# How do you show that $\det(A)$ is the product of a linear form in $x_1, x_2, \ldots, x_n$ and a linear form in $y_1, y_2, \ldots, y_n$?

Someone pointed out another solution on how to prove Showing the determinant for a specific type of matrix is $\det(A) = (-1)^n 2^{n-1} \sum_{i=1}^{n} a_1 a_2 \ldots a_{i-1} a_{i+1} \ldots a_n$ and it involves a type of cofactor expansion theorem of a specific form.

Assuming some additional conditions on the ring and using the following formula

$\det(A) = \det (\begin{pmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ . & . & . & . \\ a_{n1} & a_{n2} & \ldots & a_{nn} \\ \end{pmatrix} ) z - \sum_{i=1}^{n} \sum_{j=1}^{n} (-1)^{i + j} M_{ij}x_i y_i$

where

$$A = \begin{pmatrix} a_{11} & a_{12} & \ldots & a_{1n} & x_1 \\ a_{21} & a_{22} & \ldots & a_{2n} & x_2 \\ . & . & . & . & . \\ a_{n1} & a_{n2} & \ldots & a_{nn} & x_n \\ y_1 & y_2 & \ldots & y_n & z \end{pmatrix}$$

and $M_{ij}$ is the minor of order $n-1$.

Then if we assume all $a_{ij}$ are elements of a field and $\det (\begin{pmatrix} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ . & . & . & . \\ a_{n1} & a_{n2} & \ldots & a_{nn} \\ \end{pmatrix} ) = 0$ then

How do you show that $\det(A)$ is the product of a linear form in $x_1, x_2, \ldots, x_n$ and a linear form in $y_1, y_2, \ldots, y_n$?

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What is $z$? A field element? An indeterminant? – Bill Cook Nov 8 '11 at 11:51
@BillCook It doesn't matter, the term with $z$ disappears. – Phira Nov 26 '11 at 16:47

Since the determinant of the matrix $A$ is 0, there is one row of $A$ that can be expressed as linear combination of the other rows, $row_k=\sum_{l\neq k}c_lrow_l$.
If we now set $x_k=\sum_{l\neq k}c_lx_l$, the determinant of the whole matrix would be zero, so we found a linear form in the $x_i$ that divides the determinant.
A similar argument gives a linear form in the $y_i$ that divides the determinant.
Since the total degree in the $x_i$ is one and the total degree in the $y_i$ is one, this gives the determinant up to a constant factor and we see that the determinant is the product of two linear forms.