The given matrix is
![A= \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]](https://img.qammunity.org/2018/formulas/mathematics/college/mx1reskmtukh992x5kycia3kn02axx7hok.png)
A given vector is of the form [a b c]'.
If a given vector {x} is an eigenvector of A, then
Test the vector x₁ = [1 -2 1]'
![det \left[\begin{array}{ccc}0&2&3\\4&7&6\\7&8&8\end{array}\right] =-2(32-42)+3(32-49)=-31](https://img.qammunity.org/2018/formulas/mathematics/college/don4bpj01z8rrz1avmbtx6sovdf8lisydl.png)
This is not an eigenvector because the determinant is not zero.
Test x₂ = [-1 0 1]'
![det \left[\begin{array}{ccc}2&2&3\\4&5&6\\7&8&8\end{array}\right] =2(40-48)-2(32-42)+3(32-35)=-5](https://img.qammunity.org/2018/formulas/mathematics/college/5t1djoosgkwqf9xpappdo468h66s6ruhzs.png)
This is not an eigenvector because the determinant is not ero.
Test x₃ = [0 0 0]'
![det \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] =(45-48)-2(36-42)+3(32-35)=0](https://img.qammunity.org/2018/formulas/mathematics/college/2dyxnue2ja1sgbu7j459tf22lk86oxzjxd.png)
This vector is an eigevector because the determinant is zero.
Answer:
The eigenvector is x₃