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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

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 λ-7 -63

0 -4 -4 λ-20

This results to a polynomial of degree *n=4*

= λ^{4}-18 λ^{4}-95λ^{2}+1156 λ-1120=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)( λ-7)(-20)=0

These values translates to the possible values of x_{1} ,x_{2} x_{3} and x_{4} respectively

Therefore,

X =