Consider two distinct points \begin{bmatrix}a_1\\b_1\end{bmatrix} and \begin{bmatrix}a_2\\b_2\end{bmatrix} in the plane. Explain why the solutions \begin{bmatrix}x_1\\x_1\end{bmatrix} of the equation det(A) = 0 form a line and why this line goes through the two points \begin{bmatrix}a_1\\b_1\end{bmatrix} and \begin{bmatrix}a_2\\b_2\end{bmatrix}.

A = \begin{bmatrix}1&1&1\\x_1&a_1&b_1\\x_2&a_2&b_2\end{bmatrix}.

So far I just took the determinant of the matrix and got:

$a_1b_2 + b_1x_2 + x_1a_2 - a_1x_2 - a_2b_1 - b_2x_1$

and set it equal to zero as per the question. So I'm left with:

$a_1b_2 + b_1x_2 + x_1a_2 - a_1x_2 - a_2b_1 - b_2x_1 = 0$

I'm not sure where to go from here.

  • $\begingroup$ This is best explained considering the projective plane: the point $(x_1,x_2)$, when you embed the affine plane in the projective plane via $(x_1,x_2)\longmapsto (x_1:x_2:1)$ is on the line defined by the points $(a_1,a_2)$ and $(b_1,b_2)$ if and only if the corresponding projective points are collinear. $\endgroup$
    – Bernard
    Jun 29 '17 at 22:16

Use another characterisation of $\det A$ being zero:

\begin{align*} \det A = 0 &⇔ \text{the columns of $A$ are linearly dependent} \\ &⇔ ∃λ,μ,ν ∈ F \colon\quad λ\begin{bmatrix}x_1 \\ x_2\end{bmatrix} + μ\begin{bmatrix}a_1 \\ a_2\end{bmatrix} + ν\begin{bmatrix}b_1 \\ b_2\end{bmatrix} = 0 \quad\text{and}\quad λ + μ + ν = 0\\ &\hphantom{⇔}\text{and not all of $λ, μ, ν$ are zero}. \end{align*} Here, $F$ is the underlying field (probably $F = ℝ$) and the last equivalence comes from splitting up the nontrivial linear combinations of the columns into their first and the last two components respectively.

If $λ = 0$, then $μ = -ν$ and both are nonzero, so $\begin{bmatrix}a_1 \\ a_2\end{bmatrix} = \begin{bmatrix}b_1 \\ b_2\end{bmatrix}$, which is absurd since both points are assumed to be distinct.

What happens if $λ ≠ 0$? Furthermore, what happens if $\begin{bmatrix}x_1 \\ x_2\end{bmatrix}$ lies on the line through $\begin{bmatrix}a_1 \\ a_2\end{bmatrix}$ and $\begin{bmatrix}b_1 \\ b_2\end{bmatrix}$?

  • $\begingroup$ Hmm, I did not know that property. We are still learning about the determinant and I'm actually heading to class right now. When I get back I will see if I understand your answer more $\endgroup$
    – Hello
    Jun 29 '17 at 21:17
  • $\begingroup$ @Hello If not, you should tell us what you do know about the determinant yet. $\endgroup$
    – k.stm
    Jun 29 '17 at 21:22

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