<s>[INST] Set of linear equations Find eigenvalues and eigenvectors of the matrix:
$\begin{pmatrix} 1 & 0 & -2 \\ 1 & 3 & -1 \\ -1 & 0 & 2 \end{pmatrix}$
$\begin{pmatrix} 1-\lambda & 0 & -2 \\ 1 & 3-\lambda & -1 \\ -1 & 0 & 2-\lambda \end{pmatrix}$
I have found the eigenvalues, which are: 0 and 3.
For the 0-eig I have found their corresponding eigenvector (-6,1,-3)
and now I try to find the eigenvector for eig = 3:
$\begin{pmatrix} -2 & 0 & -2 \\ 1 & 0 & -1 \\ -1 & 0 & -1 \end{pmatrix}$~$\begin{pmatrix} 1 & 0 & -1 \\ 0 & 0 & 1 \end{pmatrix}$
and I am then supposed to find this solution: (0,1,0)
First, I don't understand how to get from the first matrix to the second matrix.
Secondly, how to find the solution. 
May you help me please.
Inverse matrix:
First step:
$$
\left[
      \begin{array}{ccc|ccc}
        1&0&0&1&0&0\\
        -1&2&1&0&1&0\\
        0&0&1&0&0&1\\
      \end{array}
    \right]
$$
Second step:
$$\left[
      \begin{array}{ccc|ccc}
        1&0&0&1&0&0\\
        0&2&1&1&1&0\\
        0&0&1&0&0&1\\
      \end{array}
    \right]
$$
Last step:
$$\left[
      \begin{array}{ccc|ccc}
        1&0&0&1&0&0\\
        0&1&0&1/2&1/2&-1/2\\
        0&0&1&0&0&1\\
      \end{array}
    \right]
$$
 [/INST] I would gladly write a comment but I have no enough rep. yet.
It seems that your matrix (3x3) has only 2 eigenvalues which means that you can not find the third eigenvector. You have to find a  generalized eigenvector, which seems to be your solution.
Regarding your second matrix, I believe that eig = 0 will give you (1 0 -1)' (its corresponding eigenvector transposed), and eig = 3 will give you (0 0 1)'
Again, apologize for using the answer window! 
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