Matrices & DeterminantshardFree

Matrices & Determinants: Let Vmatrix Vmatrix Value

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

Question

Let

\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}

where

a_{pq} = (i^p)^q

and

i = \sqrt{-1}

. The value of

\Delta

is:

AReal and positive

BReal and negative

0

DImaginarycorrect

Solution

Step 1: Write out the matrix

a_{pq} = i^{pq}

, so the matrix is:

\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

i

from

C_1

i^2

from

C_2

i^3

from

C_3

= i^6 \begin{vmatrix} 1 & 1 & 1 \\ i & i^2 & i^3 \\ i^2 & i^4 & i^6 \end{vmatrix}

Factor

i

from

R_2

and

i^2

from

R_3

= 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

R_2 \to R_2-R_1

and

R_3 \to R_3-R_1

V = \begin{vmatrix} 1 & 1 & 1 \\ 0 & i-1 & i^2-1 \\ 0 & i^2-1 & i^4-1 \end{vmatrix}

Since

i^4 = 1

, the entry

i^4-1 = 0

. Expanding along column 1:

V = 1 \cdot \begin{vmatrix} i-1 & i^2-1 \\ i^2-1 & 0 \end{vmatrix} = -(i^2-1)^2 = -(-1-1)^2 = -4

\Delta = i^9 \cdot (-4) = i \cdot (-4) = -4i

The result is purely imaginary. Answer: (4)

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