<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] 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, $\det  ( A- \lambda I) =0 \implies a+b = \lambda = c+d$ 
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