Can the determinant of a matrix can be made $0$? An entry of an $n \times n$ matrix with nonzero determinant is defined as interesting if by changing this entry (and only this entry) the determinant of the matrix can be made $0$.


*

*Is it true that each entry of every matrix with nonzero determinant is interesting?

*Is it true that there is an interesting entry in each row of a matrix with nonzero determinant?
 A: Your question 1: Let us consider an entry $a$ of a matrix $M$; if its cofactor (or minor) is non zero, it is automatically "interesting" because the determinant of $M$ is a first degree polynomial $ua+v$ in this entry with $u \neq 0$. It suffices then to take the value $a=-v/u.$
An example among billions:
$$\det\pmatrix{5&-3&-2\\8&-5&-4\\a & 3&3}=9+2a \ \ \text{with cofactor}  12-10=2.$$
Thus, in a rather vast majority of cases, any entry is "interesting".
Your question 2: Yes necessarily, because if all the cofactors of a certain row are zero, the determinant itself is zero ; contradiction.
A: A couple of hints.
For 1), consider the matrix
$$
\begin{bmatrix}
1 & 0\\
0 & 1\\
\end{bmatrix}
$$
in particular the two zeroes.
For 2), consider the expansion of the determinant with respect to a row.
A: In an attempt to show that $a_{i,j}$ is interesting, note that the $n$ column vectors of size $n-1$ obtained by dropping row $i$ are linearly dependent. This may mean that the (cropped) $j$th column is a linear combination of the other (cropped) columns. In that case, you can change $a_{i,j}$ to match the value it should have for the "same" linear combination of the full column vectors. Now contemplate, how can this fail?
A: Let $M$ be an $n \times n$ matrix.
Using Laplace's formula and expanding along  the $j^{th}$ column, $|M| = \displaystyle \sum_{i=1}^n (-1)^{i+j}m_{i,j}M_{i,j}$ where $M_{ij}$ is the determinant of the submatrix of M that results from removing the $i^{th}$ row and the $j^{th}$ column.
Let $M[i,j;x]$ be the determinant you get of the matrix $M$ with the $ij^{th}$ member replaced with $x$. Then  $M[i,j;x] = ax - b$ where $a = (-1)^{i+j}M_{ij}$ and $b = a m_{i,j} - |M|$.
Suppose $a \ne 0$, then $M[i,j;\frac ba] = 0$.
Suppose $a=0$. If $b=0$, then $|M| = 0$. So, by hypothesis, $b \ne 0$. Then $M[i,j;x] = b \ne 0$ for all $x$.
So a matrix in interesting if and only if all of its minors are non zero.
