Given the matrix A
To find real values of X such that A.X=0,
 deduce the forth row with the given rows to come up with a 4×4 matrix
 determine p(λ)=det (A λI_{n})

^{3. }Factorize p(λ) to obtain p(λ)=( λ λ_{1})^{n1}. (λ λ2)^{n2} …( λ λk)^{nk}
Where λi, i=1,…..,k.
These might be real or complex, whereby i and the powers n_{i} is referred to as the algebraic multiplicity of the eigenvalue λ_{i}
for every eigen value the associated eigenvalues are calculated.
For instance given λ_{i } as the eigenvector, the eigenvectors are represented by a linear system, given as A.X= λiX or (A λ_{i}I_{n})X)=0
We introduce a forth row to get a 4 by 4 matrix by subtracting two times row one from row two, we get
1 2 2 17
2 8 8 54
4 7 7 163
0 4 4 20
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The next step is to diagonalize the 4 by 4 matrix A and find the diagonal matrix B whereby, B=P^{1}AP. This is achieved by deducing a characteristic polynomial f_{A}=det (A_{t}I_{n})
f_{A}= λ+1 2 2 17
2 λ+8 8 54
4 7 λ7 63
0 4 4 λ20
This results to a polynomial of degree n=4
= λ^{4}18 λ^{4}95λ^{2}+1156 λ1120=0
Solving the polynomial simultaneously, we acquire the roots as the possible values of λ by taking arbitrary values from 20 to +20 through scrutinizing and examining the matrix carefully. The following values of λ satisfied the equation, 1, 8, 7, and 20.
Rewriting the polynomial equation in its factorized form
(λ +1)( λ+8)( λ7)(20)=0
These values translates to the possible values of x_{1} ,x_{2} x_{3} and x_{4} respectively
Therefore,
X =