Can the determinant of a matrix can be made $0$?

Question

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 $0$.

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

2. Is it true that there is an interesting entry in each row of a matrix with nonzero determinant?

Answer 1

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.

Answer 2

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.

Answer 3

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?

Answer 4

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.