Determining an eigenvalue of given matrix. I am trying the following question from a competitive exam :
Which one of the following is an eigenvalue of the following matrix $A$ where $a,b,c,d$ are distinct real numbers such that $a+b=c+d$
$$
        \begin{bmatrix}
        a & b & 1 \\
        c & d & 1 \\
        1 & -1 & 0 \\
        \end{bmatrix}
$$
(1) $a+c$  (2) $a+b$  (3) $a-b$  (4) $b-d$  
I went with the conventional method of equating the characteristic polynomial of $A$ to 0 but the characteristic equation is quite complicated and I'm stuck. Is there another method? Any help is appreciated. Thanks. :)

The determinant of $A - xI$ isn't too hard to compute: expand along the last column we get 
$$(-c - (d-x)) - ( - a + x  - b) + -x ((a-x)(d-x) - bc)) = -x(x^2 + (a+d)x + (ad-bc))$$
where we use $a+b = c+d$ to cancel the first two subdeterminants.
So $x=0$ is a solution (clear as the sum of the first two columns of $A$ equals $(a+b) = (c+d)$ times the last) and the two remaining eigenvalues sum to $a+d$ (which is the trace, so that checks out) and their product is $ad-bc$.
Now try the different candidates for one root, compute the other and check the relations.  
