Matrix pivots with unknowns I'm reading this series of articles on matrix determinants.
It is said:

Cofactor expansions are also very useful when computing the determinant of a matrix with unknown entries. Indeed, it is inconvenient to row reduce in this case, because one cannot be sure whether an entry containing an unknown is a pivot or not.

Could somebody elaborate this statement a bit? What does it mean if an entry containing an unknown is a pivot or not? As I understand pivots in Gaussian elimination, they are simply the first non-zero elements of rows when the matrix is in its row-echelon form. So what does it mean if a matrix element with unknowns is not a pivot? And how does it prevent or hinder us from using row reduction to find the determinant?
 A: Suppose you have a matrix like $$\begin{pmatrix} x & 1 \\ 1 & x \end{pmatrix}$$ If $x \neq 0$ then your pivot is in the first row but if $x = 0$ then the pivot is instead in the second row.
Also notice that if $x = 1$ then the matrix has rank 1 and if $x = 0$ then the matrix has rank $2$. So based on this observation we expect there to be something preventing us from row reducing to the identity matrix because if $x = 1$ that would be impossible.
I disagree a little with how the textbook is worded. You can still do some row reduction, you just can't divide by $0$ or by any quantity (like $x$) with the possibility of being $0$.
So here what I would do is subtract $x$ times the second row from the first row to get $$\begin{pmatrix} 0 & 1 - x^2 \\ 1 & x \end{pmatrix}.$$
Then swap the rows to get $$\begin{pmatrix} 1 & x \\ 0 & 1 - x^2 \end{pmatrix}.$$ Now we can see that if $1 - x^2 = 0$ then the matrix has rank $1$ and otherwise the matrix has rank $2$.
Moreover, we know the determinant is $-(1 - x^2)$ since we did one row operation which did not affect the determinant and a second which multiplied it by $-1$.
