If I have the matrix
H = [2,2,1;-4,-1,4;4,5,x]
and I'm supposed to give x when at least one of the eigenvalues of the matrix is 2, how can I proceed? I tried to simply solve it with matlab but I think the matrix form gets in the way of solve() function.
You can start calculating the eigenvalues of the given matrix
$ A:=\left( \begin{matrix} 2 & 2 & 1 \\ -4 & -1 & 4 \\ 4 & 5 & x \end{matrix} \right)$
as usual by evaluating the determinant of $\lambda I - A$ being:
$\left|\lambda I - A\right| = \left| \begin{matrix} \lambda-2 & 2 & 1 \\ -4 & \lambda+1 & 4 \\ 4 & 5 & \lambda-x \end{matrix} \right| = (\lambda -2) \left| \begin{matrix} \lambda+1 & 4 \\ 5 & \lambda -x \end{matrix}\right| + 4 \left| \begin{matrix} 2 & 1 \\ 5 & \lambda -x \end{matrix} \right| - 4 \left| \begin{matrix} 2 & 1 \\ \lambda +1 & 4 \end{matrix} \right|$
which results in
$-\lambda^3 + \lambda^2 x + \lambda^2 - \lambda x + 18 \lambda + 6 x - 24$
which should be zero for $\lambda =2$ as $2$ should be an eigenvalue. We then find $x=-1$, hence $x=-1$ should be an answer.