2012

Abstract

Notice that $a_{ij}=-a_{ji}$, $i, j=0,1,\ldots , 2012$, i.e., the matrix $A=(a_{ij})_{i, j=0}^{2012}$ is skew-symmetric, $A^\text {T}=-A.$ Therefore,

$$\begin{aligned} \det \,A=\det \, A^\text {T}=\det (-A)=(-1)^{2013}\det \,A=-\det \, A, \end{aligned}$$

whence $\det \, A=0$.