For a field $K$ and $1<n\in \mathbb{N}$ let $A\in K^{(n-1)\times n}$ aa matrix with rank $n-1$. For a row vector $z\in K^{1\times n}$ let $\left (\frac{A}{z}\right )\in K^{n\times n}$ be the matrix that we get if we add as the $n$-th row of the matrix $A$ the vector $z$.
To show that there is a column vector $s=(s_1, \ldots , s_n)^T$ such that for each row vector $z=(z_1, \ldots , z_n)$ it holds $$\det \left [\left (\frac{A}{z}\right )\right ]=\sum_{i=1}^nz_is_i=z\cdot s$$ we consider the laplace formula for the clculation of the determinant. We expand for the last row. For $i\in \{1, \ldots , n\}$ let $A_i$ be the submatrix of$A$ if we remove the $i$-th columnn and $s_i:=(-1)^{i+n}\det A_i$.
Then we get $$\det \left [\left (\frac{A}{z}\right )\right ]=\sum_{i=1}^nz_is_i$$
Is this correct? We could have expand also for an other row or column, or not?
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The vector $s$ is uniquely defined. Why does this hold?
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I want to show that the vector $s$ spans the solution space $L(A,0)$ of the linear system of equations $A\cdot x=0$ as a $K$-vector space.
How could we do that? Could you give me a hint?
Hint: Plug in the standard basis vectors for $z$ to get the coordinates of $s$.