Some properties of a $2\times 2$ matrix with repeated eigenvalues I got a problem in my exam

Consider the matrix $ A =\left(
                               \begin{array}{cc}
                                 a & b \\
                                  c & d\\
                               \end{array}
                             \right)$ with real entries. 
  Suppose it has repeated eigenvalues. Pick the correct statement:
  
  
*
  
*$bc = 0$ 
  
*$A$ is always a diagonal matrix
  
*$det(A)\geq 0$
  
*$\det(A)$ can take any real value.
  

I took up a example of matrix $ A =\left(
                               \begin{array}{cc}
                                 0 & 1 \\
                                  0 & 0\\
                               \end{array}
                             \right)$ it has repeated eigen values zero and is not a diagonal matrix. so one possibility is removed. If it has repetead eigen values then $\det(A)\geq 0$, since product of eigenvalues will be $\det(A)$. In any case product will be non negative. I am not sure about other two possibilities. Is there any better way to do this problem?
 A: All you did was right. Of course, since you've already deduced that $\det(A)\geq 0$, you already know it is not the case that $\det(A)$ can be any real value: it can't be negative! But it can be any nonnegative real value: given $r\gt 0$, take the diagonal matrix with diagonal entries $\sqrt{r}$ to get a matrix with that determinant.
So we are down to whether the matrix must have $bc=0$.
Say the characteristic polynomial is $t^2+2t+1 = (t+1)^2$. Then we can take
$$\left(\begin{array}{rr}
0 & -1\\
1 & -2
\end{array}\right)$$
and note that the characteristic polynomial is precisely $-t(-2-t)+1 = t^2 + 2t+1$, exactly what we want. However $bc=-1$. 
(How did I come up with that? It's the "companion matrix" of $t^2+2t+1$; but you can come up with such a matrix for any quadratic: if you have $t^2+at+b$, write $t^2+at+b = t(t+a)+b = -t(-t-a)+b$, so put a $0$ and a $-a$ on the diagonal, and have the other two entries multiply to $-b$ and you are done; now just pick a polynomial with a double root).
Or you can obtain an example by starting with a matrix that is not diagonal and has repeated eigenvalues different from $0$, say
$$\left(\begin{array}{cc}1&1\\0&1\end{array}\right)$$
and then conjugating by an appropriate invertible matrix, say
$$\left(\begin{array}{cr}
\frac{1}{2} & \frac{1}{2}\\
\frac{1}{2} & -\frac{1}{2}
\end{array}\right) \left(\begin{array}{cc}
\vphantom{\frac{1}{2}}1 & 1\\
\vphantom{\frac{1}{2}}0 & 1
\end{array}\right)\left(\begin{array}{cr}
\vphantom{\frac{1}{2}}1 & 1\\
\vphantom{\frac{1}{2}}1 & -1
\end{array}\right) = \left(\begin{array}{cr}
\frac{3}{2} & -\frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}\end{array}\right)$$
which again has $1$ as a repeated eigenvalue, but with $bc\neq 0$. 
A: The matrix
$$\begin{pmatrix} 2&-1 \\ 1&4 \end{pmatrix}$$
has a repeated eigenvalue 3. This excludes the answer $bc=0$ and shows that $A$ need not be diagonal.
You write that $\det(A)\geq 0$ because $\det(A)$ is the square of the repeated eigenvalue. I agree. However, you should argue why the eigenvalue is real (unless you're doing a course where complex eigenvalues are not considered). A simple argument would be that the trace of $A$ is twice the repeated eigenvalue, so the eigenvalue must indeed be real.
Remark. You may wonder how to find the matrix given above. Suppose you have a matrix $\left(\begin{smallmatrix} a&b\\c&d \end{smallmatrix}\right)$. You want it to have a repeated eigenvalue of 3. This means that $a+d=6$, so you pick for example $a=2$ and $d=4$. Moreover, you need $ad-bc=9$, so $bc=-1$. This means you can take $b=-1$ and $c=1$.
A: What you did was fine as far as it went, but since $A$ is only $2\times 2$, you can also simply solve for the eigenvalues. If you do, you find yourself solving the quadratic equation $$(a-\lambda)(d-\lambda)-bc=\lambda^2-(a+d)\lambda+(ad-bc)=0\;,\tag{1}$$ so $$\lambda=\frac{a+d\pm\sqrt{(a+d)^2-4(ad-bc)}}2\;,$$ and the eigenvalue is repeated iff 
$$\begin{align*}
0&=(a+d)^2-4(ad-bc)\\
&=a^2-2ad+d^2+4bc\\
&=(a-d)^2+4bc\;,
\end{align*}$$
i.e., iff $4bc=-(a-d)^2$. This guarantees that $bc\le 0$, but clearly $bc$ need not be $0$, and therefore $A$ need not be a diagonal matrix. Finally, it's clear from $(1)$ that $\det A$ is the product of the eigenvalues (even if you didn't know this already), so it's clear that $\det A\ge 0$ and therefore cannot assume all real values.
