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.
 A: $a+b=c+d$ 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 $a+b$.
A: Thank you all for your replies. While I was just reflecting, I thought of the following solution, so, thought of sharing the following solution :
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} $
Adding the first two columns and placing it in the first results in no change in the determinant value : 
Hence : 
$\det ( A- \lambda I) = \det  \begin{pmatrix} a+b-\lambda&b&1 \\ c+d-\lambda&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix}$
$= (a+b-\lambda) \det  \begin{pmatrix} 1&b&1 \\ 1&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix} =  (c+d-\lambda) \det  \begin{pmatrix} 1&b&1 \\ 1&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix}$
Hence, $\det  ( A- \lambda I) =0 \implies a+b = \lambda = c+d$ 
A: The characteristic polynomial is not difficult to solve. Setting $d=a+b-c$ it is 
$$
\chi (A)=t^3 + t^2( - 2a - b + c) + t(a^2 + ab - ac - bc)=t(a + b - t)(a - c - t),
$$
so that $0$, $a+b$ and $d-b$ are the eigenvalues. This solution has the advantage that you obtain all eigenvalues.
A: If you write out the cubic you will see that you can group the quantity $-(c+d)$ together. If you replace that quantity with $-(a+b)$ you can cancel it with the $(a+b)$ that appears elsewhere in the cubic. Additionally, you will be able to cancel a $\lambda$ term with a $-\lambda$. When all is said and done you should find that $$\det(A-\lambda I) = -\lambda(a-\lambda)(d-\lambda)+bc\lambda$$ and you should have a much easier time solving $$-\lambda(a-\lambda)(d-\lambda)+bc\lambda=0$$ Can you proceed from here?
