Why $\det\begin{pmatrix} 1&1&0&...&0 \\ -1 & 1 & 0 &...&0 \\ -1 & 0 & 1 &...&0 \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ -1&0&0&...&1\end{pmatrix}=2$? Let $$A = \begin{pmatrix}
1 & 1 & 0 & 0 &0 &... & 0
\\ -1 & 1 & 0 & 0 &0&...&0
\\ -1 & 0 & 1 & 0 &0&...&0
\\ -1 & 0 & 0 & 1 &0&...&0
\\ -1 & 0 & 0 & 0 &1&...&0
\\ \vdots & \vdots & \vdots & \vdots &\vdots&\ddots&\vdots
\\ -1 & 0 & 0 & 0 &0&...&1\end{pmatrix}$$
be an $n \times n$ matrix consisting of $1$'s on its diagonal, a $1$ in the entry located on row $1$ and column $2$, and $-1$'s from  the second entry of the first column all the way to $n$. Prove that the determinant of $A$ is always $2$.
How should I begin this? I tried taking the determinants of the matrix when $n=2, 3, $ and $ 4$ and saw that they were all $2$, but am not sure how to proceed.
 A: Determinants are unchanged if you add a row, or a multiple thereof, to another.
Hence, add the first row to every other row.
$$\begin{pmatrix}
1 & 1 & 0 & 0 &0 & \cdots & 0
\\\ 0 & 2 & 0 & 0 &0& \cdots&0
\\\ 0 & 1 & 1 & 0 &0& \cdots&0
\\\ 0 & 1 & 0 & 1 &0& \cdots&0
\\\ 0 & 1 & 0 & 0 &1& \cdots&0
\\\ \vdots & \vdots & \vdots& \vdots &\vdots & \ddots &\vdots
\\\ 0 & 1 & 0 & 0 &0& \cdots&1\end{pmatrix}$$
Now add $-1/2$ times the second row to the first row:
$$\begin{pmatrix}
1 & 0 & 0 & 0 &0 & \cdots & 0
\\\ 0 & 2 & 0 & 0 &0& \cdots&0
\\\ 0 & 1 & 1 & 0 &0& \cdots&0
\\\ 0 & 1 & 0 & 1 &0& \cdots&0
\\\ 0 & 1 & 0 & 0 &1& \cdots&0
\\\ \vdots & \vdots & \vdots& \vdots &\vdots & \ddots &\vdots
\\\ 0 & 1 & 0 & 0 &0& \cdots &1\end{pmatrix}$$
This is a triangular matrix; the determinant of such a matrix is the product of its diagonal entries. Hence, the determinant is $2$.

Addendum:
As noted in the comments, a simpler solution in the same spirit is to just subtract the second row from the first. This gives
$$\begin{pmatrix}
2 & 0 & 0 & 0 &0 &... & 0
\\ -1 & 1 & 0 & 0 &0&...&0
\\ -1 & 0 & 1 & 0 &0&...&0
\\ -1 & 0 & 0 & 1 &0&...&0
\\ -1 & 0 & 0 & 0 &1&...&0
\\ \vdots & \vdots & \vdots & \vdots &\vdots&\ddots&\vdots
\\ -1 & 0 & 0 & 0 &0&...&1\end{pmatrix}$$
This too is lower-triangular, so the determinant is the product of the diagonal entries, which is clearly $2$.
