# Linear combination and sign of determinant

Given $$d+2$$ vectors $$x_1,x_2,\dots,x_{d+2} \in \mathbb{R}^d$$. Clearly, they are linearly dependent. However, we have better result: Let $$y_i = \begin{bmatrix} x_i\\ 1 \end{bmatrix} \in \mathbb{R}^{d+1}.$$ They are also linearly dependent. Thus, there exists coefficients $$a_1, a_2, \dots, a_{d+2}$$ that are not simultaneously zero which satisfy $$a_1y_1 + a_2y_2 + \dots + a_{d+2}y_{d+2} = 0$$ (or in other words, $$a_1x_1+a_2x_2+\dots+a_{d+2}x_{d+2}=0$$ and $$a_1+a_2+\dots+a_{d+2}=0$$).

Question is to determine sign of $$a_i$$. Obviously, it can be anything since we can just multiply all of the coefficients by $$-1$$. Nevertheless, there is some relation between sign of $$a_i$$ and sign of determinant of $$\begin{bmatrix} y_1 & \cdots & y_{i-1} & y_{i+1} & \cdots & y_{d+2} \end{bmatrix} \in \mathbb{R}^{(d+1) \times (d+1)}.$$ For every $$1 \leq i \leq d+2$$, $$\frac{\operatorname{sign}(a_i)}{\operatorname{sign} \Bigl( \det \bigl( \begin{bmatrix} y_1 & \cdots & y_{i-1} & y_{i+1} & \cdots & y_{d+2} \end{bmatrix} \bigr) \Bigr)}$$ is same. Any idea on how to prove this result? Thanks!

For $$i=1,\dots, d+1$$, the $$a_i's$$ are the solutions of $$(y_1, y_2,\dots, y_{d+1}) \left ( \begin{array}{c} a_1\\ a_2\\ \vdots\\ a_{d+1} \end{array} \right)=-a_{d+2}y_{d+2}$$ Using Cramer's rule for solving linear systems, the solution is given by $$\begin{split} a_i&=\frac{\det(y_1, \dots, y_{i-1}, -a_{d+2}y_{d+2},y_{i+1},\dots,y_{d+1})}{\det(y_1, \dots, y_{d+1})}\\ &= (-1)^{d-i}a_{d+2}\frac{\det(y_1, \dots, y_{i-1},y_{i+1},\dots, y_{d+2})}{\det(y_1, \dots, y_{d+1})} \end{split}$$ which gives the result but with a factor of $$(-1)^i$$. I'm not sure where I messed up.