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If the characteristic values of $\begin{pmatrix} 3 & -1 \\ 5 & 6 \end{pmatrix}$

are $a$ and $b$.

And of

$\begin{pmatrix} 1 & 2 \\ -1 & 5 \end{pmatrix}$ Are $c$ and $d$. Then the equation whose roots are $\frac{1}{a} + \frac{1}{b}$ and $\frac{1}{c} + \frac{1}{d}$ is

$(a)$. $201x^{2} - 161x + 54 = 0$

$(b)$. $161x^{2} - 201x + 54 = 0$

$(c)$. $201x^{2} +161x - 54 = 0$

$(d)$. $161x^{2} + 201x - 54 = 0$

Characteristic equation for both matrices are respectively $y^{2} - 9 y+23 = 0$ and $y^{2} - 6y +7 = 0$ Now, I don't have any idea where to go. Roots of this equation can be found, but calculations become quite complicated. Is there any other way to solve$?$

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Note that $$ \frac1a+\frac1b=\frac{a+b}{ab}=\frac{9}{23} $$ and you have a similar relation for the other number.

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  • $\begingroup$ Yes, it's $\frac{trace}{det}$ and answer is b. Thanks. $\endgroup$ – Mathsaddict Aug 16 at 19:33
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    $\begingroup$ @Mathsaddict You’re welcome $\endgroup$ – egreg Aug 16 at 19:34

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