<s>[INST] An Eigen Value of $\tiny \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$ is : Let $a,b,c,d$ be distinct non zero real numbers with $a+b=c+d.$ Then, an eigen value of the matrix $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$ is :
$(i)~a+c ~~(ii)~a+b~~(iii)~a-b~~(iv)~(b-d$
Attempt:
If $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$, then :
$A - \lambda I = \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix}$
$\det ( A- \lambda I) = \det \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix} $
Solving $\det ( A - \lambda I) = 0$ results in a third degree equation which is difficult to solve. 
Could anyone tell me of an easier method to solve it. Thank you very much for your help in this regard.
 [/INST] $a+b=c+d$ implies that we should add up the first two columns of the matrix using a vector of the form $\begin{pmatrix}
1\\ 
1\\ 
t
\end{pmatrix}$. 
You can see that if $A$ is the matrix, then $A\begin{pmatrix}
1\\ 
1\\ 
t
\end{pmatrix}$=$\begin{pmatrix}
a+b+t\\ 
c+d+t\\ 
0
\end{pmatrix}$. 
Therefore, the suitable choice is of course $t=0$, and so the eigenvalue is $a+b$.
</s>