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Matrices & Determinants: Let Vmatrix Vmatrix Value

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

Question
Let Δ=∣a11a12a13a21a22a23a31a32a33∣\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} where apq=(ip)qa_{pq} = (i^p)^q and i=−1i = \sqrt{-1}. The value of Δ\Delta is:
AReal and positive
BReal and negative
C00
DImaginarycorrect
Solution
Step 1: Write out the matrix apq=ipqa_{pq} = i^{pq}, so the matrix is:
Δ=∣ii2i3i2i4i6i3i6i9∣\Delta = \begin{vmatrix} i & i^2 & i^3 \\ i^2 & i^4 & i^6 \\ i^3 & i^6 & i^9 \end{vmatrix}
 Step 2: Factor from columns Factor ii from C1C_1, i2i^2 from C2C_2, i3i^3 from C3C_3:
=i6∣111ii2i3i2i4i6∣= i^6 \begin{vmatrix} 1 & 1 & 1 \\ i & i^2 & i^3 \\ i^2 & i^4 & i^6 \end{vmatrix}
 Factor ii from R2R_2 and i2i^2 from R3R_3:
=i6⋅i⋅i2∣1111ii21i2i4∣=i9⋅V= i^6 \cdot i \cdot i^2 \begin{vmatrix} 1 & 1 & 1 \\ 1 & i & i^2 \\ 1 & i^2 & i^4 \end{vmatrix} = i^9 \cdot V
 Step 3: Evaluate the inner determinant Apply R2→R2−R1R_2 \to R_2-R_1 and R3→R3−R1R_3 \to R_3-R_1
V=∣1110i−1i2−10i2−1i4−1∣V = \begin{vmatrix} 1 & 1 & 1 \\ 0 & i-1 & i^2-1 \\ 0 & i^2-1 & i^4-1 \end{vmatrix}
 Since i4=1i^4 = 1, the entry i4−1=0i^4-1 = 0. Expanding along column 1:
V=1⋅∣i−1i2−1i2−10∣=−(i2−1)2=−(−1−1)2=−4V = 1 \cdot \begin{vmatrix} i-1 & i^2-1 \\ i^2-1 & 0 \end{vmatrix} = -(i^2-1)^2 = -(-1-1)^2 = -4
Δ=i9⋅(−4)=i⋅(−4)=−4i\Delta = i^9 \cdot (-4) = i \cdot (-4) = -4i
 The result is purely imaginary. Answer: (4)
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