Exercise 13, § 8

Algebra, III, 640

Let $\Delta = \det(\alpha_{ij})$ be a determinant of order $n$ and $\Delta_{ij}$ the minor of order $n-1$ , the complement of $\alpha_{ij}$ . Show that

\begin{vmatrix} \alpha_{11} & \alpha_{12} & \cdots & \alpha_{1n} & x_1 \\ \alpha_{21} & \alpha_{22} & \cdots & \alpha_{2n} & x_2 \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ \alpha_{n1} & \alpha_{n2} & \cdots & \alpha_{nn} & x_n \\ y_1 & y_2 & \cdots & y_n & z \end{vmatrix} = \Delta z - \sum_{i,j} (-1)^{i+j} \Delta_{ij} x_i y_j.

If $\Delta = 0$ and the $\alpha_{ij}$ belong to a field, show that the above determinant is the product of a linear form in $x_1, x_2, \ldots, x_n$ and a linear form in $y_1, y_2, \ldots, y_n$ (use Exercise 5 of § 5 and Exercise 8 of II, § 6).

Give an example where this result fails to hold when the ring of scalars $A$ is not a field (take $A$ to be the ring $\mathbf{Z}/(6)$ and $n = 2$ ).

Solution

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