Consider a triangle $PQR$, $P(0,2,1), Q(1,0,2), R(0,4,9)$. Determined the polarities if triangle $PQR$ is self polar.
By definition of self polar triangle, point $P$ gets mapped to line $QR$, point $Q$ gets mapped to line $PR$, and point $R$ gets mapped to line $PQ$.
Equation of $QR; 8x_1+9x_2-4x_3=0$
Equation of $PR; x_1=0$
Equation of $PQ; 4x_1+x_2-2x_3=0$
We have that $(0,2,1)$ gets mapped to $(8,9,-4)$, $(1,0,2)$ gets mapped to $(1,0,0)$, $(0,4,9)$ gets mapped to $(4,1,-2)$. But I cannot find any polarities because the determinant of the matrix of the polarity becomes 0. In that case, some points are singular points. But how do I know which points are singular??
Singular points are defined by to $P^tC=0$, where $C$ is the matrix of the polarity. But I don’t even have the matrix of the polarity... how do I go about doing it?
I’m not sure why you think that there’s anything singular here. The polarity constraints are captured by the matrix equation $$C \begin{bmatrix} P & Q & R \end{bmatrix} = \begin{bmatrix} s_P l_{QR} & s_Q l_{RP} & s_R l_{PQ} \end{bmatrix} \tag 1$$ with $s_i\ne0$ and $l_\alpha$ homogeneous coordinate vector representations of the lines, from which $$C = \begin{bmatrix} s_P l_{QR} & s_Q l_{RP} & s_R l_{PQ} \end{bmatrix} \begin{bmatrix} P & Q & R \end{bmatrix}^{-1}.\tag 2$$ The three point coordinate vectors are obviously linearly independent, so the matrix on the right is invertible.
The three scalars $s_\alpha$ are indeterminate. Each unique set of ratios $s_P:s_Q:s_R$ generates a distinct polarity for which $\triangle{PQR}$ is self-polar. Since they must be nonzero, you can fix one of them at $1$ to get a two-parameter family of conics. The matrix $C$ is not unique (even up to scale) because the three polarity constraints are not independent, so are not in general sufficient to determine a unique conic.