$\phi_1∧ ... ∧\phi_n = (det A) dx_1 ∧ ... dx_n$

Question

For $i=1,...,n$, we consider the 1-tensors on $\mathbb{R}^n$ given by: $$\phi_i:=\sum_{k=1}^{n} \alpha_{ik}\mathrm{d}x_k \in \Lambda^1((\mathbb{R}^n)^*),$$ where $ A=(\alpha_{ik})_{1 \le i, k\le n} \in \mathrm{Mat}(n; \mathbb{R})$. Show that $$\phi_1∧ ... ∧\phi_n = (\det A) \mathrm{d}x_1 \land \cdots \land \mathrm{d}x_n.$$ What I thought: $$\phi_1 \land \cdots \land \phi_n=\alpha_1\mathrm{d}x_1 \land \cdots \land \alpha_n\mathrm{d}x_n= \left(\sum_{k=1}^{n} \alpha_{ik}\right) \mathrm{d}x_1\land \cdots \land\mathrm{d}x_n.$$ I don't know how to continue and I don't know if it really makes sense...

Answer 1

Think in a simple example: In $\mathbb R^2$, if $\phi_1 := a_{11}dx_1 + a_{12}dx_2$ and $\phi_2 := a_{21}dx_1 + a_{22}dx_2$, then using the bilinearity of $\wedge$ together with the fact that $dx_i \wedge dx_j = -dx_j \wedge dx_i$ for any $i$ and $j$ (in particular, $dx_i \wedge dx_j = 0$ when $i=j$) we have \begin{align} \phi_1 \wedge \phi_2 &= (a_{11}dx_1 + a_{12}dx_2) \wedge (a_{21}dx_1 + a_{22}dx_2) \\ &= a_{11}a_{21}dx_1 \wedge dx_1 + a_{11}a_{22}dx_1 \wedge dx_2 \\ & \qquad \qquad + a_{12}a_{21}dx_2 \wedge dx_1 + a_{12}a_{22}dx_2 \wedge dx_2 \\ &= a_{11}a_{22}dx_1 \wedge dx_2 + a_{12}a_{21}(-dx_1 \wedge dx_2) \\ &= (a_{11}a_{22} - a_{12}a_{21})dx_1 \wedge dx_2 \\ &= \det\! \begin{pmatrix} a_{11}&a_{12} \\ a_{21}&a_{22} \end{pmatrix} dx_1 \wedge dx_2. \end{align}