Maths-
General
Easy

Question

Consider the matrix A, B, C, D with order 2 × 3, 3×4, 4×4, 4×2 respectively. Let x = (alpha A B gamma C2 D)3 where alphagamma are scalars . Let |x| = k|ABC2D|3, then k is

  1. alphagamma    
  2. alpha squared gamma squared
  3. alpha to the power of 4 gamma to the power of 4    
  4. alpha to the power of 6 gamma to the power of 6    

The correct answer is: alpha to the power of 6 gamma to the power of 6


    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell x equals open parentheses alpha gamma A B C squared D close parentheses cubed end cell row cell vertical line x vertical line equals open vertical bar alpha gamma A B C squared D close vertical bar cubed end cell row cell text  order of  end text A B C squared D text  is  end text 2 cross times 2 end cell row cell vertical line x vertical line equals open parentheses alpha squared gamma squared close parentheses cubed open vertical bar A B C squared D close vertical bar cubed end cell row cell therefore k equals alpha to the power of 6 gamma to the power of 6 end cell end table

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