<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] 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.
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