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This statement is from the book of Winitzki Linear Algebra via Exterior Products. (Section 3.4, page 123) Let $V$ be finite dimensional vector space, $\dim(V)=N$. The determinant of the matrix $v_{ij}$ is a linear function of each of the vectors $\{\mathbf{v}_i \}$. Consider $\det(v_{ij})$ as a linear function of the first vector, $\mathbf{v}_1$; this function is a covector that we may temporarily denote by $\mathbf{f}_1^{*}$. Show that $\mathbf{f}_1^{*}$ can be represented in the dual basis $\{\mathbf{e}_j^{*} \}$ as $$ \mathbf{f}_1^{*}=\sum_{i=1}^N(-1)^{i-1} B_{1i}\mathbf{e}_i^{*}, $$ where the coefficients $B_{1i}$ are minors of the matrix $v_{ij}$, that is, determinants of the matrix $v_{ij}$ from which row 1 and column $i$ have been deleted.

Solution. Consider one of the coefficients, for example $B_{11}\equiv\mathbf{f}_1^{*}(\mathbf{e}_1)$. This coefficient can be determined from the tensor equality $$ \mathbf{e}_1\wedge\mathbf{v}_2\wedge\ldots\wedge\mathbf{v}_N=B_{11}\mathbf{e}_1\wedge\ldots\wedge\mathbf{e}_N.\qquad(1) $$

Edit: the next sentence is

"We could reduce $B_{11}$ to a determinant of an $(n-1)\times(N-1)$ matrix if we could cancel $\mathbf{e}_1$on both sides of equality."

I don't understand the reason of this equality (1). Accepting this equality to be valid, the remainder part of the proof is correct. Please, give some details.

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This is fairly straightforward to check for say a $3\times 3$ matrix, but the details become messy to do this in general. This should give you enough idea of how to do the general case to convince you it is true though:

Suppose $N=3$. Suppose $v_i=\left(\begin{matrix} a_{i1} \\ a_{i2} \\ a_{i3}\end{matrix}\right)$. Then $B_{11}=det(e_1, v_2, v_3)=a_{22}a_{33}-a_{23}a_{32}$, i.e. the determinant of the lower right $(N-1)\times (N-1)$ minor.

Now just expand out in terms of basis vectors and use standard properties of wedges to see the identity.

$v_2=a_{22}e_2+a_{23}e_3$ and $v_3=a_{32}e_2+a_{33}e_3$,

Thus $e_1\wedge v_2 \wedge v_3 = e_1 \wedge (a_{22}e_2+a_{23}e_3)\wedge (a_{32}e_2+a_{33}e_3) \\ = (a_{22}e_1\wedge e_2 +a_{23}e_1\wedge e_3)\wedge (a_{32}e_2+a_{33}e_3)\\ = (a_{22}a_{33}-a_{23}a_{32})e_1\wedge e_2 \wedge e_3 \\ = B_{11} e_1 \wedge e_2 \wedge e_3$

Edit: I misunderstood the question:

Checking $B_{11}$ is the correct minor directly generalizes from what I wrote above too. There seems to be no need to go through that wedge equality (of course using the determinant interpretation it is equivalent). The definition of $f_1^*(e_1)$ is exactly placing $e_1$ in the first column and taking the determinant. So in general $B_{11}=f_1^*(e_1)=det(e_1, v_2, \ldots , v_N)$, thus computing the determinant using the cofactor expansion down the first column, the only non-zero entry is the first one and hence $B_{11}=(-1)^2$(times the correct minor).

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we want to prove that $B_{11}$ is the minor. Here you used this to prove the wedge equality. –  vesszabo Aug 5 '12 at 7:38
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