Here is the question: Find the set of all pairs $(a, b)$ of real numbers such that
$\det(A)=\left| \, \begin{matrix} a+1 & 3a & b+3a & b+1\\ 2b & b+1 & 2-b & 1\\ a +2 & 0 & 1 & a +3\\ b -1 & 1 & a + 2 & a +b\\ \end{matrix}\, \right|=0. $
I have reduced the matrix using column operations where the first column is $(a,0,0,0)^T$, but that still leaves me with a nasty polynomial $p(a,b)$ to solve. When I solve, I get something like this: $\{(0,b)\, |\, b\in \mathbb{R}\}$ or $\left\{(a,b)\,|\, b\in \mathbb{R}\setminus\{-1\},\, a=\frac{(5 b+2)\pm \sqrt{(-5 b-2)^2-4 (-b-1) (b^2-8 b-1)}}{-2(b+1)}\right\}$
I feel like there has to be a trick here that I'm missing.
By the way, this is for a graduate level linear algebra class.