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:

Show Hint

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

Updated On: Apr 11, 2025

$ x = 1 $
$ x = -1 $
\[ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \]
$ x = 2 $

Hide Solution

Verified By Collegedunia

1. The determinant given 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. Use the property of determinants: - 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. Factorize $(x - 1)$ 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. For the determinant to be zero, either:

$-x + 1 = 0 \implies x = 1$, or

The remaining determinant is zero.

6. Since $x = 1$ satisfies the condition, the correct answer is $x = 1$.

Was this answer helpful?

Let A = $\begin{bmatrix} 1 & 2 & 1 \\ 1 & 3 & 2 \\ 2 & 4 & 1 \end{bmatrix}$ and Mij, Aij respectively denote the minor, co-factor of an element aij of matrix A, then which of the following are true?
(A) M22 = -1
(B) A23 = 0
(C) A32 = 3
(D) M23 = 1
(E) M32 = -3
Choose the correct answer from the options given below:
- CUET (UG) - 2025
- Mathematics
- Matrices and Determinants
View Solution

Match List-I with List-II

List-I (Matrix)	List-II (Inverse of the Matrix)
(A) $\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}$	(I) $\begin{bmatrix} \tfrac{2}{15} & \tfrac{1}{10} \\[6pt] -\tfrac{1}{15} & \tfrac{1}{5} \end{bmatrix}$
(B) $\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}$	(II) $\begin{bmatrix} \tfrac{1}{5} & -\tfrac{2}{15} \\[6pt] -\tfrac{1}{10} & \tfrac{7}{30} \end{bmatrix}$
(C) $\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}$	(III) $\begin{bmatrix} \tfrac{1}{15} & \tfrac{7}{30} \\[6pt] \tfrac{2}{15} & -\tfrac{1}{30} \end{bmatrix}$
(D) $\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}$	(IV) $\begin{bmatrix} \tfrac{2}{15} & -\tfrac{1}{15} \\[6pt] \tfrac{1}{6} & \tfrac{1}{6} \end{bmatrix}$

List-I (Matrix)	List-II (Inverse of the Matrix)
(A) \(\begin{bmatrix} 1 & 7 \\ 4 & -2 \end{bmatrix}\)	(I) \(\begin{bmatrix} \tfrac{2}{15} & \tfrac{1}{10} \\[6pt] -\tfrac{1}{15} & \tfrac{1}{5} \end{bmatrix}\)
(B) \(\begin{bmatrix} 6 & -3 \\ 2 & 4 \end{bmatrix}\)	(II) \(\begin{bmatrix} \tfrac{1}{5} & -\tfrac{2}{15} \\[6pt] -\tfrac{1}{10} & \tfrac{7}{30} \end{bmatrix}\)
(C) \(\begin{bmatrix} 5 & 2 \\ -5 & 4 \end{bmatrix}\)	(III) \(\begin{bmatrix} \tfrac{1}{15} & \tfrac{7}{30} \\[6pt] \tfrac{2}{15} & -\tfrac{1}{30} \end{bmatrix}\)
(D) \(\begin{bmatrix} 7 & 4 \\ 3 & 6 \end{bmatrix}\)	(IV) \(\begin{bmatrix} \tfrac{2}{15} & -\tfrac{1}{15} \\[6pt] \tfrac{1}{6} & \tfrac{1}{6} \end{bmatrix}\)