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Let $a,b,c,d$ be distinct non-zero real numbers with $a+b=c+d$. Find the eigenvalues of the matrix

$$M=\left(\begin{array}{ccc} a&b&1\\ c&d&1\\ 1&-1&0\end{array}\right).$$

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    http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-02-05
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    What have you tried? Where does this problem come from? What difficulties did you face? – 2017-02-05
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    I have found the characteristic equation but am not and to solve it – 2017-02-05
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    Hint: what happens when you multiply $M$ by $(1 1 0)$? This gives one of the 3 eigenvalues, which should help in factoring the characteristic equation. – 2017-02-05
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    @dxiv Do you mean $0$ is an eigenvalue? – 2017-02-05
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    @ErickWong Right, of course. I removed my previous comment, it was a mental typo. – 2017-02-05

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Let $M$ be that $3\times3$ matrix and $\lambda=a+b=c+d$.

The vector $(1,1,0)$ is an eigenvector associated with the eigenvalue $\lambda$.

It remains to find two other eigenvalues $\mu$ and $\nu$.

We know that $\lambda+\mu+\nu=tr(M)=a+d$ and $\lambda\mu\nu=\det(M)$

This should give the solution ...

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    The matrix is singular so another eigenvalue is easy to obtain. – 2017-02-05
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    @ErickWong: that's true, 0 is an eigenvalue. But what if $a+b=c+d=0$ ? – 2017-02-05
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    Yeah, then I guess $\det(M)$ doesn't yield enough information since $\lambda = 0$. – 2017-02-05
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    Thank you for the solution..... I have got a+b as an eigen value is there any way to check if it's correct – 2017-02-05
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    The characteristic polynomial is easy to compute. I get $P(x)=-x(x+b-d)(x-a-b)$. So, it is correct to say that $a+b$ is an eigenvalue. – 2017-02-05