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Matrices & Determinants: Vmatrix Vmatrix

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

Question
If ∣x2+xx+1x−22x2+3x−13x3x−3x2+2x+32x−12x−1∣=Px−12\begin{vmatrix} x^2+x & x+1 & x-2 \\ 2x^2+3x-1 & 3x & 3x-3 \\ x^2+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = Px - 12, then:
AP=24P = 24correct
BP=−24P = -24
CP=0P = 0
DP=12P = 12
Solution
Step 1: Apply R2→R2−(R1+R3)R_2 \to R_2 - (R_1 + R_3) Each entry in R1+R3R_1+R_3: (x2+x)+(x2+2x+3)=2x2+3x+3(x^2+x)+(x^2+2x+3)=2x^2+3x+3; (x+1)+(2x−1)=3x(x+1)+(2x-1)=3x; (x−2)+(2x−1)=3x−3(x-2)+(2x-1)=3x-3. The new R2R_2 becomes (2x2+3x−1)−(2x2+3x+3),3x−3x,(3x−3)−(3x−3)(2x^2+3x-1)-(2x^2+3x+3), 3x-3x, (3x-3)-(3x-3) = (−4, 0, 0)(-4,\ 0,\ 0). Step 2: Expand along R2R_2
Δ=−4⋅(−1)2+1∣x+1x−22x−12x−1∣\Delta = -4 \cdot (-1)^{2+1}\begin{vmatrix} x+1 & x-2 \\ 2x-1 & 2x-1 \end{vmatrix}
=4[(x+1)(2x−1)−(x−2)(2x−1)]=4(2x−1)(x+1−x+2)=4(2x−1)(3)= 4\bigl[(x+1)(2x-1)-(x-2)(2x-1)\bigr] = 4(2x-1)(x+1-x+2) = 4(2x-1)(3)
=12(2x−1)=24x−12= 12(2x-1) = 24x-12
 Therefore P=24P = 24. Answer: (1)
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