# Conjecture about determinant of matrix with column $j$ with all numbers not divisible by $j+1$

Consider an $$n \times n$$ matrix $$A_n$$ with elements $$a_{i,j}$$ such that $$a_{1,j},a_{2,j},a_{3,j},\ldots$$ is the sequence of numbers not divisible by $$j+1$$ in increasing order starting from $$1$$ (e.g. $$1,2,4,5,7,8,10,\ldots$$ for $$j = 2$$).

For fun, I have computed the determinant for $$n \le 8$$ and then conjectured that:

$$\lvert A_n \rvert = (-1)^{n+1}$$

Is the conjecture true? How would you prove it?

• To elaborate on the answer below, the sequence $$a_{1,j} - 1, \quad a_{2,j} - 2, \quad a_{3,j} - 3, \quad \dots \quad a_{i,j} - i,\quad \dots$$ will consist of $j$ 0's, followed by $j$ 1's, followed by $j$ 2's, and so forth. The first non-zero entry of this sequence is a 1, and this 1 will always occur where $i = j$, corresponding to the diagonal entry of the column. Jul 18, 2023 at 19:24

The last column of the matrix is always $$(1,2,3,\dots, n)^T$$. Subtracting this from every other column, we obtain a matrix of the following form:
$$\begin{bmatrix} 0,...,0 & 1 \\ L_n & *\end{bmatrix}$$
where $$L_n$$ is lower-triangular $$(n-1)\times(n-1)$$ with ones on the diagonal. (We can be more explicit if we like, but there is no need to.) Cofactor expansion along the first row shows that $$\det(A_n) = (-1)^{n+1}\det(L_n)$$, and $$\det(L_n)=1$$.
• What you describe as the bottom row is actually the bottom column; note that $a_{1,j}, a_{2,j}, \dots$ is a single column of the matrix. Of course, the answer is still basically correct because the transpose of a matrix has the same determinant Jul 18, 2023 at 19:24