The determinant $\begin{vmatrix}y^2&-xy&x^2\\a&b&c\\a'&b'&c'\end{vmatrix}$ is equal to
Answer & explanation
Correct answer: option 2
Let $Δ =\begin{vmatrix}y^2&-xy&x^2\\a&b&c\\a'&b'&c'\end{vmatrix}$. Then,
$Δ =\frac{1}{xy}\begin{vmatrix}xy^2&-xy&x^2y\\ax&b&cy\\a'x&b'&c'y\end{vmatrix}$ [Applying $C_1→C_1 +(x),C_3→C_3 +(y)$]
$=\frac{1}{xy}\begin{vmatrix}0&-xy&0\\ax+by&b&bx+cy\\a'x+b'y&b'&b'x+c'y\end{vmatrix}$ [Applying $C_1→C_1 + y C_2, C_3 → C_3+xC_2$]
$=\frac{1}{xy}.xy\begin{vmatrix}ax+by&bx + cy\\a'x+b'y &b'x+c' y\end{vmatrix}$ [Expanding along $R_1$]
$=\begin{vmatrix}ax+by&bx + cy\\a'x+b'y &b'x+c' y\end{vmatrix}$