Coefficient of $x^{5}$ in expansion of determinant How to find the coefficient of $x^{5}$ in expansion of determinant $\begin{vmatrix}
 7 & 2 & 1 & 3 & x & 7 & 2\\ 
 2 & 8 & 7 & x & 5 & 2 & 8\\ 
 1 & 7 & x & 8 & 4 & 1 & 7\\ 
 3 & x & 8 & 10 & 6 & 3 & 9\\ 
 x & 5 & 4 & 6 & 2 & 10 & x\\ 
 7 & 2 & 1 & 3 & 10 & x & 2\\ 
 2 & 8 & 7 & 9 & x & 2 & 8
\end{vmatrix}$?
The only way I see is to write all possible combinations with $x^{5}$ like $-(2*2)*x^{5} + (7 * 2)*x^{5} ...$
 A: The way I would do this is to use the definition of the determinant as a sum of terms over the permutation group $S_7$.  You can think of this the following way:
(1) overlap a $7$ by $7$ permutation matrix over your matrix, using the $1$s in the permutation matrix to pick out entries.  For a smaller illustrative example if you had $ \left( \begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i \end{array} \right)$ and used the permutation $\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \end{array} \right)$ you would end up with $ \left( \begin{array}{ccc}
a & \cdot & \cdot \\
\cdot & \cdot & f \\
\cdot & h & \cdot \end{array} \right)$.
(2) Multiply these entries together and multiply by the sign of the permutation you used.  In the example above you get $-afh$ since the permutation had an odd number of two-cycles (one).
(3) If you do this for every permutation and add the results you get the determinant.  
In your case there are $7!$ permutations, but the situation isn't bad since you are only looking for the coefficient of $x^5$.  You can determine the coefficient of $x^5$ by looking for all permutation matrices that would pick out exactly $5$ $x$'s in your matrix, and adding together the resulting terms.
For your problem you can only pick at most one of the $x$'s in row five and one of the $x$'s in column five.  You have only four other $x$s besides these to choose from, so there aren't many permutations that pick out exactly $5$ $x$'s.
A: Here is one approach which should work:
Let 
$$A(X) =\begin{vmatrix}
 7 & 2 & 1 & 3 & x & 7 & 2\\ 
 2 & 8 & 7 & x & 5 & 2 & 8\\ 
 1 & 7 & x & 8 & 4 & 1 & 7\\ 
 3 & x & 8 & 10 & 6 & 3 & 9\\ 
 x & 5 & 4 & 6 & 2 & 10 & x\\ 
 7 & 2 & 1 & 3 & 10 & x & 2\\ 
 2 & 8 & 7 & 9 & x & 2 & 8
\end{vmatrix}$$
Then, you are looking for $\frac{A^{5}(0)}{5!}$.
Since determinants are defined by products, the rule of differentiating a determinant is easy to find and prove: to differentiate a determinant, you differentiate a column, and keep the rest unchanged, and add all matrices together.
In this case, differentiating a column twice will give you zero. Because of this you get
$$A^{(5)}(x) = \sum_{K} A_K$$ 
where $K$ runs over all $5$ elements subset of $\{ 1,2,3,4,5,6,7 \}$ and 
$A_K$ is the determinant obtained from $A$ by differentiating the columns corresponding to $K$.
For example, if $K= \{ 1,2 ,4,5,7\}$ we have
$$A_K= \begin{vmatrix}
 0 & 0 & 1 & 0 & 1 & 7 & 0\\ 
 0 & 0 & 7 & 1 & 0 & 2 & 0\\ 
 0 & 0 & x & 0 & 0 & 1 & 0\\ 
 0 & 1 & 8 & 0 & 0 & 3 & 0\\ 
 1 & 0 & 4 & 0 & 0 & 10 & 1\\ 
 0 & 0 & 1 & 0 & 0 & x & 0\\ 
 0 & 0 & 7 & 0 & 1 & 2 & 0
\end{vmatrix}$$
As at least $4$ of the five columns will have six zeroes, and the last one has at least 5 zeroes, your determinants will be easy to calculate. Esspecially if you plug in $x=0$ before calculating each $A_K$. 
Unfortunately there are 21 of them, so the solution will be very vert long but simple.
