Find the set of pairs such that determinant is 0. Question Find the set of all pairs (a,b) of real numbers such that
        \begin{vmatrix}
    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{vmatrix}
is equal to zero. (This is determinant=0)
I attempted to use the Method of Condensation but after about 4 pages of equations realized I missed a step in the beginning (twice....). SO before I waste anymore time, my question is this: 


*

*In order to solve this problem do I use method of condensation or just work through the vary nasty determinants? What is the best method to solve this? 

*Also, once I have computed the determinant equation and set it equal to zero, will it be just guess work to find the pairs or is there a method I should use?
Method of Condensation: http://www.maa.org/sites/default/files/pdf/Mathhorizons/pdfs/nov_2006_pp.12-15.pdf
 A: Calculating the determinant using Mathematica one gets 
$$-a (1 + 2 a + a^2 + 8 b + 5 a b + a^2 b - b^2)$$
Without more structure known about this matrix (perhaps it arrises in some natural context?), to compute the determinant symbolically you should probably use a computer algebra system. The method of condensation will not work, it seems to be a purely numerical algorithm, and since you have indeterminants to use row reduction you would need to work in the field of rational functions over a and b. In practice a better quick method for determining the determinant is to query an appropriate amount of points and use polynomial interpolation to recover the determinant. In the case each of your entries is linear one would expect this to run in $O(n^{2+\omega})$.
Since $a$ is a factor if you set $a = 0$ and $b$ to be anything the determinant will vanish.
A: As Nick R. wrote, the determinant is $$-a( 1+2a+a^2+8b+5ab+a^2b−b^2)$$
You could use expansion by minors to calculate it, though that's rather tedious for a $4 \times 4$.  Some preliminary row or column operations might help a bit, but maybe not a whole lot.  I agree that in this day and age a Computer Algebra system is the most sensible tool to use for something like this.
The second factor is irreducible, and describes a curve in the $ab$ plane: it turns out to be an elliptic curve. Since it's a quadratic in $b$, you could solve for $b$ in terms of $a$:
$$ b = \frac{{a}^{2}+5\,a+8  \pm \sqrt {{a}^{4}+10\,{a}^{3}+45\,{a}^{2}+88\,
a+68}}{2}
$$
The graph looks like this:

(note that the $b$ axis should also be included here). 
