If a_i, b_i, c_i in mathbb{R} for i=1, 2, 3 a | vedclass

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(C) Let the given determinant be $\Delta$. Applying the column operation $C_1 	o C_1 - x C_2$:$\Delta = \begin{vmatrix} a_1(1-x^2) & a_1 x+b_1 & c_1

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$\begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} = 0$

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(C) Let the given determinant be $\Delta$. Applying the column operation $C_1 	o C_1 - x C_2$:$\Delta = \begin{vmatrix} a_1(1-x^2) & a_1 x+b_1 & c_1 \ a_2(1-x^2) & a_2 x+b_2 & c_2 \ a_3(1-x^2) & a_3 x+b_3 & c_3 \end{vmatrix} = 0$Taking $(1-x^2)$ common from the first column:$(1-x^2) \begin{vmatrix} a_1 & a_1 x+b_1 & c_1 \ a_2 & a_2 x+b_2 & c_2 \ a_3 & a_3 x+b_2 & c_3 \end{vmatrix} = 0$Alternatively, applying $C_1 	o C_1 + C_2$ and $C_1 	o C_1 - C_2$ shows that the determinant is proportional to $(1-x^2) 	imes \Delta_0$ where $\Delta_0$ is the determinant of the matrix with columns $a_i, b_i, c_i$.Thus, $(1-x^2) \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} = 0$.This implies that either $x^2 = 1$ or the determinant of the matrix formed by $a_i, b_i, c_i$ is $0$. Given the options, the correct condition is $\begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix} = 0$.

If $a_i, b_i, c_i \in \mathbb{R}$ for $i=1, 2, 3$ and $x \in \mathbb{R}$ and $\begin{vmatrix} a_1+b_1 x & a_1 x+b_1 & c_1 \\ a_2+b_2 x & a_2 x+b_2 & c_2 \\ a_3+b_3 x & a_3 x+b_3 & c_3 \end{vmatrix} = 0$, then:

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