Find the determinant of a "piecewise" matrix

Question

Let $n$ be a positive integer, and $A_n$ is a $n \times n$ matrix, where the $(i,j)^\text{th}$ entry is $1$, if $i \leq j$ and it is $-1$ otherwise. What is the determinant of $A_n$?

My first thought is that it's similar to a piecewise function, and I found out that generally it's a weird "upper triangular" shape, where the diagonal and the lower part of the matrix is $1$, and the upper part of the matrix is $-1$. Where should I proceed? I plan on doing some row manipulation and maybe have a matrix in RREF, which is easy since it's all just one. Should the determinant be $1$?

Answer 1

Let $A$ be the $n \times n$ matrix given by \begin{align*} a_{ij} = \begin{cases} 1 & \text{if } i \leq j \\ -1 & \text{otherwise} \end{cases}. \end{align*} Then, $\det{A} = \det{B}$ where the matrix $B$ is such that \begin{align*} b_{ij} = \begin{cases} 1 & \text{if } i = 1 \\ 2 & \text{if } i \neq 1 \wedge i \leq j \\ 0 & \text{otherwise} \end{cases}. \end{align*} This is true because you can add the first row to all the other $n-1$ rows and the determinant will be the same. Next, note that $B$ is a triangular matrix and the entries of the diagonal are such that \begin{align*} b_{ii} = \begin{cases} 1 & \text{if } i = 1 \\ 2 &\text{otherwise} \end{cases}. \end{align*} Therefore, \begin{align*} \det{A} & = \det{B} \\ & = 1 \times \prod_{i=2}^n 2 \\ & = 2^{n-1}. & \square \end{align*}

Answer 2

I think it would be good to do row manipulations. If I am not wrong, the upper part of the matrix (including the diagonal) has the positive $1$ elements. Then we can add the first row to the $i$-th row for $i=2, \dots, n$. We will get a matrix of following form: $$\begin{pmatrix} 1 & 1&1&...&1\\ 0 & 2&2&...&2\\ \vdots & \ddots& \ddots &...&\vdots\\ 0 & 0&0&...&2\\ \end{pmatrix}$$ These row manipulations don't change the determinant and the determinant of the resulting matrix is obviously $\det(A) = 2^{n-1}$.