What is the value of the determinant \(\left| {\begin{array}{*{20}{c}} {\rm{i}}&{{{\rm{i}}^2}}&{{{\rm{i}}^3}}\\ {{{\rm{i}}^4}}&{{{\rm{i}}^6}}&{{{\rm{i}}^8}}\\ {{{\rm{i}}^9}}&{{{\rm{i}}^{12}}}&{{{\rm{i}}^{15}}} \end{array}} \right|\) where \(\rm i = \sqrt {-1}\) ?

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NDA (Held On: 6 Sep 2020) Maths Previous Year paper
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  1. 0
  2. -2
  3. 4i
  4. -4i

Answer (Detailed Solution Below)

Option 4 : -4i
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Detailed Solution

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Concept:

\(\rm i = \sqrt {-1}\)

i= -1 , i= - i, i4 = 1, i6 = - 1 , i= 1 , i= i, i 12 = 1, and i15 = - i

 

Calculations: 

Given determinant is \(\left| {\begin{array}{*{20}{c}} {\rm{i}}&{{{\rm{i}}^2}}&{{{\rm{i}}^3}}\\ {{{\rm{i}}^4}}&{{{\rm{i}}^6}}&{{{\rm{i}}^8}}\\ {{{\rm{i}}^9}}&{{{\rm{i}}^{12}}}&{{{\rm{i}}^{15}}} \end{array}} \right|\)

Since, we have, 

\(\rm i = \sqrt {-1}\)

i= -1 , i= - i, i4 = 1, i6 = - 1 , i= 1 , i= i, i 12 = 1, and i15 = - i

=\(\left| {\begin{array}{*{20}{c}} {\rm{i}}&{{{\rm{-1}}}}&{{{\rm{-i}}}}\\ {{{\rm{1}}}}&{{{\rm{-1}}}}&{{{\rm{1}}}}\\ {{{\rm{i}}}}&{{{\rm{1}}}}&{{{\rm{-i}}}} \end{array}} \right|\)

=i(i - 1) + 1(-i - i) - i (1 + i)

= i- i - 2i - i - i2

= - 4i

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