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Matrices & Determinants: Vmatrix Vmatrix Value

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

Question
If ∣32+k4232+3+k42+k5242+4+k52+k6252+5+k∣=0\begin{vmatrix} 3^2+k & 4^2 & 3^2+3+k \\ 4^2+k & 5^2 & 4^2+4+k \\ 5^2+k & 6^2 & 5^2+5+k \end{vmatrix} = 0, then the value of kk is:
A22
B11correct
C−1-1
D00
Solution
Step 1: Split the determinant Write C3=C1+{3,4,5}C_3 = C_1 + \{3, 4, 5\}, separating the kk-dependence:
∣9+k169+k16+k2516+k25+k3625+k∣+∣9+k16316+k25425+k365∣=0\begin{vmatrix} 9+k & 16 & 9+k \\ 16+k & 25 & 16+k \\ 25+k & 36 & 25+k \end{vmatrix} + \begin{vmatrix} 9+k & 16 & 3 \\ 16+k & 25 & 4 \\ 25+k & 36 & 5 \end{vmatrix} = 0
 The first determinant has C1=C3C_1 = C_3, so it vanishes. Step 2: Apply R2→R2−R1R_2 \to R_2-R_1 and R3→R3−R2R_3 \to R_3-R_2 to the second determinant
∣9+k1637919111∣=0\begin{vmatrix} 9+k & 16 & 3 \\ 7 & 9 & 1 \\ 9 & 11 & 1 \end{vmatrix} = 0
 Step 3: Apply C2→C2−C1C_2 \to C_2-C_1 then R3→R3−R2R_3 \to R_3-R_2
∣9+k7−k3721200∣=0\begin{vmatrix} 9+k & 7-k & 3 \\ 7 & 2 & 1 \\ 2 & 0 & 0 \end{vmatrix} = 0
 Expanding along R3R_3: 2[(7−k)(1)−3(2)]=2(1−k)=0  ⟹  k=12\bigl[(7-k)(1) - 3(2)\bigr] = 2(1-k) = 0 \implies k = 1. Answer: (2)
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