I have a question regarding finding the eigenvalues of this determinant.
$$\small \left| \begin{matrix}\dfrac{3}{4} - \lambda & \dfrac{1}{3}\\ \dfrac{1}{4} & \dfrac{2}{3} - \lambda \end{matrix}\right| = \left(\dfrac{3}{4} - \lambda\right) \left(\dfrac{2}{3} - \lambda\right) - \dfrac{1}{12} = \lambda^2\ \boxed{-\dfrac{17}{12}\lambda + \dfrac{5}{12}} = (\lambda - 1)\left(\lambda - \dfrac{5}{12}\right) = 0. $$
I don't quite understand where these two boxed values are coming from. It's probably something really simple but I can't for the life of me figure it out. I've looked at the equation multiple times and something just isn't clicking. Needless to say I am very stuck and frustrated! Any help is greatly appreciated.
$$\dfrac{3}{4} + \dfrac{2}{3} = \dfrac{3\times 3 + 2 \times 4}{3 \times 4} = \dfrac{9 + 8}{12} = \boxed{\dfrac{17}{12}}\\ \dfrac{3}{4}\times\dfrac{2}{3} - \dfrac{1}{12} = \dfrac{6}{12} - \dfrac{1}{12} = \boxed{\dfrac{5}{12}}$$ $\begin{align} \left( \dfrac{3}{4} - \lambda \right) \left( \dfrac{2}{3} - \lambda \right) - \dfrac{1}{12} &= \dfrac{3}{4} \times \dfrac{2}{3} - \lambda \left( \dfrac{3}{4} + \dfrac{2}{3}\right) + \lambda^2 - \dfrac{1}{12}\\ & = \dfrac{5}{12} - \dfrac{17}{12}\lambda + \lambda^2 \end{align}$