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Matrices & Determinants: Total Number Roots Vmatrix Vmatrix

JEE Maths question with a full step-by-step solution.

Question
The total number of roots of ∣x−1111x−1111x−1∣=0\begin{vmatrix} x-1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{vmatrix} = 0 is:
A11
B22correct
C33
D44
Solution
Step 1: Apply R1→R1+R2+R3R_1 \to R_1+R_2+R_3 Each entry in R1R_1 becomes x+1x+1:
(x+1)∣1111x−1111x−1∣=0(x+1)\begin{vmatrix} 1 & 1 & 1 \\ 1 & x-1 & 1 \\ 1 & 1 & x-1 \end{vmatrix} = 0
 Step 2: Apply C2→C2−C1C_2 \to C_2-C_1 and C3→C3−C1C_3 \to C_3-C_1 inside the determinant
(x+1)∣1001x−2010x−2∣=(x+1)(x−2)2=0(x+1)\begin{vmatrix} 1 & 0 & 0 \\ 1 & x-2 & 0 \\ 1 & 0 & x-2 \end{vmatrix} = (x+1)(x-2)^2 = 0
 Step 3: State the roots x=−1x = -1 or x=2x = 2. These are 22 distinct roots. Answer: (2)
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