Arbitrary Eigenvalues So we're given $$C_A(x) = \det(xI-A)$$ and the matrix $$A=\begin{bmatrix}a_1 & a_2\\a_3 & a_4\end{bmatrix}$$
In the previous question, we were told to prove that $$C_A(x) = x^2-\operatorname{tr}(A)+\det(A)$$
This questions says "Let $λ_1$, $λ_2$, not necessarily distinct, be the eigenvalues of $A$. Show that
$$C_A (x) = x^ 2 - (λ_1 + λ_2 ) x + λ_1 λ_2$$
I know that the way to do this is to make $a_1 = \lambda_1$ and $a_4 = \lambda_2$ $a_2=0 $and $a_3 =0$. But that's where I'm having trouble. I don't know how to use the information given regarding the Eigenvalues to make all the $a$'s what I need them to be. Help would be much appreciated.

Eigenvalues are roots of the characteristic polynomial. On the other hand, you have Vieta's formula.
edit
if you don't know the Vieta's formula, then just write explicitly the characteristic polynomial:
$$x^2 - (a_1+a_4)x+(a_1a_4-a_2a_3).$$ It has roots $$\lambda_\pm = \frac{a_1+a_4\pm\sqrt{(a_1+a_4)^2 - 4 (a_1a_4-a_2a_3)}}{2}.$$
Now it is straightforward to check that $$\lambda_++\lambda_- = \operatorname{tr}(A),\\\lambda_+\lambda_-=\det(A).$$
