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Matrices & Determinants: Value Determinant Vmatrix Vmatrix

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

Question
The value of the determinant ∣a+b+2cabcb+c+2abcac+a+2b∣\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} is:
A2(a+b+c)2(a+b+c)
B2(a+b+c)22(a+b+c)^2
C2(a+b+c)32(a+b+c)^3correct
D(2a+2b+2c)3(2a+2b+2c)^3
Solution
Step 1: Apply C1→C1+C2+C3C_1 \to C_1+C_2+C_3 Each entry in the new C1C_1 equals 2(a+b+c)2(a+b+c):
=2(a+b+c)∣1ab1b+c+2ab1ac+a+2b∣= 2(a+b+c)\begin{vmatrix} 1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix}
 Step 2: Apply R2→R2−R1R_2 \to R_2-R_1 and R3→R3−R1R_3 \to R_3-R_1
=2(a+b+c)∣1ab0a+b+c000a+b+c∣= 2(a+b+c)\begin{vmatrix} 1 & a & b \\ 0 & a+b+c & 0 \\ 0 & 0 & a+b+c \end{vmatrix}
 Step 3: Evaluate the upper triangular determinant
=2(a+b+c)⋅1⋅(a+b+c)⋅(a+b+c)=2(a+b+c)3= 2(a+b+c)\cdot 1\cdot(a+b+c)\cdot(a+b+c) = 2(a+b+c)^3
 Answer: (3)
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