How to prove the cofactor of a matrix is related to it's minor? How do I prove $A_{ij}$, the cofactor of $a_{ij}$ of a matrix $D$, has the property $A_{ij}=(-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor obtained by deleting the $i$th row and $j$th column? 

For the case of $i=j=1$, $A_11$ will be the sum of the terms $(1)^{N(\alpha_1=1,\alpha_2,\dots,\alpha_n)} a_{\alpha_2 2} \cdots a_{\alpha_n n}$. Observe that there is no transposition (or inversion) that contains $a_{11}$. So $N(\alpha_1=1,\alpha_2,\dots,\alpha_n)=N(\alpha_2,\dots,\alpha_n)$. Then we have the sum of terms like $(-1)^{N(\alpha_2,\dots,\alpha_n) } a_{\alpha_2 2} \cdots a_{\alpha_n n}$. The sum of these term, however, is the determinant of the following matrix
$$
        \begin{pmatrix}
        a_{22} & \cdots & a_{2 n} \\
        \vdots & \vdots & \vdots \\
        a_{n 2} & \cdots & a_{nn} \\
        \end{pmatrix}
$$
and clearly this matrix is the minor $M_{11}$ of the original matrix. Thus $A_{11}=M_{11}$. This shows the claim $A_{ij}=(-1)^{i+j} M_{ij}$ when $i=j=1$. In the general case you just move the element from $i$th row $j$th column to the first row first column, and follow a similar argument.
