# A problem on characteristic values.

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$$?$$

Note that $$\frac1a+\frac1b=\frac{a+b}{ab}=\frac{9}{23}$$ and you have a similar relation for the other number.
• Yes, it's $\frac{trace}{det}$ and answer is b. Thanks. – Mathsaddict Aug 16 at 19:33