# Finding the kernel of linear determinant transform

Consider three distinct points on a plane $[a_1, a_2]^T$, $[b_1, b_2]^T$, $[c_1, c_2]^T$. Describe the set of all points $[x_1, x_2]^T$ satisfying the equation $$\det \begin{bmatrix} 1 & 1 & 1 & 1\\ x_1 & a_1 & b_1& c_1\\ x_2 & a_2 & b_2 & c_2\\ x_1^2 + x_2^2 & a_1^2 + a_2^2 & b_1^2 + b_2^2 & c_1^2 + c_2^2 \end{bmatrix} = 0$$

My Attempt: By Theorem 6.2.2 (this is a theorem that states the linearity of the determinant when you hold all but one row/column fixed which is the imput of the transformation; $T(x): \Bbb R^4 \to \Bbb R$), we have that this this is now a linear transform from $\Bbb R^4 \to \Bbb R$. Now we want to know what the kernel of this transform is. Because the given points are all distinct, we observe that that the vectors $[1, a_1, a_2, a_1^2 + a_2^2]$, $[1, b_1, b_2, b_1^2 + b_2^2]$, $[1, c_1, c_2, c_1^2 + c_2^2]$ are all linearly independent. Thus, the points $x$, such that $[1, x_1, x_2, x_1^2 + x_2^2]$ is in the kernel, make up only the points given.

Is this correct/close at all?

Define $f:\mathbb R^2\to\mathbb R^3$ by $f(x,y) = (x,y,x^2+y^2)$. The set $\{f(x,y) : x,y\in\mathbb R\}$ is the surface of a paraboloid in $\mathbb R^3$. The determinant is zero if and only if there are $s,t,u\in\mathbb R$ such that $f(x_1,x_2) = sf(a_1,a_2) + tf(b_1,b_2) + uf(c_1,c_2)$, where $s+t+u = 1$. This means that $f(x_1,x_2)$ is contained in the affine hull $A$ of the other three points, which is just the (maybe degenerate) surface spanned by those. In other words, the points $(x_1,x_2)$ for which the determinant is zero can be written as $\{(x_1,x_2) : f(x_1,x_2)\in A\}$ or (maybe better) $\pi(A\cap P)$, where $P$ is the paraboloid and $\pi$ the orthogonal projection onto the $x_1$-$x_2$-plane.