Help to evaluate determinant I want to evaluate the determinant of the $n \times n$ matrix
$\left|\begin{array}{ccccc} 1 & 0 & \ldots & 0& 0 \\
0 & 0 & \ldots & 0 & -a\\
0 & 0 & \ldots & -a & 0\\
&&&\vdots \\
0 & -a & 0 &\dots & 0
\end{array}\right|.$
So I try to say that it is $(-1)^{ n + (n-1) + \ldots n-(n-2)}(-a)^{n-1}$. So power of -1 should be $\frac{(n-1)(n+2)}{2} + n-1$. However answer given is $n(n-1)/2$. Where is wrong?

Using cofactor expansion, we have:
$$
\left|\begin{array}{ccccc} 1 & 0 & \ldots & 0& 0 \\
0 & 0 & \ldots & 0 & -a\\
0 & 0 & \ldots & -a & 0\\
&&&\vdots \\
0 & -a & 0 &\dots & 0
\end{array}\right|
=
\left|\begin{array}{ccccc}
0 & \ldots & 0 & -a\\
0 & \ldots & -a & 0\\
&&&\vdots \\
-a & 0 &\dots & 0
\end{array}\right|
.
$$
The determinant of the anti-diagonal matrix in the RHS is give by (note that RHS is $n-1\times n-1$):
$$ \det = \dfrac{(-1)^{\frac{n(n-1)}{2}} (-a)^n }{(-a)} = (-1)^{\frac{n(n-1)}{2}}(-a)^{n-1}.$$
