \[ \left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&3&5&2&4&0\\ 5&-13&4&1&3&0\\ 4&1&-12&5&2&0\\ 3&5&2&-11&1&0\\ 2&4&1&3&-10&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}+\frac {5 R_{1}}{14} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&3&5&2&4&0\\ 0&-{\frac {167}{14}}&{\frac {81}{14}}&{\frac {12}{7}}&{\frac {31}{7}}&0\\ 4&1&-12&5&2&0\\ 3&5&2&-11&1&0\\ 2&4&1&3&-10&0 \end {array} \right ] \end {align*}

\begin {align*} R_{3} = R_{3}+\frac {2 R_{1}}{7} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&3&5&2&4&0\\ 0&-{\frac {167}{14}}&{\frac {81}{14}}&{\frac {12}{7}}&{\frac {31}{7}}&0\\ 0&{\frac {13}{7}}&-{\frac {74}{7}}&{\frac {39}{7}}&{\frac {22}{7}}&0\\ 3&5&2&-11&1&0\\ 2&4&1&3&-10&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}+\frac {3 R_{1}}{14} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&3&5&2&4&0\\ 0&-{\frac {167}{14}}&{\frac {81}{14}}&{\frac {12}{7}}&{\frac {31}{7}}&0\\ 0&{\frac {13}{7}}&-{\frac {74}{7}}&{\frac {39}{7}}&{\frac {22}{7}}&0\\ 0&{\frac {79}{14}}&{\frac {43}{14}}&-{\frac {74}{7}}&{\frac {13}{7}}&0\\ 2&4&1&3&-10&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}+\frac {R_{1}}{7} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&3&5&2&4&0\\ 0&-{\frac {167}{14}}&{\frac {81}{14}}&{\frac {12}{7}}&{\frac {31}{7}}&0\\ 0&{\frac {13}{7}}&-{\frac {74}{7}}&{\frac {39}{7}}&{\frac {22}{7}}&0\\ 0&{\frac {79}{14}}&{\frac {43}{14}}&-{\frac {74}{7}}&{\frac {13}{7}}&0\\ 0&{\frac {31}{7}}&{\frac {12}{7}}&{\frac {23}{7}}&-{\frac {66}{7}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{3} = R_{3}+\frac {26 R_{2}}{167} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&3&5&2&4&0\\ 0&-{\frac {167}{14}}&{\frac {81}{14}}&{\frac {12}{7}}&{\frac {31}{7}}&0\\ 0&0&-{\frac {1615}{167}}&{\frac {975}{167}}&{\frac {640}{167}}&0\\ 0&{\frac {79}{14}}&{\frac {43}{14}}&-{\frac {74}{7}}&{\frac {13}{7}}&0\\ 0&{\frac {31}{7}}&{\frac {12}{7}}&{\frac {23}{7}}&-{\frac {66}{7}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}+\frac {79 R_{2}}{167} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&3&5&2&4&0\\ 0&-{\frac {167}{14}}&{\frac {81}{14}}&{\frac {12}{7}}&{\frac {31}{7}}&0\\ 0&0&-{\frac {1615}{167}}&{\frac {975}{167}}&{\frac {640}{167}}&0\\ 0&0&{\frac {970}{167}}&-{\frac {1630}{167}}&{\frac {660}{167}}&0\\ 0&{\frac {31}{7}}&{\frac {12}{7}}&{\frac {23}{7}}&-{\frac {66}{7}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}+\frac {62 R_{2}}{167} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&3&5&2&4&0\\ 0&-{\frac {167}{14}}&{\frac {81}{14}}&{\frac {12}{7}}&{\frac {31}{7}}&0\\ 0&0&-{\frac {1615}{167}}&{\frac {975}{167}}&{\frac {640}{167}}&0\\ 0&0&{\frac {970}{167}}&-{\frac {1630}{167}}&{\frac {660}{167}}&0\\ 0&0&{\frac {645}{167}}&{\frac {655}{167}}&-{\frac {1300}{167}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}+\frac {194 R_{3}}{323} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&3&5&2&4&0\\ 0&-{\frac {167}{14}}&{\frac {81}{14}}&{\frac {12}{7}}&{\frac {31}{7}}&0\\ 0&0&-{\frac {1615}{167}}&{\frac {975}{167}}&{\frac {640}{167}}&0\\ 0&0&0&-{\frac {2020}{323}}&{\frac {2020}{323}}&0\\ 0&0&{\frac {645}{167}}&{\frac {655}{167}}&-{\frac {1300}{167}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}+\frac {129 R_{3}}{323} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&3&5&2&4&0\\ 0&-{\frac {167}{14}}&{\frac {81}{14}}&{\frac {12}{7}}&{\frac {31}{7}}&0\\ 0&0&-{\frac {1615}{167}}&{\frac {975}{167}}&{\frac {640}{167}}&0\\ 0&0&0&-{\frac {2020}{323}}&{\frac {2020}{323}}&0\\ 0&0&0&{\frac {2020}{323}}&-{\frac {2020}{323}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}+R_{4} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} -14&3&5&2&4&0\\ 0&-{\frac {167}{14}}&{\frac {81}{14}}&{\frac {12}{7}}&{\frac {31}{7}}&0\\ 0&0&-{\frac {1615}{167}}&{\frac {975}{167}}&{\frac {640}{167}}&0\\ 0&0&0&-{\frac {2020}{323}}&{\frac {2020}{323}}&0\\ 0&0&0&0&0&0 \end {array} \right ] \end {align*}

Therefore the system in Echelon form is \[ \left [\begin {array}{ccccc} -14 & 3 & 5 & 2 & 4 \\ 0 & -\frac {167}{14} & \frac {81}{14} & \frac {12}{7} & \frac {31}{7} \\ 0 & 0 & -\frac {1615}{167} & \frac {975}{167} & \frac {640}{167} \\ 0 & 0 & 0 & -\frac {2020}{323} & \frac {2020}{323} \\ 0 & 0 & 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{5}\}\) and the leading variables are \(\{v_{1}, v_{2}, v_{3}, v_{4}\}\). Let \(v_{5} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\{v_{1} = t, v_{2} = t, v_{3} = t, v_{4} = t\}\)

Hence the solution is \[ \left [\begin {array}{c} t \\ t \\ t \\ t \\ t \end {array}\right ] = \left [\begin {array}{c} t \\ t \\ t \\ t \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} t \\ t \\ t \\ t \\ t \end {array}\right ] = t \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ] \] Or, by letting \(t = 1\) then the eigenvector is \[ \left [\begin {array}{c} t \\ t \\ t \\ t \\ t \end {array}\right ] = \left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ] \] Considering \(\lambda = 5^{\frac {3}{4}}\)

We need now to determine the eigenvector \(\boldsymbol {v}\) where \begin {align*} A \boldsymbol {v} &= \lambda \boldsymbol {v} \\ A \boldsymbol {v} - \lambda \boldsymbol {v} &= \boldsymbol {0} \\ (A - \lambda I ) \boldsymbol {v} &= \boldsymbol {0} \\ \left (\left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] - \left (5^{\frac {3}{4}}\right ) \left [\begin {array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end {array}\right ] \right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \left (\left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] - \left [\begin {array}{ccccc} 5^{\frac {3}{4}} & 0 & 0 & 0 & 0 \\ 0 & 5^{\frac {3}{4}} & 0 & 0 & 0 \\ 0 & 0 & 5^{\frac {3}{4}} & 0 & 0 \\ 0 & 0 & 0 & 5^{\frac {3}{4}} & 0 \\ 0 & 0 & 0 & 0 & 5^{\frac {3}{4}} \end {array}\right ] \right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \left [\begin {array}{ccccc} 1-5^{\frac {3}{4}} & 3 & 5 & 2 & 4 \\ 5 & 2-5^{\frac {3}{4}} & 4 & 1 & 3 \\ 4 & 1 & 3-5^{\frac {3}{4}} & 5 & 2 \\ 3 & 5 & 2 & 4-5^{\frac {3}{4}} & 1 \\ 2 & 4 & 1 & 3 & 5-5^{\frac {3}{4}} \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}

We will now do Gaussian elimination in order to solve for the eigenvector. The augmented matrix is \[ \left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-5^{\frac {3}{4}}&3&5&2&4&0\\ 5&2-5^{\frac {3}{4}}&4&1&3&0\\ 4&1&3-5^{\frac {3}{4}}&5&2&0\\ 3&5&2&4-5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5-5^{\frac {3}{4}}&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}-\frac {5 R_{1}}{1-5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {4 \,5^{\frac {3}{4}}+21}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&0\\ 4&1&3-5^{\frac {3}{4}}&5&2&0\\ 3&5&2&4-5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5-5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{3} = R_{3}-\frac {4 R_{1}}{1-5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {4 \,5^{\frac {3}{4}}+21}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&0\\ 0&\frac {5^{\frac {3}{4}}+11}{5^{\frac {3}{4}}-1}&\frac {-5 \sqrt {5}+4 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&\frac {5 \,5^{\frac {3}{4}}+3}{5^{\frac {3}{4}}-1}&\frac {2 \,5^{\frac {3}{4}}+14}{5^{\frac {3}{4}}-1}&0\\ 3&5&2&4-5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5-5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {3 R_{1}}{1-5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {4 \,5^{\frac {3}{4}}+21}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&0\\ 0&\frac {5^{\frac {3}{4}}+11}{5^{\frac {3}{4}}-1}&\frac {-5 \sqrt {5}+4 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&\frac {5 \,5^{\frac {3}{4}}+3}{5^{\frac {3}{4}}-1}&\frac {2 \,5^{\frac {3}{4}}+14}{5^{\frac {3}{4}}-1}&0\\ 0&\frac {5 \,5^{\frac {3}{4}}+4}{5^{\frac {3}{4}}-1}&\frac {2 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {-5 \sqrt {5}+5 \,5^{\frac {3}{4}}+2}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+11}{5^{\frac {3}{4}}-1}&0\\ 2&4&1&3&5-5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {2 R_{1}}{1-5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {4 \,5^{\frac {3}{4}}+21}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&0\\ 0&\frac {5^{\frac {3}{4}}+11}{5^{\frac {3}{4}}-1}&\frac {-5 \sqrt {5}+4 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&\frac {5 \,5^{\frac {3}{4}}+3}{5^{\frac {3}{4}}-1}&\frac {2 \,5^{\frac {3}{4}}+14}{5^{\frac {3}{4}}-1}&0\\ 0&\frac {5 \,5^{\frac {3}{4}}+4}{5^{\frac {3}{4}}-1}&\frac {2 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {-5 \sqrt {5}+5 \,5^{\frac {3}{4}}+2}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+11}{5^{\frac {3}{4}}-1}&0\\ 0&\frac {4 \,5^{\frac {3}{4}}+2}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+1}{5^{\frac {3}{4}}-1}&\frac {-5 \sqrt {5}+6 \,5^{\frac {3}{4}}+3}{5^{\frac {3}{4}}-1}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{3} = R_{3}-\frac {\left (5^{\frac {3}{4}}+11\right ) R_{2}}{-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {4 \,5^{\frac {3}{4}}+21}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&0\\ 0&0&\frac {-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {55 \sqrt {5}-125 \,5^{\frac {1}{4}}+54 \,5^{\frac {3}{4}}-60}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-55 \sqrt {5}-50 \,5^{\frac {1}{4}}+18 \,5^{\frac {3}{4}}-5}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0\\ 0&\frac {5 \,5^{\frac {3}{4}}+4}{5^{\frac {3}{4}}-1}&\frac {2 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {-5 \sqrt {5}+5 \,5^{\frac {3}{4}}+2}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+11}{5^{\frac {3}{4}}-1}&0\\ 0&\frac {4 \,5^{\frac {3}{4}}+2}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+1}{5^{\frac {3}{4}}-1}&\frac {-5 \sqrt {5}+6 \,5^{\frac {3}{4}}+3}{5^{\frac {3}{4}}-1}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {\left (5 \,5^{\frac {3}{4}}+4\right ) R_{2}}{-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {4 \,5^{\frac {3}{4}}+21}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&0\\ 0&0&\frac {-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {55 \sqrt {5}-125 \,5^{\frac {1}{4}}+54 \,5^{\frac {3}{4}}-60}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-55 \sqrt {5}-50 \,5^{\frac {1}{4}}+18 \,5^{\frac {3}{4}}-5}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0\\ 0&0&\frac {-135 \sqrt {5}-50 \,5^{\frac {1}{4}}-56 \,5^{\frac {3}{4}}+85}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-25 \sqrt {5}-200 \,5^{\frac {1}{4}}+22 \,5^{\frac {3}{4}}+115}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-115 \sqrt {5}-25 \,5^{\frac {1}{4}}-51 \,5^{\frac {3}{4}}+75}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0\\ 0&\frac {4 \,5^{\frac {3}{4}}+2}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+1}{5^{\frac {3}{4}}-1}&\frac {-5 \sqrt {5}+6 \,5^{\frac {3}{4}}+3}{5^{\frac {3}{4}}-1}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {\left (4 \,5^{\frac {3}{4}}+2\right ) R_{2}}{-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {4 \,5^{\frac {3}{4}}+21}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&0\\ 0&0&\frac {-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {55 \sqrt {5}-125 \,5^{\frac {1}{4}}+54 \,5^{\frac {3}{4}}-60}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-55 \sqrt {5}-50 \,5^{\frac {1}{4}}+18 \,5^{\frac {3}{4}}-5}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0\\ 0&0&\frac {-135 \sqrt {5}-50 \,5^{\frac {1}{4}}-56 \,5^{\frac {3}{4}}+85}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-25 \sqrt {5}-200 \,5^{\frac {1}{4}}+22 \,5^{\frac {3}{4}}+115}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-115 \sqrt {5}-25 \,5^{\frac {1}{4}}-51 \,5^{\frac {3}{4}}+75}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0\\ 0&0&\frac {-110 \sqrt {5}-25 \,5^{\frac {1}{4}}-52 \,5^{\frac {3}{4}}+75}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {20 \sqrt {5}-75 \,5^{\frac {1}{4}}+4 \,5^{\frac {3}{4}}-5}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-50 \sqrt {5}-225 \,5^{\frac {1}{4}}+13 \,5^{\frac {3}{4}}+130}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {\left (-135 \sqrt {5}-50 \,5^{\frac {1}{4}}-56 \,5^{\frac {3}{4}}+85\right ) R_{3}}{-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {4 \,5^{\frac {3}{4}}+21}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&0\\ 0&0&\frac {-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {55 \sqrt {5}-125 \,5^{\frac {1}{4}}+54 \,5^{\frac {3}{4}}-60}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-55 \sqrt {5}-50 \,5^{\frac {1}{4}}+18 \,5^{\frac {3}{4}}-5}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0\\ 0&0&0&\frac {19750 \sqrt {5}-9550-500 \,5^{\frac {1}{4}}+11200 \,5^{\frac {3}{4}}}{\left (-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115\right ) \left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-20250 \sqrt {5}[phone] \,5^{\frac {1}{4}}+8550 \,5^{\frac {3}{4}}}{\left (-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115\right ) \left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0\\ 0&0&\frac {-110 \sqrt {5}-25 \,5^{\frac {1}{4}}-52 \,5^{\frac {3}{4}}+75}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {20 \sqrt {5}-75 \,5^{\frac {1}{4}}+4 \,5^{\frac {3}{4}}-5}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-50 \sqrt {5}-225 \,5^{\frac {1}{4}}+13 \,5^{\frac {3}{4}}+130}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {\left (-110 \sqrt {5}-25 \,5^{\frac {1}{4}}-52 \,5^{\frac {3}{4}}+75\right ) R_{3}}{-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {4 \,5^{\frac {3}{4}}+21}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&0\\ 0&0&\frac {-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {55 \sqrt {5}-125 \,5^{\frac {1}{4}}+54 \,5^{\frac {3}{4}}-60}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-55 \sqrt {5}-50 \,5^{\frac {1}{4}}+18 \,5^{\frac {3}{4}}-5}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0\\ 0&0&0&\frac {19750 \sqrt {5}-9550-500 \,5^{\frac {1}{4}}+11200 \,5^{\frac {3}{4}}}{\left (-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115\right ) \left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-20250 \sqrt {5}[phone] \,5^{\frac {1}{4}}+8550 \,5^{\frac {3}{4}}}{\left (-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115\right ) \left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0\\ 0&0&0&-\frac {25 \left (581 \,5^{\frac {3}{4}}-1829 \,5^{\frac {1}{4}}-677 \sqrt {5}+813\right )}{\left (-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115\right ) \left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {28800 \sqrt {5}-52300-66050 \,5^{\frac {1}{4}}+31450 \,5^{\frac {3}{4}}}{\left (-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115\right ) \left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}+\frac {\left (581 \,5^{\frac {3}{4}}-1829 \,5^{\frac {1}{4}}-677 \sqrt {5}+813\right ) R_{4}}{448 \,5^{\frac {3}{4}}-20 \,5^{\frac {1}{4}}+790 \sqrt {5}-382} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1}&\frac {4 \,5^{\frac {3}{4}}+21}{5^{\frac {3}{4}}-1}&\frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1}&\frac {3 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1}&0\\ 0&0&\frac {-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {55 \sqrt {5}-125 \,5^{\frac {1}{4}}+54 \,5^{\frac {3}{4}}-60}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-55 \sqrt {5}-50 \,5^{\frac {1}{4}}+18 \,5^{\frac {3}{4}}-5}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0\\ 0&0&0&\frac {19750 \sqrt {5}-9550-500 \,5^{\frac {1}{4}}+11200 \,5^{\frac {3}{4}}}{\left (-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115\right ) \left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&\frac {-20250 \sqrt {5}[phone] \,5^{\frac {1}{4}}+8550 \,5^{\frac {3}{4}}}{\left (-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115\right ) \left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )}&0\\ 0&0&0&0&0&0 \end {array} \right ] \end {align*}

Therefore the system in Echelon form is \[ \left [\begin {array}{ccccc} 1-5^{\frac {3}{4}} & 3 & 5 & 2 & 4 \\ 0 & \frac {-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13}{5^{\frac {3}{4}}-1} & \frac {4 \,5^{\frac {3}{4}}+21}{5^{\frac {3}{4}}-1} & \frac {5^{\frac {3}{4}}+9}{5^{\frac {3}{4}}-1} & \frac {3 \,5^{\frac {3}{4}}+17}{5^{\frac {3}{4}}-1} \\ 0 & 0 & \frac {-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )} & \frac {55 \sqrt {5}-125 \,5^{\frac {1}{4}}+54 \,5^{\frac {3}{4}}-60}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )} & \frac {-55 \sqrt {5}-50 \,5^{\frac {1}{4}}+18 \,5^{\frac {3}{4}}-5}{\left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )} \\ 0 & 0 & 0 & \frac {19750 \sqrt {5}-9550-500 \,5^{\frac {1}{4}}+11200 \,5^{\frac {3}{4}}}{\left (-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115\right ) \left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )} & \frac {-20250 \sqrt {5}[phone] \,5^{\frac {1}{4}}+8550 \,5^{\frac {3}{4}}}{\left (-110 \sqrt {5}-175 \,5^{\frac {1}{4}}+38 \,5^{\frac {3}{4}}+115\right ) \left (-5 \sqrt {5}+3 \,5^{\frac {3}{4}}+13\right ) \left (5^{\frac {3}{4}}-1\right )} \\ 0 & 0 & 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{5}\}\) and the leading variables are \(\{v_{1}, v_{2}, v_{3}, v_{4}\}\). Let \(v_{5} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\left \{v_{1} = -\left (2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}-3 \sqrt {5}-6\right ) t, v_{2} = \left (5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}-\sqrt {5}-4\right ) t, v_{3} = \left (5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}-2 \sqrt {5}-4\right ) t, v_{4} = -\left (5^{\frac {1}{4}}-1\right ) t\right \}\)

Hence the solution is \[ \left [\begin {array}{c} -\left (2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}-3 \sqrt {5}-6\right ) t \\ \left (5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}-\sqrt {5}-4\right ) t \\ \left (5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}-2 \sqrt {5}-4\right ) t \\ -\left (5^{\frac {1}{4}}-1\right ) t \\ t \end {array}\right ] = \left [\begin {array}{c} -\left (2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}-3 \sqrt {5}-6\right ) t \\ \left (5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}-\sqrt {5}-4\right ) t \\ \left (5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}-2 \sqrt {5}-4\right ) t \\ -\left (5^{\frac {1}{4}}-1\right ) t \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} -\left (2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}-3 \sqrt {5}-6\right ) t \\ \left (5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}-\sqrt {5}-4\right ) t \\ \left (5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}-2 \sqrt {5}-4\right ) t \\ -\left (5^{\frac {1}{4}}-1\right ) t \\ t \end {array}\right ] = t \left [\begin {array}{c} -2 \,5^{\frac {3}{4}}-4 \,5^{\frac {1}{4}}+3 \sqrt {5}+6 \\ 5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}-\sqrt {5}-4 \\ 5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}-2 \sqrt {5}-4 \\ -5^{\frac {1}{4}}+1 \\ 1 \end {array}\right ] \] Or, by letting \(t = 1\) then the eigenvector is \[ \left [\begin {array}{c} -\left (2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}-3 \sqrt {5}-6\right ) t \\ \left (5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}-\sqrt {5}-4\right ) t \\ \left (5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}-2 \sqrt {5}-4\right ) t \\ -\left (5^{\frac {1}{4}}-1\right ) t \\ t \end {array}\right ] = \left [\begin {array}{c} -2 \,5^{\frac {3}{4}}-4 \,5^{\frac {1}{4}}+3 \sqrt {5}+6 \\ 5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}-\sqrt {5}-4 \\ 5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}-2 \sqrt {5}-4 \\ -5^{\frac {1}{4}}+1 \\ 1 \end {array}\right ] \] Considering \(\lambda = -i 5^{\frac {3}{4}}\)

We need now to determine the eigenvector \(\boldsymbol {v}\) where \begin {align*} A \boldsymbol {v} &= \lambda \boldsymbol {v} \\ A \boldsymbol {v} - \lambda \boldsymbol {v} &= \boldsymbol {0} \\ (A - \lambda I ) \boldsymbol {v} &= \boldsymbol {0} \\ \left (\left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] - \left (-i 5^{\frac {3}{4}}\right ) \left [\begin {array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end {array}\right ] \right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \left (\left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] - \left [\begin {array}{ccccc} -i 5^{\frac {3}{4}} & 0 & 0 & 0 & 0 \\ 0 & -i 5^{\frac {3}{4}} & 0 & 0 & 0 \\ 0 & 0 & -i 5^{\frac {3}{4}} & 0 & 0 \\ 0 & 0 & 0 & -i 5^{\frac {3}{4}} & 0 \\ 0 & 0 & 0 & 0 & -i 5^{\frac {3}{4}} \end {array}\right ] \right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \left [\begin {array}{ccccc} 1+i 5^{\frac {3}{4}} & 3 & 5 & 2 & 4 \\ 5 & 2+i 5^{\frac {3}{4}} & 4 & 1 & 3 \\ 4 & 1 & 3+i 5^{\frac {3}{4}} & 5 & 2 \\ 3 & 5 & 2 & 4+i 5^{\frac {3}{4}} & 1 \\ 2 & 4 & 1 & 3 & 5+i 5^{\frac {3}{4}} \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}

We will now do Gaussian elimination in order to solve for the eigenvector. The augmented matrix is \[ \left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+i 5^{\frac {3}{4}}&3&5&2&4&0\\ 5&2+i 5^{\frac {3}{4}}&4&1&3&0\\ 4&1&3+i 5^{\frac {3}{4}}&5&2&0\\ 3&5&2&4+i 5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5+i 5^{\frac {3}{4}}&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}-\frac {5 R_{1}}{1+i 5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13}{1+i 5^{\frac {3}{4}}}&\frac {-4 \,5^{\frac {3}{4}}-21 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-17 i}{i-5^{\frac {3}{4}}}&0\\ 4&1&3+i 5^{\frac {3}{4}}&5&2&0\\ 3&5&2&4+i 5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5+i 5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{3} = R_{3}-\frac {4 R_{1}}{1+i 5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13}{1+i 5^{\frac {3}{4}}}&\frac {-4 \,5^{\frac {3}{4}}-21 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-17 i}{i-5^{\frac {3}{4}}}&0\\ 0&\frac {-5^{\frac {3}{4}}-11 i}{i-5^{\frac {3}{4}}}&\frac {-5 \sqrt {5}+4 i 5^{\frac {3}{4}}-17}{1+i 5^{\frac {3}{4}}}&\frac {-5 \,5^{\frac {3}{4}}-3 i}{i-5^{\frac {3}{4}}}&\frac {-2 \,5^{\frac {3}{4}}-14 i}{i-5^{\frac {3}{4}}}&0\\ 3&5&2&4+i 5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5+i 5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {3 R_{1}}{1+i 5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13}{1+i 5^{\frac {3}{4}}}&\frac {-4 \,5^{\frac {3}{4}}-21 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-17 i}{i-5^{\frac {3}{4}}}&0\\ 0&\frac {-5^{\frac {3}{4}}-11 i}{i-5^{\frac {3}{4}}}&\frac {-5 \sqrt {5}+4 i 5^{\frac {3}{4}}-17}{1+i 5^{\frac {3}{4}}}&\frac {-5 \,5^{\frac {3}{4}}-3 i}{i-5^{\frac {3}{4}}}&\frac {-2 \,5^{\frac {3}{4}}-14 i}{i-5^{\frac {3}{4}}}&0\\ 0&\frac {-5 \,5^{\frac {3}{4}}-4 i}{i-5^{\frac {3}{4}}}&\frac {-2 \,5^{\frac {3}{4}}-13 i}{i-5^{\frac {3}{4}}}&\frac {-5 \sqrt {5}+5 i 5^{\frac {3}{4}}-2}{1+i 5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-11 i}{i-5^{\frac {3}{4}}}&0\\ 2&4&1&3&5+i 5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {2 R_{1}}{1+i 5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13}{1+i 5^{\frac {3}{4}}}&\frac {-4 \,5^{\frac {3}{4}}-21 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-17 i}{i-5^{\frac {3}{4}}}&0\\ 0&\frac {-5^{\frac {3}{4}}-11 i}{i-5^{\frac {3}{4}}}&\frac {-5 \sqrt {5}+4 i 5^{\frac {3}{4}}-17}{1+i 5^{\frac {3}{4}}}&\frac {-5 \,5^{\frac {3}{4}}-3 i}{i-5^{\frac {3}{4}}}&\frac {-2 \,5^{\frac {3}{4}}-14 i}{i-5^{\frac {3}{4}}}&0\\ 0&\frac {-5 \,5^{\frac {3}{4}}-4 i}{i-5^{\frac {3}{4}}}&\frac {-2 \,5^{\frac {3}{4}}-13 i}{i-5^{\frac {3}{4}}}&\frac {-5 \sqrt {5}+5 i 5^{\frac {3}{4}}-2}{1+i 5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-11 i}{i-5^{\frac {3}{4}}}&0\\ 0&\frac {-4 \,5^{\frac {3}{4}}-2 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-i}{i-5^{\frac {3}{4}}}&\frac {-5 \sqrt {5}+6 i 5^{\frac {3}{4}}-3}{1+i 5^{\frac {3}{4}}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{3} = R_{3}-\frac {\left (-5^{\frac {3}{4}}-11 i\right ) \left (1+i 5^{\frac {3}{4}}\right ) R_{2}}{\left (i-5^{\frac {3}{4}}\right ) \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13}{1+i 5^{\frac {3}{4}}}&\frac {-4 \,5^{\frac {3}{4}}-21 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-17 i}{i-5^{\frac {3}{4}}}&0\\ 0&0&\frac {885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}}{\left (i-5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )}&\frac {114 \,5^{\frac {3}{4}}+400 \,5^{\frac {1}{4}}+215 i \sqrt {5}+565 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&\frac {23 \,5^{\frac {3}{4}}-225 \,5^{\frac {1}{4}}+145 i \sqrt {5}+245 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&0\\ 0&\frac {-5 \,5^{\frac {3}{4}}-4 i}{i-5^{\frac {3}{4}}}&\frac {-2 \,5^{\frac {3}{4}}-13 i}{i-5^{\frac {3}{4}}}&\frac {-5 \sqrt {5}+5 i 5^{\frac {3}{4}}-2}{1+i 5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-11 i}{i-5^{\frac {3}{4}}}&0\\ 0&\frac {-4 \,5^{\frac {3}{4}}-2 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-i}{i-5^{\frac {3}{4}}}&\frac {-5 \sqrt {5}+6 i 5^{\frac {3}{4}}-3}{1+i 5^{\frac {3}{4}}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {\left (-5 \,5^{\frac {3}{4}}-4 i\right ) \left (1+i 5^{\frac {3}{4}}\right ) R_{2}}{\left (i-5^{\frac {3}{4}}\right ) \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13}{1+i 5^{\frac {3}{4}}}&\frac {-4 \,5^{\frac {3}{4}}-21 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-17 i}{i-5^{\frac {3}{4}}}&0\\ 0&0&\frac {885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}}{\left (i-5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )}&\frac {114 \,5^{\frac {3}{4}}+400 \,5^{\frac {1}{4}}+215 i \sqrt {5}+565 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&\frac {23 \,5^{\frac {3}{4}}-225 \,5^{\frac {1}{4}}+145 i \sqrt {5}+245 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&0\\ 0&0&\frac {-625 \,5^{\frac {1}{4}}-145 i \sqrt {5}+335 i-141 \,5^{\frac {3}{4}}}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&\frac {1208 \,5^{\frac {3}{4}}+600 \,5^{\frac {1}{4}}+330 i \sqrt {5}-1490 i}{\left (i-5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )}&\frac {-126 \,5^{\frac {3}{4}}-550 \,5^{\frac {1}{4}}-140 i \sqrt {5}+200 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&0\\ 0&\frac {-4 \,5^{\frac {3}{4}}-2 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-i}{i-5^{\frac {3}{4}}}&\frac {-5 \sqrt {5}+6 i 5^{\frac {3}{4}}-3}{1+i 5^{\frac {3}{4}}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {\left (-4 \,5^{\frac {3}{4}}-2 i\right ) \left (1+i 5^{\frac {3}{4}}\right ) R_{2}}{\left (i-5^{\frac {3}{4}}\right ) \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13}{1+i 5^{\frac {3}{4}}}&\frac {-4 \,5^{\frac {3}{4}}-21 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-17 i}{i-5^{\frac {3}{4}}}&0\\ 0&0&\frac {885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}}{\left (i-5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )}&\frac {114 \,5^{\frac {3}{4}}+400 \,5^{\frac {1}{4}}+215 i \sqrt {5}+565 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&\frac {23 \,5^{\frac {3}{4}}-225 \,5^{\frac {1}{4}}+145 i \sqrt {5}+245 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&0\\ 0&0&\frac {-625 \,5^{\frac {1}{4}}-145 i \sqrt {5}+335 i-141 \,5^{\frac {3}{4}}}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&\frac {1208 \,5^{\frac {3}{4}}+600 \,5^{\frac {1}{4}}+330 i \sqrt {5}-1490 i}{\left (i-5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )}&\frac {-126 \,5^{\frac {3}{4}}-550 \,5^{\frac {1}{4}}-140 i \sqrt {5}+200 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&0\\ 0&0&\frac {-525 \,5^{\frac {1}{4}}-150 i \sqrt {5}+200 i-127 \,5^{\frac {3}{4}}}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&\frac {9 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}+370 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (-i+5^{\frac {3}{4}}\right )^{2}}&\frac {1372 \,5^{\frac {3}{4}}+600 \,5^{\frac {1}{4}}+470 i \sqrt {5}-1130 i}{\left (i-5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {\left (-625 \,5^{\frac {1}{4}}-145 i \sqrt {5}+335 i-141 \,5^{\frac {3}{4}}\right ) \left (i-5^{\frac {3}{4}}\right ) \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right ) R_{3}}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13}{1+i 5^{\frac {3}{4}}}&\frac {-4 \,5^{\frac {3}{4}}-21 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-17 i}{i-5^{\frac {3}{4}}}&0\\ 0&0&\frac {885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}}{\left (i-5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )}&\frac {114 \,5^{\frac {3}{4}}+400 \,5^{\frac {1}{4}}+215 i \sqrt {5}+565 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&\frac {23 \,5^{\frac {3}{4}}-225 \,5^{\frac {1}{4}}+145 i \sqrt {5}+245 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&0\\ 0&0&0&\frac {946983800 \,5^{\frac {3}{4}}[phone] \,5^{\frac {1}{4}}[phone] i \sqrt {5}[phone] i}{\left (1+i 5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )^{3} \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )}&\frac {1273400 \,5^{\frac {3}{4}}[phone] i \sqrt {5}[phone] i+2218600 \,5^{\frac {1}{4}}}{\left (5 \sqrt {5}-3 i 5^{\frac {3}{4}}+13\right )^{2} \left (-i+5^{\frac {3}{4}}\right )^{2} \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )}&0\\ 0&0&\frac {-525 \,5^{\frac {1}{4}}-150 i \sqrt {5}+200 i-127 \,5^{\frac {3}{4}}}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&\frac {9 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}+370 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (-i+5^{\frac {3}{4}}\right )^{2}}&\frac {1372 \,5^{\frac {3}{4}}+600 \,5^{\frac {1}{4}}+470 i \sqrt {5}-1130 i}{\left (i-5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {\left (-525 \,5^{\frac {1}{4}}-150 i \sqrt {5}+200 i-127 \,5^{\frac {3}{4}}\right ) \left (i-5^{\frac {3}{4}}\right ) \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right ) R_{3}}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13}{1+i 5^{\frac {3}{4}}}&\frac {-4 \,5^{\frac {3}{4}}-21 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-17 i}{i-5^{\frac {3}{4}}}&0\\ 0&0&\frac {885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}}{\left (i-5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )}&\frac {114 \,5^{\frac {3}{4}}+400 \,5^{\frac {1}{4}}+215 i \sqrt {5}+565 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&\frac {23 \,5^{\frac {3}{4}}-225 \,5^{\frac {1}{4}}+145 i \sqrt {5}+245 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&0\\ 0&0&0&\frac {946983800 \,5^{\frac {3}{4}}[phone] \,5^{\frac {1}{4}}[phone] i \sqrt {5}[phone] i}{\left (1+i 5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )^{3} \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )}&\frac {1273400 \,5^{\frac {3}{4}}[phone] i \sqrt {5}[phone] i+2218600 \,5^{\frac {1}{4}}}{\left (5 \sqrt {5}-3 i 5^{\frac {3}{4}}+13\right )^{2} \left (-i+5^{\frac {3}{4}}\right )^{2} \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )}&0\\ 0&0&0&\frac {15103000 \,5^{\frac {3}{4}}[phone] i \sqrt {5}[phone] i+36190200 \,5^{\frac {1}{4}}}{\left (1+i 5^{\frac {3}{4}}\right )^{2} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )^{2} \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )}&\frac {2797850800 \,5^{\frac {3}{4}}[phone] \,5^{\frac {1}{4}}[phone] i \sqrt {5}[phone] i}{\left (1+i 5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )^{3} \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {\left (1+i 5^{\frac {3}{4}}\right ) \left (75515 \,5^{\frac {3}{4}}+115138 i \sqrt {5}+247067 i+180951 \,5^{\frac {1}{4}}\right ) \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right ) R_{4}}{4734919 \,5^{\frac {3}{4}}[phone] \,5^{\frac {1}{4}}+1922330 i \sqrt {5}+3668376 i} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13}{1+i 5^{\frac {3}{4}}}&\frac {-4 \,5^{\frac {3}{4}}-21 i}{i-5^{\frac {3}{4}}}&\frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}}&\frac {-3 \,5^{\frac {3}{4}}-17 i}{i-5^{\frac {3}{4}}}&0\\ 0&0&\frac {885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}}{\left (i-5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )}&\frac {114 \,5^{\frac {3}{4}}+400 \,5^{\frac {1}{4}}+215 i \sqrt {5}+565 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&\frac {23 \,5^{\frac {3}{4}}-225 \,5^{\frac {1}{4}}+145 i \sqrt {5}+245 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}}&0\\ 0&0&0&\frac {946983800 \,5^{\frac {3}{4}}[phone] \,5^{\frac {1}{4}}[phone] i \sqrt {5}[phone] i}{\left (1+i 5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )^{3} \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )}&\frac {1273400 \,5^{\frac {3}{4}}[phone] i \sqrt {5}[phone] i+2218600 \,5^{\frac {1}{4}}}{\left (5 \sqrt {5}-3 i 5^{\frac {3}{4}}+13\right )^{2} \left (-i+5^{\frac {3}{4}}\right )^{2} \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )}&0\\ 0&0&0&0&0&0 \end {array} \right ] \end {align*}

Therefore the system in Echelon form is \[ \left [\begin {array}{ccccc} 1+i 5^{\frac {3}{4}} & 3 & 5 & 2 & 4 \\ 0 & \frac {-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13}{1+i 5^{\frac {3}{4}}} & \frac {-4 \,5^{\frac {3}{4}}-21 i}{i-5^{\frac {3}{4}}} & \frac {-5^{\frac {3}{4}}-9 i}{i-5^{\frac {3}{4}}} & \frac {-3 \,5^{\frac {3}{4}}-17 i}{i-5^{\frac {3}{4}}} \\ 0 & 0 & \frac {885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}}{\left (i-5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )} & \frac {114 \,5^{\frac {3}{4}}+400 \,5^{\frac {1}{4}}+215 i \sqrt {5}+565 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}} & \frac {23 \,5^{\frac {3}{4}}-225 \,5^{\frac {1}{4}}+145 i \sqrt {5}+245 i}{\left (5 i \sqrt {5}+3 \,5^{\frac {3}{4}}+13 i\right ) \left (i-5^{\frac {3}{4}}\right )^{2}} \\ 0 & 0 & 0 & \frac {946983800 \,5^{\frac {3}{4}}[phone] \,5^{\frac {1}{4}}[phone] i \sqrt {5}[phone] i}{\left (1+i 5^{\frac {3}{4}}\right )^{3} \left (-5 \sqrt {5}+3 i 5^{\frac {3}{4}}-13\right )^{3} \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )} & \frac {1273400 \,5^{\frac {3}{4}}[phone] i \sqrt {5}[phone] i+2218600 \,5^{\frac {1}{4}}}{\left (5 \sqrt {5}-3 i 5^{\frac {3}{4}}+13\right )^{2} \left (-i+5^{\frac {3}{4}}\right )^{2} \left (885 i+1875 \,5^{\frac {1}{4}}+1067 \,5^{\frac {3}{4}}+85 i \sqrt {5}\right )} \\ 0 & 0 & 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{5}\}\) and the leading variables are \(\{v_{1}, v_{2}, v_{3}, v_{4}\}\). Let \(v_{5} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\left \{v_{1} = \frac {t \left (-7542856792418920-3370828393023080 \sqrt {5}-2398617049157928 i 5^{\frac {3}{4}}-5360068694461640 i 5^{\frac {1}{4}}\right )}{-323060820625160-142295694614440 \sqrt {5}+7355700331154200 i 5^{\frac {1}{4}}+3290903801182872 i 5^{\frac {3}{4}}}, v_{2} = -\frac {t \left (-903502865289325283095578240-404060124039924116964757120 \sqrt {5}-415477534769715915186883584 i 5^{\frac {3}{4}}-929038567351137060812826880 i 5^{\frac {1}{4}}\right )}{\left (40 i \sqrt {5}-125 \,5^{\frac {1}{4}}+5^{\frac {3}{4}}+2 i\right ) \left (-323060820625160-142295694614440 \sqrt {5}+7355700331154200 i 5^{\frac {1}{4}}+3290903801182872 i 5^{\frac {3}{4}}\right ) \left (126419610 \,5^{\frac {1}{4}}-67376776 i-30929640 i \sqrt {5}[phone] \,5^{\frac {3}{4}}\right )}, v_{3} = \frac {t \left (448381104560849079638085120+200525260687192269352684800 \sqrt {5}-313035771894228876273258496 i 5^{\frac {3}{4}}-699968001936593435483505920 i 5^{\frac {1}{4}}\right )}{\left (40 i \sqrt {5}-125 \,5^{\frac {1}{4}}+5^{\frac {3}{4}}+2 i\right ) \left (-323060820625160-142295694614440 \sqrt {5}+7355700331154200 i 5^{\frac {1}{4}}+3290903801182872 i 5^{\frac {3}{4}}\right ) \left (126419610 \,5^{\frac {1}{4}}-67376776 i-30929640 i \sqrt {5}[phone] \,5^{\frac {3}{4}}\right )}, v_{4} = -\frac {t \left (-27555518 \,5^{\frac {3}{4}}[phone] i \sqrt {5}[phone] i-59042834 \,5^{\frac {1}{4}}\right )}{126419610 \,5^{\frac {1}{4}}-67376776 i-30929640 i \sqrt {5}[phone] \,5^{\frac {3}{4}}}\right \}\)

Hence the solution is \[ \text {Expression too large to display} \] 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 {t \left (-7542856792418920-3370828393023080 \sqrt {5}-2398617049157928 \,\operatorname {I} \,5^{\frac {3}{4}}-5360068694461640 \,\operatorname {I} \,5^{\frac {1}{4}}\right )}{-323060820625160-142295694614440 \sqrt {5}+7355700331154200 \,\operatorname {I} \,5^{\frac {1}{4}}+3290903801182872 \,\operatorname {I} \,5^{\frac {3}{4}}} \\ -\frac {t \left (-903502865289325283095578240-404060124039924116964757120 \sqrt {5}-415477534769715915186883584 \,\operatorname {I} \,5^{\frac {3}{4}}-929038567351137060812826880 \,\operatorname {I} \,5^{\frac {1}{4}}\right )}{\left (40 \,\operatorname {I} \sqrt {5}-125 \,5^{\frac {1}{4}}+5^{\frac {3}{4}}+2 \,\operatorname {I}\right ) \left (-323060820625160-142295694614440 \sqrt {5}+7355700331154200 \,\operatorname {I} \,5^{\frac {1}{4}}+3290903801182872 \,\operatorname {I} \,5^{\frac {3}{4}}\right ) \left (126419610 \,5^{\frac {1}{4}}-67376776 \,\operatorname {I}-30929640 \,\operatorname {I} \sqrt {5}[phone] \,5^{\frac {3}{4}}\right )} \\ \frac {t \left (448381104560849079638085120+200525260687192269352684800 \sqrt {5}-313035771894228876273258496 \,\operatorname {I} \,5^{\frac {3}{4}}-699968001936593435483505920 \,\operatorname {I} \,5^{\frac {1}{4}}\right )}{\left (40 \,\operatorname {I} \sqrt {5}-125 \,5^{\frac {1}{4}}+5^{\frac {3}{4}}+2 \,\operatorname {I}\right ) \left (-323060820625160-142295694614440 \sqrt {5}+7355700331154200 \,\operatorname {I} \,5^{\frac {1}{4}}+3290903801182872 \,\operatorname {I} \,5^{\frac {3}{4}}\right ) \left (126419610 \,5^{\frac {1}{4}}-67376776 \,\operatorname {I}-30929640 \,\operatorname {I} \sqrt {5}[phone] \,5^{\frac {3}{4}}\right )} \\ -\frac {t \left (-27555518 \,5^{\frac {3}{4}}[phone] \,\operatorname {I} \sqrt {5}[phone] \,\operatorname {I}-59042834 \,5^{\frac {1}{4}}\right )}{126419610 \,5^{\frac {1}{4}}-67376776 \,\operatorname {I}-30929640 \,\operatorname {I} \sqrt {5}[phone] \,5^{\frac {3}{4}}} \\ t \end {array}\right ] = t \left [\begin {array}{c} \frac {-7542856792418920-3370828393023080 \sqrt {5}-2398617049157928 i 5^{\frac {3}{4}}-5360068694461640 i 5^{\frac {1}{4}}}{-323060820625160-142295694614440 \sqrt {5}+7355700331154200 i 5^{\frac {1}{4}}+3290903801182872 i 5^{\frac {3}{4}}} \\ -\frac {-903502865289325283095578240-404060124039924116964757120 \sqrt {5}-415477534769715915186883584 i 5^{\frac {3}{4}}-929038567351137060812826880 i 5^{\frac {1}{4}}}{\left (40 i \sqrt {5}-125 \,5^{\frac {1}{4}}+5^{\frac {3}{4}}+2 i\right ) \left (-323060820625160-142295694614440 \sqrt {5}+7355700331154200 i 5^{\frac {1}{4}}+3290903801182872 i 5^{\frac {3}{4}}\right ) \left (126419610 \,5^{\frac {1}{4}}-67376776 i-30929640 i \sqrt {5}[phone] \,5^{\frac {3}{4}}\right )} \\ \frac {448381104560849079638085120+200525260687192269352684800 \sqrt {5}-313035771894228876273258496 i 5^{\frac {3}{4}}-699968001936593435483505920 i 5^{\frac {1}{4}}}{\left (40 i \sqrt {5}-125 \,5^{\frac {1}{4}}+5^{\frac {3}{4}}+2 i\right ) \left (-323060820625160-142295694614440 \sqrt {5}+7355700331154200 i 5^{\frac {1}{4}}+3290903801182872 i 5^{\frac {3}{4}}\right ) \left (126419610 \,5^{\frac {1}{4}}-67376776 i-30929640 i \sqrt {5}[phone] \,5^{\frac {3}{4}}\right )} \\ -\frac {-27555518 \,5^{\frac {3}{4}}[phone] i \sqrt {5}[phone] i-59042834 \,5^{\frac {1}{4}}}{126419610 \,5^{\frac {1}{4}}-67376776 i-30929640 i \sqrt {5}[phone] \,5^{\frac {3}{4}}} \\ 1 \end {array}\right ] \] Or, by letting \(t = 1\) then the eigenvector is \[ \text {Expression too large to display} \] Which can be normalized to \[ \text {Expression too large to display} \] Considering \(\lambda = -5^{\frac {3}{4}}\)

We need now to determine the eigenvector \(\boldsymbol {v}\) where \begin {align*} A \boldsymbol {v} &= \lambda \boldsymbol {v} \\ A \boldsymbol {v} - \lambda \boldsymbol {v} &= \boldsymbol {0} \\ (A - \lambda I ) \boldsymbol {v} &= \boldsymbol {0} \\ \left (\left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] - \left (-5^{\frac {3}{4}}\right ) \left [\begin {array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end {array}\right ] \right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \left (\left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] - \left [\begin {array}{ccccc} -5^{\frac {3}{4}} & 0 & 0 & 0 & 0 \\ 0 & -5^{\frac {3}{4}} & 0 & 0 & 0 \\ 0 & 0 & -5^{\frac {3}{4}} & 0 & 0 \\ 0 & 0 & 0 & -5^{\frac {3}{4}} & 0 \\ 0 & 0 & 0 & 0 & -5^{\frac {3}{4}} \end {array}\right ] \right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \left [\begin {array}{ccccc} 1+5^{\frac {3}{4}} & 3 & 5 & 2 & 4 \\ 5 & 2+5^{\frac {3}{4}} & 4 & 1 & 3 \\ 4 & 1 & 3+5^{\frac {3}{4}} & 5 & 2 \\ 3 & 5 & 2 & 4+5^{\frac {3}{4}} & 1 \\ 2 & 4 & 1 & 3 & 5+5^{\frac {3}{4}} \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}

We will now do Gaussian elimination in order to solve for the eigenvector. The augmented matrix is \[ \left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+5^{\frac {3}{4}}&3&5&2&4&0\\ 5&2+5^{\frac {3}{4}}&4&1&3&0\\ 4&1&3+5^{\frac {3}{4}}&5&2&0\\ 3&5&2&4+5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5+5^{\frac {3}{4}}&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}-\frac {5 R_{1}}{1+5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {4 \,5^{\frac {3}{4}}-21}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&0\\ 4&1&3+5^{\frac {3}{4}}&5&2&0\\ 3&5&2&4+5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5+5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{3} = R_{3}-\frac {4 R_{1}}{1+5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {4 \,5^{\frac {3}{4}}-21}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&0\\ 0&\frac {5^{\frac {3}{4}}-11}{1+5^{\frac {3}{4}}}&\frac {5 \sqrt {5}+4 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&\frac {5 \,5^{\frac {3}{4}}-3}{1+5^{\frac {3}{4}}}&\frac {2 \,5^{\frac {3}{4}}-14}{1+5^{\frac {3}{4}}}&0\\ 3&5&2&4+5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5+5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {3 R_{1}}{1+5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {4 \,5^{\frac {3}{4}}-21}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&0\\ 0&\frac {5^{\frac {3}{4}}-11}{1+5^{\frac {3}{4}}}&\frac {5 \sqrt {5}+4 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&\frac {5 \,5^{\frac {3}{4}}-3}{1+5^{\frac {3}{4}}}&\frac {2 \,5^{\frac {3}{4}}-14}{1+5^{\frac {3}{4}}}&0\\ 0&\frac {5 \,5^{\frac {3}{4}}-4}{1+5^{\frac {3}{4}}}&\frac {2 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {5 \sqrt {5}+5 \,5^{\frac {3}{4}}-2}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-11}{1+5^{\frac {3}{4}}}&0\\ 2&4&1&3&5+5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {2 R_{1}}{1+5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {4 \,5^{\frac {3}{4}}-21}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&0\\ 0&\frac {5^{\frac {3}{4}}-11}{1+5^{\frac {3}{4}}}&\frac {5 \sqrt {5}+4 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&\frac {5 \,5^{\frac {3}{4}}-3}{1+5^{\frac {3}{4}}}&\frac {2 \,5^{\frac {3}{4}}-14}{1+5^{\frac {3}{4}}}&0\\ 0&\frac {5 \,5^{\frac {3}{4}}-4}{1+5^{\frac {3}{4}}}&\frac {2 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {5 \sqrt {5}+5 \,5^{\frac {3}{4}}-2}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-11}{1+5^{\frac {3}{4}}}&0\\ 0&\frac {4 \,5^{\frac {3}{4}}-2}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-1}{1+5^{\frac {3}{4}}}&\frac {5 \sqrt {5}+6 \,5^{\frac {3}{4}}-3}{1+5^{\frac {3}{4}}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{3} = R_{3}-\frac {\left (5^{\frac {3}{4}}-11\right ) R_{2}}{5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {4 \,5^{\frac {3}{4}}-21}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&0\\ 0&0&\frac {-110 \sqrt {5}+175 \,5^{\frac {1}{4}}-38 \,5^{\frac {3}{4}}+115}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {55 \sqrt {5}+125 \,5^{\frac {1}{4}}-54 \,5^{\frac {3}{4}}-60}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-55 \sqrt {5}+50 \,5^{\frac {1}{4}}-18 \,5^{\frac {3}{4}}-5}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0\\ 0&\frac {5 \,5^{\frac {3}{4}}-4}{1+5^{\frac {3}{4}}}&\frac {2 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {5 \sqrt {5}+5 \,5^{\frac {3}{4}}-2}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-11}{1+5^{\frac {3}{4}}}&0\\ 0&\frac {4 \,5^{\frac {3}{4}}-2}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-1}{1+5^{\frac {3}{4}}}&\frac {5 \sqrt {5}+6 \,5^{\frac {3}{4}}-3}{1+5^{\frac {3}{4}}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {\left (5 \,5^{\frac {3}{4}}-4\right ) R_{2}}{5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {4 \,5^{\frac {3}{4}}-21}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&0\\ 0&0&\frac {-110 \sqrt {5}+175 \,5^{\frac {1}{4}}-38 \,5^{\frac {3}{4}}+115}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {55 \sqrt {5}+125 \,5^{\frac {1}{4}}-54 \,5^{\frac {3}{4}}-60}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-55 \sqrt {5}+50 \,5^{\frac {1}{4}}-18 \,5^{\frac {3}{4}}-5}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0\\ 0&0&\frac {-135 \sqrt {5}+50 \,5^{\frac {1}{4}}+56 \,5^{\frac {3}{4}}+85}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-25 \sqrt {5}+200 \,5^{\frac {1}{4}}-22 \,5^{\frac {3}{4}}+115}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-115 \sqrt {5}+25 \,5^{\frac {1}{4}}+51 \,5^{\frac {3}{4}}+75}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0\\ 0&\frac {4 \,5^{\frac {3}{4}}-2}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-1}{1+5^{\frac {3}{4}}}&\frac {5 \sqrt {5}+6 \,5^{\frac {3}{4}}-3}{1+5^{\frac {3}{4}}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {\left (4 \,5^{\frac {3}{4}}-2\right ) R_{2}}{5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {4 \,5^{\frac {3}{4}}-21}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&0\\ 0&0&\frac {-110 \sqrt {5}+175 \,5^{\frac {1}{4}}-38 \,5^{\frac {3}{4}}+115}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {55 \sqrt {5}+125 \,5^{\frac {1}{4}}-54 \,5^{\frac {3}{4}}-60}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-55 \sqrt {5}+50 \,5^{\frac {1}{4}}-18 \,5^{\frac {3}{4}}-5}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0\\ 0&0&\frac {-135 \sqrt {5}+50 \,5^{\frac {1}{4}}+56 \,5^{\frac {3}{4}}+85}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-25 \sqrt {5}+200 \,5^{\frac {1}{4}}-22 \,5^{\frac {3}{4}}+115}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-115 \sqrt {5}+25 \,5^{\frac {1}{4}}+51 \,5^{\frac {3}{4}}+75}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0\\ 0&0&\frac {-110 \sqrt {5}+25 \,5^{\frac {1}{4}}+52 \,5^{\frac {3}{4}}+75}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {20 \sqrt {5}+75 \,5^{\frac {1}{4}}-4 \,5^{\frac {3}{4}}-5}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-50 \sqrt {5}+225 \,5^{\frac {1}{4}}-13 \,5^{\frac {3}{4}}+130}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {\left (-135 \sqrt {5}+50 \,5^{\frac {1}{4}}+56 \,5^{\frac {3}{4}}+85\right ) R_{3}}{-110 \sqrt {5}+175 \,5^{\frac {1}{4}}-38 \,5^{\frac {3}{4}}+115} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {4 \,5^{\frac {3}{4}}-21}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&0\\ 0&0&\frac {-110 \sqrt {5}+175 \,5^{\frac {1}{4}}-38 \,5^{\frac {3}{4}}+115}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {55 \sqrt {5}+125 \,5^{\frac {1}{4}}-54 \,5^{\frac {3}{4}}-60}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-55 \sqrt {5}+50 \,5^{\frac {1}{4}}-18 \,5^{\frac {3}{4}}-5}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0\\ 0&0&0&\frac {-19750 \sqrt {5}+9550-500 \,5^{\frac {1}{4}}+11200 \,5^{\frac {3}{4}}}{\left (38 \,5^{\frac {3}{4}}-175 \,5^{\frac {1}{4}}+110 \sqrt {5}-115\right ) \left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {20250 \sqrt {5}-65550-9050 \,5^{\frac {1}{4}}+8550 \,5^{\frac {3}{4}}}{\left (38 \,5^{\frac {3}{4}}-175 \,5^{\frac {1}{4}}+110 \sqrt {5}-115\right ) \left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0\\ 0&0&\frac {-110 \sqrt {5}+25 \,5^{\frac {1}{4}}+52 \,5^{\frac {3}{4}}+75}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {20 \sqrt {5}+75 \,5^{\frac {1}{4}}-4 \,5^{\frac {3}{4}}-5}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-50 \sqrt {5}+225 \,5^{\frac {1}{4}}-13 \,5^{\frac {3}{4}}+130}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {\left (-110 \sqrt {5}+25 \,5^{\frac {1}{4}}+52 \,5^{\frac {3}{4}}+75\right ) R_{3}}{-110 \sqrt {5}+175 \,5^{\frac {1}{4}}-38 \,5^{\frac {3}{4}}+115} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {4 \,5^{\frac {3}{4}}-21}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&0\\ 0&0&\frac {-110 \sqrt {5}+175 \,5^{\frac {1}{4}}-38 \,5^{\frac {3}{4}}+115}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {55 \sqrt {5}+125 \,5^{\frac {1}{4}}-54 \,5^{\frac {3}{4}}-60}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-55 \sqrt {5}+50 \,5^{\frac {1}{4}}-18 \,5^{\frac {3}{4}}-5}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0\\ 0&0&0&\frac {-19750 \sqrt {5}+9550-500 \,5^{\frac {1}{4}}+11200 \,5^{\frac {3}{4}}}{\left (38 \,5^{\frac {3}{4}}-175 \,5^{\frac {1}{4}}+110 \sqrt {5}-115\right ) \left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {20250 \sqrt {5}-65550-9050 \,5^{\frac {1}{4}}+8550 \,5^{\frac {3}{4}}}{\left (38 \,5^{\frac {3}{4}}-175 \,5^{\frac {1}{4}}+110 \sqrt {5}-115\right ) \left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0\\ 0&0&0&-\frac {25 \left (581 \,5^{\frac {3}{4}}-1829 \,5^{\frac {1}{4}}+677 \sqrt {5}-813\right )}{\left (38 \,5^{\frac {3}{4}}-175 \,5^{\frac {1}{4}}+110 \sqrt {5}-115\right ) \left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-28800 \sqrt {5}[phone] \,5^{\frac {1}{4}}+31450 \,5^{\frac {3}{4}}}{\left (38 \,5^{\frac {3}{4}}-175 \,5^{\frac {1}{4}}+110 \sqrt {5}-115\right ) \left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}+\frac {\left (581 \,5^{\frac {3}{4}}-1829 \,5^{\frac {1}{4}}+677 \sqrt {5}-813\right ) R_{4}}{448 \,5^{\frac {3}{4}}-20 \,5^{\frac {1}{4}}-790 \sqrt {5}+382} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1+5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}}&\frac {4 \,5^{\frac {3}{4}}-21}{1+5^{\frac {3}{4}}}&\frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}}&\frac {3 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}}&0\\ 0&0&\frac {-110 \sqrt {5}+175 \,5^{\frac {1}{4}}-38 \,5^{\frac {3}{4}}+115}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {55 \sqrt {5}+125 \,5^{\frac {1}{4}}-54 \,5^{\frac {3}{4}}-60}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {-55 \sqrt {5}+50 \,5^{\frac {1}{4}}-18 \,5^{\frac {3}{4}}-5}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0\\ 0&0&0&\frac {-19750 \sqrt {5}+9550-500 \,5^{\frac {1}{4}}+11200 \,5^{\frac {3}{4}}}{\left (38 \,5^{\frac {3}{4}}-175 \,5^{\frac {1}{4}}+110 \sqrt {5}-115\right ) \left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&\frac {20250 \sqrt {5}-65550-9050 \,5^{\frac {1}{4}}+8550 \,5^{\frac {3}{4}}}{\left (38 \,5^{\frac {3}{4}}-175 \,5^{\frac {1}{4}}+110 \sqrt {5}-115\right ) \left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )}&0\\ 0&0&0&0&0&0 \end {array} \right ] \end {align*}

Therefore the system in Echelon form is \[ \left [\begin {array}{ccccc} 1+5^{\frac {3}{4}} & 3 & 5 & 2 & 4 \\ 0 & \frac {5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13}{1+5^{\frac {3}{4}}} & \frac {4 \,5^{\frac {3}{4}}-21}{1+5^{\frac {3}{4}}} & \frac {5^{\frac {3}{4}}-9}{1+5^{\frac {3}{4}}} & \frac {3 \,5^{\frac {3}{4}}-17}{1+5^{\frac {3}{4}}} \\ 0 & 0 & \frac {-110 \sqrt {5}+175 \,5^{\frac {1}{4}}-38 \,5^{\frac {3}{4}}+115}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )} & \frac {55 \sqrt {5}+125 \,5^{\frac {1}{4}}-54 \,5^{\frac {3}{4}}-60}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )} & \frac {-55 \sqrt {5}+50 \,5^{\frac {1}{4}}-18 \,5^{\frac {3}{4}}-5}{\left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )} \\ 0 & 0 & 0 & \frac {-19750 \sqrt {5}+9550-500 \,5^{\frac {1}{4}}+11200 \,5^{\frac {3}{4}}}{\left (38 \,5^{\frac {3}{4}}-175 \,5^{\frac {1}{4}}+110 \sqrt {5}-115\right ) \left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )} & \frac {20250 \sqrt {5}-65550-9050 \,5^{\frac {1}{4}}+8550 \,5^{\frac {3}{4}}}{\left (38 \,5^{\frac {3}{4}}-175 \,5^{\frac {1}{4}}+110 \sqrt {5}-115\right ) \left (5 \sqrt {5}+3 \,5^{\frac {3}{4}}-13\right ) \left (1+5^{\frac {3}{4}}\right )} \\ 0 & 0 & 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{5}\}\) and the leading variables are \(\{v_{1}, v_{2}, v_{3}, v_{4}\}\). Let \(v_{5} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\left \{v_{1} = \left (2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}+3 \sqrt {5}+6\right ) t, v_{2} = -\left (5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}+\sqrt {5}+4\right ) t, v_{3} = -\left (5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}+2 \sqrt {5}+4\right ) t, v_{4} = \left (5^{\frac {1}{4}}+1\right ) t\right \}\)

Hence the solution is \[ \left [\begin {array}{c} \left (2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}+3 \sqrt {5}+6\right ) t \\ -\left (5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}+\sqrt {5}+4\right ) t \\ -\left (5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}+2 \sqrt {5}+4\right ) t \\ \left (5^{\frac {1}{4}}+1\right ) t \\ t \end {array}\right ] = \left [\begin {array}{c} \left (2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}+3 \sqrt {5}+6\right ) t \\ -\left (5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}+\sqrt {5}+4\right ) t \\ -\left (5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}+2 \sqrt {5}+4\right ) t \\ \left (5^{\frac {1}{4}}+1\right ) t \\ t \end {array}\right ] \] Since there is one free Variable, we have found one eigenvector associated with this eigenvalue. The above can be written as \[ \left [\begin {array}{c} \left (2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}+3 \sqrt {5}+6\right ) t \\ -\left (5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}+\sqrt {5}+4\right ) t \\ -\left (5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}+2 \sqrt {5}+4\right ) t \\ \left (5^{\frac {1}{4}}+1\right ) t \\ t \end {array}\right ] = t \left [\begin {array}{c} 2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}+3 \sqrt {5}+6 \\ -5^{\frac {3}{4}}-2 \,5^{\frac {1}{4}}-\sqrt {5}-4 \\ -5^{\frac {3}{4}}-3 \,5^{\frac {1}{4}}-2 \sqrt {5}-4 \\ 5^{\frac {1}{4}}+1 \\ 1 \end {array}\right ] \] Or, by letting \(t = 1\) then the eigenvector is \[ \left [\begin {array}{c} \left (2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}+3 \sqrt {5}+6\right ) t \\ -\left (5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}+\sqrt {5}+4\right ) t \\ -\left (5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}+2 \sqrt {5}+4\right ) t \\ \left (5^{\frac {1}{4}}+1\right ) t \\ t \end {array}\right ] = \left [\begin {array}{c} 2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}+3 \sqrt {5}+6 \\ -5^{\frac {3}{4}}-2 \,5^{\frac {1}{4}}-\sqrt {5}-4 \\ -5^{\frac {3}{4}}-3 \,5^{\frac {1}{4}}-2 \sqrt {5}-4 \\ 5^{\frac {1}{4}}+1 \\ 1 \end {array}\right ] \] Considering \(\lambda = i 5^{\frac {3}{4}}\)

We need now to determine the eigenvector \(\boldsymbol {v}\) where \begin {align*} A \boldsymbol {v} &= \lambda \boldsymbol {v} \\ A \boldsymbol {v} - \lambda \boldsymbol {v} &= \boldsymbol {0} \\ (A - \lambda I ) \boldsymbol {v} &= \boldsymbol {0} \\ \left (\left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] - \left (i 5^{\frac {3}{4}}\right ) \left [\begin {array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end {array}\right ] \right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \left (\left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ] - \left [\begin {array}{ccccc} i 5^{\frac {3}{4}} & 0 & 0 & 0 & 0 \\ 0 & i 5^{\frac {3}{4}} & 0 & 0 & 0 \\ 0 & 0 & i 5^{\frac {3}{4}} & 0 & 0 \\ 0 & 0 & 0 & i 5^{\frac {3}{4}} & 0 \\ 0 & 0 & 0 & 0 & i 5^{\frac {3}{4}} \end {array}\right ] \right ) \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \left [\begin {array}{ccccc} 1-i 5^{\frac {3}{4}} & 3 & 5 & 2 & 4 \\ 5 & 2-i 5^{\frac {3}{4}} & 4 & 1 & 3 \\ 4 & 1 & 3-i 5^{\frac {3}{4}} & 5 & 2 \\ 3 & 5 & 2 & 4-i 5^{\frac {3}{4}} & 1 \\ 2 & 4 & 1 & 3 & 5-i 5^{\frac {3}{4}} \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] &= \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \end {align*}

We will now do Gaussian elimination in order to solve for the eigenvector. The augmented matrix is \[ \left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-i 5^{\frac {3}{4}}&3&5&2&4&0\\ 5&2-i 5^{\frac {3}{4}}&4&1&3&0\\ 4&1&3-i 5^{\frac {3}{4}}&5&2&0\\ 3&5&2&4-i 5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5-i 5^{\frac {3}{4}}&0 \end {array} \right ] \] \begin {align*} R_{2} = R_{2}-\frac {5 R_{1}}{1-i 5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-21 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-17 i}{5^{\frac {3}{4}}+i}&0\\ 4&1&3-i 5^{\frac {3}{4}}&5&2&0\\ 3&5&2&4-i 5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5-i 5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{3} = R_{3}-\frac {4 R_{1}}{1-i 5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-21 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-17 i}{5^{\frac {3}{4}}+i}&0\\ 0&\frac {5^{\frac {3}{4}}-11 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-5 i \sqrt {5}-17 i}{5^{\frac {3}{4}}+i}&\frac {5 \,5^{\frac {3}{4}}-3 i}{5^{\frac {3}{4}}+i}&\frac {2 \,5^{\frac {3}{4}}-14 i}{5^{\frac {3}{4}}+i}&0\\ 3&5&2&4-i 5^{\frac {3}{4}}&1&0\\ 2&4&1&3&5-i 5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {3 R_{1}}{1-i 5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-21 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-17 i}{5^{\frac {3}{4}}+i}&0\\ 0&\frac {5^{\frac {3}{4}}-11 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-5 i \sqrt {5}-17 i}{5^{\frac {3}{4}}+i}&\frac {5 \,5^{\frac {3}{4}}-3 i}{5^{\frac {3}{4}}+i}&\frac {2 \,5^{\frac {3}{4}}-14 i}{5^{\frac {3}{4}}+i}&0\\ 0&\frac {5 \,5^{\frac {3}{4}}-4 i}{5^{\frac {3}{4}}+i}&\frac {2 \,5^{\frac {3}{4}}-13 i}{5^{\frac {3}{4}}+i}&\frac {5 \,5^{\frac {3}{4}}-5 i \sqrt {5}-2 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-11 i}{5^{\frac {3}{4}}+i}&0\\ 2&4&1&3&5-i 5^{\frac {3}{4}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {2 R_{1}}{1-i 5^{\frac {3}{4}}} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-21 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-17 i}{5^{\frac {3}{4}}+i}&0\\ 0&\frac {5^{\frac {3}{4}}-11 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-5 i \sqrt {5}-17 i}{5^{\frac {3}{4}}+i}&\frac {5 \,5^{\frac {3}{4}}-3 i}{5^{\frac {3}{4}}+i}&\frac {2 \,5^{\frac {3}{4}}-14 i}{5^{\frac {3}{4}}+i}&0\\ 0&\frac {5 \,5^{\frac {3}{4}}-4 i}{5^{\frac {3}{4}}+i}&\frac {2 \,5^{\frac {3}{4}}-13 i}{5^{\frac {3}{4}}+i}&\frac {5 \,5^{\frac {3}{4}}-5 i \sqrt {5}-2 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-11 i}{5^{\frac {3}{4}}+i}&0\\ 0&\frac {4 \,5^{\frac {3}{4}}-2 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-i}{5^{\frac {3}{4}}+i}&\frac {6 \,5^{\frac {3}{4}}-5 i \sqrt {5}-3 i}{5^{\frac {3}{4}}+i}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{3} = R_{3}-\frac {\left (5^{\frac {3}{4}}-11 i\right ) R_{2}}{3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-21 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-17 i}{5^{\frac {3}{4}}+i}&0\\ 0&0&\frac {38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i}{25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i}&\frac {54 \,5^{\frac {3}{4}}+125 \,5^{\frac {1}{4}}+55 i \sqrt {5}+60 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {18 \,5^{\frac {3}{4}}+50 \,5^{\frac {1}{4}}-55 i \sqrt {5}+5 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&0\\ 0&\frac {5 \,5^{\frac {3}{4}}-4 i}{5^{\frac {3}{4}}+i}&\frac {2 \,5^{\frac {3}{4}}-13 i}{5^{\frac {3}{4}}+i}&\frac {5 \,5^{\frac {3}{4}}-5 i \sqrt {5}-2 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-11 i}{5^{\frac {3}{4}}+i}&0\\ 0&\frac {4 \,5^{\frac {3}{4}}-2 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-i}{5^{\frac {3}{4}}+i}&\frac {6 \,5^{\frac {3}{4}}-5 i \sqrt {5}-3 i}{5^{\frac {3}{4}}+i}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {\left (5 \,5^{\frac {3}{4}}-4 i\right ) R_{2}}{3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-21 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-17 i}{5^{\frac {3}{4}}+i}&0\\ 0&0&\frac {38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i}{25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i}&\frac {54 \,5^{\frac {3}{4}}+125 \,5^{\frac {1}{4}}+55 i \sqrt {5}+60 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {18 \,5^{\frac {3}{4}}+50 \,5^{\frac {1}{4}}-55 i \sqrt {5}+5 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&0\\ 0&0&\frac {-135 i \sqrt {5}+50 \,5^{\frac {1}{4}}-56 \,5^{\frac {3}{4}}-85 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {22 \,5^{\frac {3}{4}}+200 \,5^{\frac {1}{4}}-25 i \sqrt {5}-115 i}{25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i}&\frac {-115 i \sqrt {5}+25 \,5^{\frac {1}{4}}-51 \,5^{\frac {3}{4}}-75 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&0\\ 0&\frac {4 \,5^{\frac {3}{4}}-2 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-i}{5^{\frac {3}{4}}+i}&\frac {6 \,5^{\frac {3}{4}}-5 i \sqrt {5}-3 i}{5^{\frac {3}{4}}+i}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {\left (4 \,5^{\frac {3}{4}}-2 i\right ) R_{2}}{3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-21 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-17 i}{5^{\frac {3}{4}}+i}&0\\ 0&0&\frac {38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i}{25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i}&\frac {54 \,5^{\frac {3}{4}}+125 \,5^{\frac {1}{4}}+55 i \sqrt {5}+60 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {18 \,5^{\frac {3}{4}}+50 \,5^{\frac {1}{4}}-55 i \sqrt {5}+5 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&0\\ 0&0&\frac {-135 i \sqrt {5}+50 \,5^{\frac {1}{4}}-56 \,5^{\frac {3}{4}}-85 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {22 \,5^{\frac {3}{4}}+200 \,5^{\frac {1}{4}}-25 i \sqrt {5}-115 i}{25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i}&\frac {-115 i \sqrt {5}+25 \,5^{\frac {1}{4}}-51 \,5^{\frac {3}{4}}-75 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&0\\ 0&0&\frac {-110 i \sqrt {5}+25 \,5^{\frac {1}{4}}-52 \,5^{\frac {3}{4}}-75 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {4 \,5^{\frac {3}{4}}+75 \,5^{\frac {1}{4}}+20 i \sqrt {5}+5 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {13 \,5^{\frac {3}{4}}+225 \,5^{\frac {1}{4}}-50 i \sqrt {5}-130 i}{25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{4} = R_{4}-\frac {\left (-135 i \sqrt {5}+50 \,5^{\frac {1}{4}}-56 \,5^{\frac {3}{4}}-85 i\right ) \left (25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i\right ) R_{3}}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right ) \left (38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-21 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-17 i}{5^{\frac {3}{4}}+i}&0\\ 0&0&\frac {38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i}{25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i}&\frac {54 \,5^{\frac {3}{4}}+125 \,5^{\frac {1}{4}}+55 i \sqrt {5}+60 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {18 \,5^{\frac {3}{4}}+50 \,5^{\frac {1}{4}}-55 i \sqrt {5}+5 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&0\\ 0&0&0&\frac {60360800 i 5^{\frac {3}{4}}[phone] i 5^{\frac {1}{4}}[phone] \sqrt {5}[phone]}{\left (38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i\right ) \left (25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i\right ) \left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right )^{2} \left (5^{\frac {3}{4}}+i\right )^{2}}&-\frac {200 \left (4348 i 5^{\frac {3}{4}}-11140 \sqrt {5}-21992+8393 i 5^{\frac {1}{4}}\right )}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right )^{2} \left (5^{\frac {3}{4}}+i\right )^{2} \left (38 i 5^{\frac {3}{4}}+175 i 5^{\frac {1}{4}}+110 \sqrt {5}+115\right )}&0\\ 0&0&\frac {-110 i \sqrt {5}+25 \,5^{\frac {1}{4}}-52 \,5^{\frac {3}{4}}-75 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {4 \,5^{\frac {3}{4}}+75 \,5^{\frac {1}{4}}+20 i \sqrt {5}+5 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {13 \,5^{\frac {3}{4}}+225 \,5^{\frac {1}{4}}-50 i \sqrt {5}-130 i}{25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {\left (-110 i \sqrt {5}+25 \,5^{\frac {1}{4}}-52 \,5^{\frac {3}{4}}-75 i\right ) \left (25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i\right ) R_{3}}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right ) \left (38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-21 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-17 i}{5^{\frac {3}{4}}+i}&0\\ 0&0&\frac {38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i}{25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i}&\frac {54 \,5^{\frac {3}{4}}+125 \,5^{\frac {1}{4}}+55 i \sqrt {5}+60 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {18 \,5^{\frac {3}{4}}+50 \,5^{\frac {1}{4}}-55 i \sqrt {5}+5 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&0\\ 0&0&0&\frac {60360800 i 5^{\frac {3}{4}}[phone] i 5^{\frac {1}{4}}[phone] \sqrt {5}[phone]}{\left (38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i\right ) \left (25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i\right ) \left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right )^{2} \left (5^{\frac {3}{4}}+i\right )^{2}}&-\frac {200 \left (4348 i 5^{\frac {3}{4}}-11140 \sqrt {5}-21992+8393 i 5^{\frac {1}{4}}\right )}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right )^{2} \left (5^{\frac {3}{4}}+i\right )^{2} \left (38 i 5^{\frac {3}{4}}+175 i 5^{\frac {1}{4}}+110 \sqrt {5}+115\right )}&0\\ 0&0&0&\frac {476950 i 5^{\frac {3}{4}}+2495900 \sqrt {5}+6058600+692550 i 5^{\frac {1}{4}}}{\left (i 5^{\frac {3}{4}}-1\right )^{2} \left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right )^{2} \left (38 i 5^{\frac {3}{4}}+175 i 5^{\frac {1}{4}}+110 \sqrt {5}+115\right )}&\frac {91847800 i 5^{\frac {3}{4}}[phone] i 5^{\frac {1}{4}}[phone] \sqrt {5}[phone]}{\left (38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i\right ) \left (25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i\right ) \left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right )^{2} \left (5^{\frac {3}{4}}+i\right )^{2}}&0 \end {array} \right ] \end {align*}

\begin {align*} R_{5} = R_{5}-\frac {\left (9539 i 5^{\frac {3}{4}}+49918 \sqrt {5}+121172+13851 i 5^{\frac {1}{4}}\right ) \left (38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i\right ) \left (25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i\right ) \left (5^{\frac {3}{4}}+i\right )^{2} R_{4}}{4 \left (i 5^{\frac {3}{4}}-1\right )^{2} \left (38 i 5^{\frac {3}{4}}+175 i 5^{\frac {1}{4}}+110 \sqrt {5}+115\right ) \left (301804 i 5^{\frac {3}{4}}+578435 i 5^{\frac {1}{4}}+421000 \sqrt {5}+1072274\right )} &\Longrightarrow \hspace {5pt}\left [\begin {array}{@{}ccccc!{\ifdefined \HCode |\else \color {red}\vline width 0.6pt\fi }c@{}} 1-i 5^{\frac {3}{4}}&3&5&2&4&0\\ 0&\frac {3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i}{5^{\frac {3}{4}}+i}&\frac {4 \,5^{\frac {3}{4}}-21 i}{5^{\frac {3}{4}}+i}&\frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i}&\frac {3 \,5^{\frac {3}{4}}-17 i}{5^{\frac {3}{4}}+i}&0\\ 0&0&\frac {38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i}{25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i}&\frac {54 \,5^{\frac {3}{4}}+125 \,5^{\frac {1}{4}}+55 i \sqrt {5}+60 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&\frac {18 \,5^{\frac {3}{4}}+50 \,5^{\frac {1}{4}}-55 i \sqrt {5}+5 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )}&0\\ 0&0&0&\frac {60360800 i 5^{\frac {3}{4}}[phone] i 5^{\frac {1}{4}}[phone] \sqrt {5}[phone]}{\left (38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i\right ) \left (25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i\right ) \left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right )^{2} \left (5^{\frac {3}{4}}+i\right )^{2}}&-\frac {200 \left (4348 i 5^{\frac {3}{4}}-11140 \sqrt {5}-21992+8393 i 5^{\frac {1}{4}}\right )}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right )^{2} \left (5^{\frac {3}{4}}+i\right )^{2} \left (38 i 5^{\frac {3}{4}}+175 i 5^{\frac {1}{4}}+110 \sqrt {5}+115\right )}&0\\ 0&0&0&0&0&0 \end {array} \right ] \end {align*}

Therefore the system in Echelon form is \[ \left [\begin {array}{ccccc} 1-i 5^{\frac {3}{4}} & 3 & 5 & 2 & 4 \\ 0 & \frac {3 \,5^{\frac {3}{4}}-5 i \sqrt {5}-13 i}{5^{\frac {3}{4}}+i} & \frac {4 \,5^{\frac {3}{4}}-21 i}{5^{\frac {3}{4}}+i} & \frac {5^{\frac {3}{4}}-9 i}{5^{\frac {3}{4}}+i} & \frac {3 \,5^{\frac {3}{4}}-17 i}{5^{\frac {3}{4}}+i} \\ 0 & 0 & \frac {38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i}{25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i} & \frac {54 \,5^{\frac {3}{4}}+125 \,5^{\frac {1}{4}}+55 i \sqrt {5}+60 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )} & \frac {18 \,5^{\frac {3}{4}}+50 \,5^{\frac {1}{4}}-55 i \sqrt {5}+5 i}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right ) \left (5^{\frac {3}{4}}+i\right )} \\ 0 & 0 & 0 & \frac {60360800 i 5^{\frac {3}{4}}[phone] i 5^{\frac {1}{4}}[phone] \sqrt {5}[phone]}{\left (38 \,5^{\frac {3}{4}}+175 \,5^{\frac {1}{4}}-110 i \sqrt {5}-115 i\right ) \left (25 \,5^{\frac {1}{4}}+10 \,5^{\frac {3}{4}}+20 i \sqrt {5}+13 i\right ) \left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right )^{2} \left (5^{\frac {3}{4}}+i\right )^{2}} & -\frac {200 \left (4348 i 5^{\frac {3}{4}}-11140 \sqrt {5}-21992+8393 i 5^{\frac {1}{4}}\right )}{\left (3 i 5^{\frac {3}{4}}+5 \sqrt {5}+13\right )^{2} \left (5^{\frac {3}{4}}+i\right )^{2} \left (38 i 5^{\frac {3}{4}}+175 i 5^{\frac {1}{4}}+110 \sqrt {5}+115\right )} \\ 0 & 0 & 0 & 0 & 0 \end {array}\right ] \left [\begin {array}{c} v_{1} \\ v_{2} \\ v_{3} \\ v_{4} \\ v_{5} \end {array}\right ] = \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \] The free variables are \(\{v_{5}\}\) and the leading variables are \(\{v_{1}, v_{2}, v_{3}, v_{4}\}\). Let \(v_{5} = t\). Now we start back substitution. Solving the above equation for the leading variables in terms of free variables gives equation \(\left \{v_{1} = -\frac {t \left (-535254249891900 i-239503585554020 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+2130940072176550 \,5^{\frac {1}{4}}\right )}{-2632502774254600 i-1173944063380280 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}}, v_{2} = -\frac {t \left (-279647606192588874424118000-125061156607937918635653000 \sqrt {5}+275916074995980288803494400 i 5^{\frac {1}{4}}+123392714550224695810605800 i 5^{\frac {3}{4}}\right )}{\left (-3 \,5^{\frac {3}{4}}+25 i \sqrt {5}+2 i\right ) \left (-2632502774254600 i-1173944063380280 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}\right ) \left (7796255 \sqrt {5}[phone] i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-19119640\right )}, v_{3} = \frac {t \left (128836653867695981523763000+57624394302796252937306600 \sqrt {5}+212210843887447208725771600 i 5^{\frac {1}{4}}+94898966142959340055582000 i 5^{\frac {3}{4}}\right )}{\left (-3 \,5^{\frac {3}{4}}+25 i \sqrt {5}+2 i\right ) \left (-2632502774254600 i-1173944063380280 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}\right ) \left (7796255 \sqrt {5}[phone] i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-19119640\right )}, v_{4} = \frac {t \left (481030535 i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-1082903250-492353920 \sqrt {5}\right )}{7796255 \sqrt {5}[phone] i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-19119640}\right \}\)

Hence the solution is \[ \text {Expression too large to display} \] 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 {t \left (-535254249891900 \,\operatorname {I}-239503585554020 \,\operatorname {I} \sqrt {5}[phone] \,5^{\frac {3}{4}}+2130940072176550 \,5^{\frac {1}{4}}\right )}{-2632502774254600 \,\operatorname {I}-1173944063380280 \,\operatorname {I} \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}} \\ -\frac {t \left (-279647606192588874424118000-125061156607937918635653000 \sqrt {5}+275916074995980288803494400 \,\operatorname {I} \,5^{\frac {1}{4}}+123392714550224695810605800 \,\operatorname {I} \,5^{\frac {3}{4}}\right )}{\left (-3 \,5^{\frac {3}{4}}+25 \,\operatorname {I} \sqrt {5}+2 \,\operatorname {I}\right ) \left (-2632502774254600 \,\operatorname {I}-1173944063380280 \,\operatorname {I} \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}\right ) \left (7796255 \sqrt {5}[phone] \,\operatorname {I} \,5^{\frac {1}{4}}[phone] \,\operatorname {I} \,5^{\frac {3}{4}}-19119640\right )} \\ \frac {t \left (128836653867695981523763000+57624394302796252937306600 \sqrt {5}+212210843887447208725771600 \,\operatorname {I} \,5^{\frac {1}{4}}+94898966142959340055582000 \,\operatorname {I} \,5^{\frac {3}{4}}\right )}{\left (-3 \,5^{\frac {3}{4}}+25 \,\operatorname {I} \sqrt {5}+2 \,\operatorname {I}\right ) \left (-2632502774254600 \,\operatorname {I}-1173944063380280 \,\operatorname {I} \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}\right ) \left (7796255 \sqrt {5}[phone] \,\operatorname {I} \,5^{\frac {1}{4}}[phone] \,\operatorname {I} \,5^{\frac {3}{4}}-19119640\right )} \\ \frac {t \left (481030535 \,\operatorname {I} \,5^{\frac {1}{4}}[phone] \,\operatorname {I} \,5^{\frac {3}{4}}-1082903250-492353920 \sqrt {5}\right )}{7796255 \sqrt {5}[phone] \,\operatorname {I} \,5^{\frac {1}{4}}[phone] \,\operatorname {I} \,5^{\frac {3}{4}}-19119640} \\ t \end {array}\right ] = t \left [\begin {array}{c} -\frac {-535254249891900 i-239503585554020 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+2130940072176550 \,5^{\frac {1}{4}}}{-2632502774254600 i-1173944063380280 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}} \\ -\frac {-279647606192588874424118000-125061156607937918635653000 \sqrt {5}+275916074995980288803494400 i 5^{\frac {1}{4}}+123392714550224695810605800 i 5^{\frac {3}{4}}}{\left (-3 \,5^{\frac {3}{4}}+25 i \sqrt {5}+2 i\right ) \left (-2632502774254600 i-1173944063380280 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}\right ) \left (7796255 \sqrt {5}[phone] i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-19119640\right )} \\ \frac {128836653867695981523763000+57624394302796252937306600 \sqrt {5}+212210843887447208725771600 i 5^{\frac {1}{4}}+94898966142959340055582000 i 5^{\frac {3}{4}}}{\left (-3 \,5^{\frac {3}{4}}+25 i \sqrt {5}+2 i\right ) \left (-2632502774254600 i-1173944063380280 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}\right ) \left (7796255 \sqrt {5}[phone] i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-19119640\right )} \\ \frac {481030535 i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-1082903250-492353920 \sqrt {5}}{7796255 \sqrt {5}[phone] i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-19119640} \\ 1 \end {array}\right ] \] Or, by letting \(t = 1\) then the eigenvector is \[ \text {Expression too large to display} \] Which can be normalized to \[ \text {Expression too large to display} \] The following table summarises the result found above.

\(\lambda \)	algebraic	geometric	defective	associated
	multiplicity	multiplicity	eigenvalue?	eigenvectors
\(15\)	\(1\)	\(5\)	No	\(\left [\begin {array}{c} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end {array}\right ]\)
\(5^{\frac {3}{4}}\)	\(1\)	\(5\)	No	\(\left [\begin {array}{c} -2 \,5^{\frac {3}{4}}-4 \,5^{\frac {1}{4}}+3 \sqrt {5}+6 \\ 5^{\frac {3}{4}}+2 \,5^{\frac {1}{4}}-\sqrt {5}-4 \\ 5^{\frac {3}{4}}+3 \,5^{\frac {1}{4}}-2 \sqrt {5}-4 \\ -5^{\frac {1}{4}}+1 \\ 1 \end {array}\right ]\)
\(-i 5^{\frac {3}{4}}\)	\(1\)	\(5\)	No	\(\left [\begin {array}{c} \frac {-7542856792418920-3370828393023080 \sqrt {5}-2398617049157928 i 5^{\frac {3}{4}}-5360068694461640 i 5^{\frac {1}{4}}}{-323060820625160-142295694614440 \sqrt {5}+7355700331154200 i 5^{\frac {1}{4}}+3290903801182872 i 5^{\frac {3}{4}}} \\ -\frac {-903502865289325283095578240-404060124039924116964757120 \sqrt {5}-415477534769715915186883584 i 5^{\frac {3}{4}}-929038567351137060812826880 i 5^{\frac {1}{4}}}{\left (40 i \sqrt {5}-125 \,5^{\frac {1}{4}}+5^{\frac {3}{4}}+2 i\right ) \left (-323060820625160-142295694614440 \sqrt {5}+7355700331154200 i 5^{\frac {1}{4}}+3290903801182872 i 5^{\frac {3}{4}}\right ) \left (126419610 \,5^{\frac {1}{4}}-67376776 i-30929640 i \sqrt {5}[phone] \,5^{\frac {3}{4}}\right )} \\ \frac {448381104560849079638085120+200525260687192269352684800 \sqrt {5}-313035771894228876273258496 i 5^{\frac {3}{4}}-699968001936593435483505920 i 5^{\frac {1}{4}}}{\left (40 i \sqrt {5}-125 \,5^{\frac {1}{4}}+5^{\frac {3}{4}}+2 i\right ) \left (-323060820625160-142295694614440 \sqrt {5}+7355700331154200 i 5^{\frac {1}{4}}+3290903801182872 i 5^{\frac {3}{4}}\right ) \left (126419610 \,5^{\frac {1}{4}}-67376776 i-30929640 i \sqrt {5}[phone] \,5^{\frac {3}{4}}\right )} \\ -\frac {-27555518 \,5^{\frac {3}{4}}[phone] i \sqrt {5}[phone] i-59042834 \,5^{\frac {1}{4}}}{126419610 \,5^{\frac {1}{4}}-67376776 i-30929640 i \sqrt {5}[phone] \,5^{\frac {3}{4}}} \\ 1 \end {array}\right ]\)
\(-5^{\frac {3}{4}}\)	\(1\)	\(5\)	No	\(\left [\begin {array}{c} 2 \,5^{\frac {3}{4}}+4 \,5^{\frac {1}{4}}+3 \sqrt {5}+6 \\ -5^{\frac {3}{4}}-2 \,5^{\frac {1}{4}}-\sqrt {5}-4 \\ -5^{\frac {3}{4}}-3 \,5^{\frac {1}{4}}-2 \sqrt {5}-4 \\ 5^{\frac {1}{4}}+1 \\ 1 \end {array}\right ]\)
\(i 5^{\frac {3}{4}}\)	\(1\)	\(5\)	No	\(\left [\begin {array}{c} -\frac {-535254249891900 i-239503585554020 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+2130940072176550 \,5^{\frac {1}{4}}}{-2632502774254600 i-1173944063380280 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}} \\ -\frac {-279647606192588874424118000-125061156607937918635653000 \sqrt {5}+275916074995980288803494400 i 5^{\frac {1}{4}}+123392714550224695810605800 i 5^{\frac {3}{4}}}{\left (-3 \,5^{\frac {3}{4}}+25 i \sqrt {5}+2 i\right ) \left (-2632502774254600 i-1173944063380280 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}\right ) \left (7796255 \sqrt {5}[phone] i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-19119640\right )} \\ \frac {128836653867695981523763000+57624394302796252937306600 \sqrt {5}+212210843887447208725771600 i 5^{\frac {1}{4}}+94898966142959340055582000 i 5^{\frac {3}{4}}}{\left (-3 \,5^{\frac {3}{4}}+25 i \sqrt {5}+2 i\right ) \left (-2632502774254600 i-1173944063380280 i \sqrt {5}[phone] \,5^{\frac {3}{4}}+1253785384570450 \,5^{\frac {1}{4}}\right ) \left (7796255 \sqrt {5}[phone] i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-19119640\right )} \\ \frac {481030535 i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-1082903250-492353920 \sqrt {5}}{7796255 \sqrt {5}[phone] i 5^{\frac {1}{4}}[phone] i 5^{\frac {3}{4}}-19119640} \\ 1 \end {array}\right ]\)

Since the matrix is not defective, then it is diagonalizable. Let \(P\) the matrix whose columns are the eigenvectors found, and let \(D\) be diagonal matrix with the eigenvalues at its diagonal. Then we can write \[ A = P D P^{-1} \] Where \begin {align*} D &= \left [\begin {array}{ccccc} 15 & 0 & 0 & 0 & 0 \\ 0 & 5^{\frac {3}{4}} & 0 & 0 & 0 \\ 0 & 0 & -i 5^{\frac {3}{4}} & 0 & 0 \\ 0 & 0 & 0 & -5^{\frac {3}{4}} & 0 \\ 0 & 0 & 0 & 0 & i 5^{\frac {3}{4}} \end {array}\right ]\\ P &= \text {Expression too large to display} \end {align*}

Therefore \[ \left [\begin {array}{ccccc} 1 & 3 & 5 & 2 & 4 \\ 5 & 2 & 4 & 1 & 3 \\ 4 & 1 & 3 & 5 & 2 \\ 3 & 5 & 2 & 4 & 1 \\ 2 & 4 & 1 & 3 & 5 \end {array}\right ]=\text {Expression too large to display} \left [\begin {array}{ccccc} 15 & 0 & 0 & 0 & 0 \\ 0 & 5^{\frac {3}{4}} & 0 & 0 & 0 \\ 0 & 0 & -i 5^{\frac {3}{4}} & 0 & 0 \\ 0 & 0 & 0 & -5^{\frac {3}{4}} & 0 \\ 0 & 0 & 0 & 0 & i 5^{\frac {3}{4}} \end {array}\right ] \text {Expression too large to display}^{-1} \]