# Difference between revisions of "Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 2 (a)/Solution 1"

Since an eigenvector  of  of eigenvalue  satisfies , i.e., , all we need to do is solve the following system.


we obtain 
Thus  and ; in vector form we get  where  is free. So an eigenvector corresponding to  is