Is there a visual proof for why this property of a matrix is true?

Question

Let's say we have three equations written in their standard form:

\begin{align} a_1x + b_1y + c_1 = 0 && (l_1) \\ a_2x + b_2y + c_2 = 0 && (l_2) \\ a_3x + b_3y + c_3 = 0 && (l_3) \end{align}

Then if we consider the matrix $$ M = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix} $$ the following is true: $$ \det M \begin{cases} \lt 0 & \text{, if the intersection point of $l_1$ and $l_2$ lies on the "positive" side of $l_3$}\\ = 0 & \text{, if the intersection point of $l_1$ and $l_2$ lies exactly on $l_3$}\\ \gt 0 & \text{, if the intersection point of $l_1$ and $l_2$ lies on the "negative" side of $l_3$} \end{cases} $$

I understand why this works mathematically, but I can't find a way that explains it more intuitively. Ideally, I'm looking for a visual proof (e.g. involving some signed volume maybe ?).

Answer 1

You're asking for a visual proof of something that's not true.

There's a problem with talking about "the intersection point of $l_1$ and $l_2$" or "the positive side" of $l_3$. For example, if $a_1=a_2=b_1=b_2=c_1=c_2=0$, $l_1,l_2$ are just the whole of $\mathbb{R}^2$. If $a_3=b_3=0$ and $c_3>0$, everything is on the "positive side" of $l_3$.

However, let's ignore that. Consider $$y=0 \tag{$l_1$}$$ $$x=0 \tag{$l_2$} $$ and $$x+y+1=0 \tag{$l_3$}$$ Then we get $$M=\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1\end{pmatrix}.$$ Then $l_1$ is just the $x$-axis, $l_2$ is just the $y$-axis, and the intersection is the origin. This is on the positive side of $l_3$, since $1\cdot 0+1\cdot 0+1=1>0$. But $\det(M)=-1$.

However, if we just switch $l_1$ and $l_2$, we get the matrix $$\begin{pmatrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix},$$ which has $\det(M)=1$. But we didn't change $l_1,l_2$, or $l_3$, and the origin (intersection of $l_1,l_2$) still lies on the positive side of $l_3$.

Answer 2

As the question is worded, we're looking at the case wherein $l_1,l_2,l_3$ are lines (so situations like $a_1=b_1=c_1=0$ are omitted), and the case wherein $l_1\cap l_2$ is a single point (so $l_1,l_2$ are not the same line).

Let $$h_i=\begin{pmatrix} a_i \\ b_i \\ c_i\end{pmatrix}$$ and let $M$ be the matrix with columns $h_1,h_2,h_3$. Our assumptions that $l_1,l_2,l_3$ are lines and that $l_1\cap l_2$ is a single point imply that $h_1,h_2$ are linearly independent.

Recall that a square matrix has determinant $0$ iff the columns are linearly dependent. With our assumptions above (which imply that $\{h_1,h_2\}$ is a linearly independent set), the columns of $M$ will be linearly dependent iff $h_3\in M:= \text{span}\{h_1,h_2\}$.

Let $(p,q)$ be the intersection point of $l_1,l_2$, so $n=(p,q,1)\in M^\perp$. By considering dimensions, $M^\perp = \text{span}\{n\}$.

We note that $h_3\in M$ iff $h_3^\perp \supset M^\perp=\text{span}\{n\}$ (which is the crux of the argument, I think). But since $h_3^\perp$ is a vector space, it contains $M^\perp=\text{span}\{n\}$ iff it contains $n$, which occurs iff $h_3\cdot n=0$, which occurs iff $$a_3p+b_3q+c_3=0,$$ which occurs iff the intersection point $(p,q)$ lies on the line $l_3$.