Question

If $a_i, b_i, c_i \in \mathbb{R}$ ($i = 1, 2, 3$) and $x \in \mathbb{R}$, and
$\begin{vmatrix} a_1 + b_1x & a_1x + b_1 & c_1 \\ a_2 + b_2x & a_2x + b_2 & c_2 \\ a_3 + b_3x & a_3x + b_3 & c_3 \end{vmatrix} = 0$,
then:

• A. \( x = 1 \)
• B. \( x = -1 \)
• 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. \( x = 2 \)

Hint:

For determinants with parameters, simplify using column or row operations to extract factors systematically.

Answer 1

The Correct Option is A, B

1. The initial determinant is:

\[\begin{vmatrix} a_1 + b_1x & a_1x + b_1 & c_1 \\ a_2 + b_2x & a_2x + b_2 & c_2 \\ a_3 + b_3x & a_3x + b_3 & c_3 \end{vmatrix}\]

2. Apply the determinant property: Subtract column 2 from column 1:

\[C_1 \rightarrow C_1 - C_2.\]

3. The determinant simplifies to:

\[\begin{vmatrix} b_1(x - 1) & a_1x + b_1 & c_1 \\ b_2(x - 1) & a_2x + b_2 & c_2 \\ b_3(x - 1) & a_3x + b_3 & c_3 \end{vmatrix}\]

4. Factor out \((x - 1)\) from column 1:

\[(x - 1) \cdot \begin{vmatrix} b_1 & a_1x + b_1 & c_1 \\ b_2 & a_2x + b_2 & c_2 \\ b_3 & a_3x + b_3 & c_3 \end{vmatrix}.\]

5. The determinant equals zero if:

\(-x + 1 = 0 \implies x = 1\), or

The remaining determinant is zero.

6. Since \(x = 1\) satisfies the condition, the solution is \(x = 1\).