# Determining complex eigenvalues of a real $3\times 3$ matrix

If the following matrix
$$\begin{bmatrix} 40 & -29 & -11 \\ -18 & 30 & -12 \\ 26 & 24 & -50 \end{bmatrix}$$ has a complex eigen value $\lambda\neq 0$, then matrix has must have an eigenvalue

a. $\lambda +20$

b. $\lambda-20$

c. $20- \lambda$

d. $-20- \lambda$

Pay attention that because sum of the third column with $(-1)\times$first column is zero so one of the eigen values is $0$. We have three eigen values. Let them be $x,y$ and zero. We know that the sum of eigen values will be trace so $x+y+0=40+30-50=20$ so $x=20-y$ or $y=20-x$ so by calling one of nonzero eigen value $\lambda$ the other will be $20-\lambda$. And the correct answer is "c".