Question

The roots of the equation \( \begin{vmatrix} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{vmatrix} = 0 \) are:

• A. \( 1, 2 \)
• B. \( -1, 2 \)
• C. \( -1, -2 \)
• D. \( 1, -2 \)
• E. \( 1, 1 \)

Hint:

Determinants with symmetric diagonal elements and constant off-diagonal elements often have a root found by summing a row, which makes all elements in that row identical.

Answer 1

The Correct Option is B

Concept: To find the roots of a determinant equation, we apply row or column operations to simplify the determinant and reveal factors. Common operations include adding all rows to the first row to find a common factor involving \( x \).

Step 1: Perform row operation \( R_1 \to R_1 + R_2 + R_3 \).
The first row becomes: \( (x-1+1+1), (1+x-1+1), (1+1+x-1) \), which is \( (x+1), (x+1), (x+1) \). \[ \begin{vmatrix} x+1 & x+1 & x+1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{vmatrix} = 0 \] Factor out \( (x+1) \) from the first row: \[ (x+1) \begin{vmatrix} 1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{vmatrix} = 0 \implies \text{One root is } x = -1. \]

Step 2: Simplify further using column operations \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \).
\[ (x+1) \begin{vmatrix} 1 & 0 & 0 \\ 1 & x-2 & 0 \\ 1 & 0 & x-2 \end{vmatrix} = 0 \] Expanding the determinant along the first row: \[ (x+1) [1 \cdot ((x-2)(x-2) - 0)] = 0 \] \[ (x+1)(x-2)^2 = 0 \]

Step 3: Identify the distinct roots.
The roots are \( x = -1 \) and \( x = 2 \) (repeated). Matching with the options, the roots are \( -1, 2 \).