Given matrix A: $$ A\in\mathbb{R}^{n \times n} $$ $$ A_{ij}=\begin{cases} 0 ~\text{if}~ i =j\\ 2i ~\text{if}~ i\neq j \end{cases} $$ Find the determinant of $A$.
How should I approach this? I tried rearranging the rows to get a triangular matrix but didn't do well.
So, the matrix looks like $A = \begin{pmatrix} 0 & 2 & 2 & \cdots & 2 \\ 4 & 0 & 4 & \cdots & 4 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 2n & 2n & 2n & \cdots & 0 \end{pmatrix}$.
So $\det A = 2.4.\cdots2n \det B$, where $B = \begin{pmatrix} 0 & 1 & 1 & \cdots & 1 \\ 1 & 0 & 1 & \cdots & 1 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 1 & 1 & 1 & \cdots & 0 \end{pmatrix}$.
Now, note that $n-1$ is an eigenvalue of $B$. Moreover, the Characteristic polynomial of $B$ is $(x+1)^{n-1}(x - (n-1))$. If you want the proof you can see this or simply for the computation of the determinant see this.
So $\det B = (-1)^{n-1}(n-1)$.
Finally $\det A = (-1)^{n-1}2^{n}(n-1) n!$.
