Correct Answer - Option 3 : ± 4
Concept:
If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}} \end{array}} \right]\) then determinant of A is given by:
|A| = a11 × a22 – a21 × a12
|An| = |A|n
Calculation:
Given that,
\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\rm{x}}&2\\ 4&{\rm{x }} \end{array}} \right]\) and |A2| = 64
⇒ |A| = x2 - 8 .... (1)
Given |A2| = 64
⇒ |A|2 = 64 [∵ |An| = |A|n]
⇒ |A| = (64)1/2 = 8 ....(2)
From equation 1 and 2
⇒ x2 - 8 = 8
⇒ x2 = 16
⇒
x = ± 4