# How do I prove this determinant reduction?

How do I prove the following ?

$$(\alpha \delta - \beta \gamma) \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} =\begin{vmatrix} \alpha x_1 + \beta y_1 & \gamma x_1+ \delta y_1 & 1 \\ \alpha x_2 + \beta y_2 &\gamma x_2+ \delta y_2 & 1 \\ \alpha x_3+ \beta y_3 & \gamma x_3+ \delta y_3 & 1 \end{vmatrix}$$

I am getting $(\alpha \delta + \beta \gamma)$ instead of $(\alpha \delta - \beta \gamma)$. Any hints ?

Let $$A=\begin{pmatrix} x_1&y_1&1\cr x_2&y_2&1\cr x_3&y_3&1\cr\end{pmatrix}$$
and $$B=\begin{pmatrix} \alpha& \gamma&0\cr \beta&\delta&0\cr 0&0&1\cr \end{pmatrix}$$
Compute $AB$.