Show that $\xi (\lambda) \equiv \left | A - \lambda I \right | $, and find the eigenvalues

Question

Let a $2 \times 2$ matrix $$A = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}.$$

Show that then $\xi (\lambda) \equiv \left | A - \lambda I \right | = \lambda^{2} - (\alpha + \delta) \lambda + (\alpha \delta - \beta \gamma)$, and find the eigenvalues.

My attempt: I have found that $$ A - \lambda I = \begin{bmatrix} \alpha - \lambda &\beta \\ \gamma & \delta - \lambda \end{bmatrix},$$

and then solved for its determinant to obtain $$\left | A - \lambda I \right | = (\alpha - \lambda)(\delta - \lambda) - \beta \lambda $$ $$= \lambda^{2} - (\alpha + \delta) \lambda + (\alpha \delta - \beta \gamma).$$

Then I have set the determinant to zero and need to solve the following equation: $$ \lambda^{2} - (\alpha + \delta) \lambda + (\alpha \delta - \beta \gamma) = 0.$$

EDIT: Typo fixed

I have tried the quadratic formula and have been getting something quite messy $$ \lambda = \frac{\alpha + \delta \pm \sqrt{\alpha^{2}+\delta^{2}-2\alpha \delta + 4 \beta \gamma} }{2}.$$

Answer 1

The correct answer is

$$\lambda=\frac{\alpha+\delta\pm\sqrt{(\alpha-\delta)^2+4\beta\gamma}}2.$$

Answer 2

The zeros of the quadratic you obtained are the eigenvalues.

Answer 3

Note that when you apply the quadratic formula, you should find that $$ b^2 - 4ac = \\ [-(\alpha + \delta)]^2 - 4(\alpha \delta - \beta \gamma) = \\ \alpha^2 + 2 \alpha \delta + \delta^2 - 4 \alpha \delta + 4 \beta \gamma =\\ \alpha^2 - 2 \alpha \delta + \delta^2 + 4 \beta \gamma =\\ (\alpha - \delta)^2 + 4 \beta \gamma $$