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Let X= open square brackets table row x row y row z end table close square brackets, D = open square brackets table row 3 row 5 row 11 end table close square bracketsand A = open square brackets table row 1 cell negative 1 end cell cell negative 2 end cell row 2 1 1 row 4 cell negative 1 end cell cell negative 2 end cell end table close square brackets. If X = A–1 D, then X is equal to

  1. open square brackets table row 1 row 0 row 2 end table close square brackets    
  2. open square brackets table row cell 8 divided by 3 end cell row cell negative 1 divided by 3 end cell row 0 end table close square brackets    
  3. open square brackets table row cell negative 8 divided by 3 end cell row 1 row 0 end table close square brackets    
  4. open square brackets table row cell 8 divided by 3 end cell row cell 1 divided by 3 end cell row cell negative 1 end cell end table close square brackets    

The correct answer is: open square brackets table row cell 8 divided by 3 end cell row cell negative 1 divided by 3 end cell row 0 end table close square brackets


    Since A = open square brackets table row 1 cell negative 1 end cell cell negative 2 end cell row 2 1 1 row 4 cell negative 1 end cell cell negative 2 end cell end table close square brackets
    thereforeA–1 =1 thirdopen square brackets table row cell negative 1 end cell 0 1 row 8 6 cell negative 5 end cell row cell negative 6 end cell cell negative 3 end cell 3 end table close square brackets
    A–1 D = 1 thirdopen square brackets table row 8 row cell negative 1 end cell row 0 end table close square brackets
    rightwards double arrow open square brackets table row x row y row z end table close square brackets = open square brackets table row cell 8 divided by 3 end cell row cell negative 1 divided by 3 end cell row 0 end table close square brackets

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