Matrices & Determinants
Determinant Properties
Grade Class 12
Question:
5. $\begin{vmatrix} a_1+b_1x & a_1x+b_1 & c_1 \\ a_2+b_2x & a_2x+b_2 & c_2 \\ a_3+b_3x & a_3x+b_3 & c_3 \end{vmatrix} = 0$, then possible conditions is/are -
(A) $x = 1 \; \forall \; a_i, b_i$, where $1 \le i, j \le 3$
(B) $x = -1 \; \forall \; a_i, b_i$, where $1 \le i, j \le 3$
(C) $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$
(D) $x = \pm 2 \; \forall \; a_i, b_i$, where $1 \le i, j \le 3$
Step-by-Step Solution
Key Concept: Perform column operations C1 -> C1 + C2 to factor out (1+x) or C1 -> C1 - C2 to factor out (1-x).
Applying $C_1 \to C_1 + C_2$, we get $\begin{vmatrix} (a_1+b_1)(1+x) & a_1x+b_1 & c_1 \\ (a_2+b_2)(1+x) & a_2x+b_2 & c_2 \\ (a_3+b_3)(1+x) & a_3x+b_3 & c_3 \end{vmatrix} = (1+x) \begin{vmatrix} a_1+b_1 & a_1x+b_1 & c_1 \\ a_2+b_2 & a_2x+b_2 & c_2 \\ a_3+b_3 & a_3x+b_3 & c_3 \end{vmatrix} = 0$. Thus $x=-1$ is a condition. Similarly, applying $C_1 \to C_1 - C_2$ gives $(1-x) \begin{vmatrix} a_1-b_1 & a_1x+b_1 & c_1 \\ a_2-b_2 & a_2x+b_2 & c_2 \\ a_3-b_3 & a_3x+b_3 & c_3 \end{vmatrix} = 0$, so $x=1$ is a condition. If $x \neq \pm 1$, the determinant can be zero if the matrix of coefficients is singular, i.e., the determinant of the matrix $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$.
Correct Answer: A,B,C