# Evaluating determinant [duplicate]

This question already has an answer here:

Let $\{\alpha_{i}\}_{i=1}^{n}$ be distinct numbers. What is the determinant of the $n$ by $n$ matrix \begin{gather} \begin{pmatrix} \alpha_{1}^{n-1} & \alpha_{1}^{n-2} & \cdots & 1 \\ \alpha_{2}^{n-1} & \alpha_{2}^{n-2} & \cdots & 1 \\ \vdots & \vdots & \ddots & \vdots \\ \alpha_{n}^{n-1} & \alpha_{n}^{n-2} & \cdots & 1 \end{pmatrix} \end{gather} ?

## marked as duplicate by user1551 linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 25 '15 at 11:12

• @BasicUser if $P$ is a permutation matrix that switch two columns, then $\det(P)=-1$. Now if you have $n$ columns, you need $\nu=\lfloor n/2\rfloor$ switch $P_1,\ldots,P_\nu$ to transform your matrix in a Vandermonde matrix. The you can use $\det(V)=\det(AP_1\ldots P_{\nu})=\det(A)(-1)^{\nu}$. – Surb Aug 25 '15 at 10:59