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    Maths-
    General
    Easy

    Question

    Inverse of the matrix open square brackets table row 3 cell – 2 end cell cell – 1 end cell row cell – 4 end cell 1 cell – 1 end cell row 2 0 1 end table close square brackets is

    1. open square brackets table row 1 2 3 row 3 3 7 row cell – 2 end cell cell – 4 end cell cell – 5 end cell end table close square brackets    
    2. open square brackets table row 1 2 3 row 2 5 7 row cell – 2 end cell cell – 4 end cell cell – 5 end cell end table close square brackets    
    3. open square brackets table row 1 cell – 3 end cell 5 row 7 4 6 row 4 2 7 end table close square brackets    
    4. open square brackets table row 1 2 cell – 4 end cell row 8 cell – 4 end cell cell – 5 end cell row 3 5 2 end table close square brackets    

    The correct answer is: open square brackets table row 1 2 3 row 2 5 7 row cell – 2 end cell cell – 4 end cell cell – 5 end cell end table close square brackets

      |A| = open vertical bar table row 3 cell – 2 end cell cell – 1 end cell row cell – 4 end cell 1 cell – 1 end cell row 2 0 1 end table close vertical bar
      = 3(1) + 2 (– 4 + 2) – 1 (0 – 2)
      = 3 – 4 + 2 = 1
      adj A = open square brackets table row 1 2 cell – 2 end cell row 2 5 cell – 4 end cell row 3 7 cell – 5 end cell end table close square brackets to the power of T end exponent = open square brackets table row 1 2 3 row 2 5 7 row cell – 2 end cell cell – 4 end cell cell – 5 end cell end table close square brackets
      A–1 = fraction numerator A d j A over denominator vertical line A vertical line end fraction = open square brackets table row 1 2 3 row 2 5 7 row cell – 2 end cell cell – 4 end cell cell – 5 end cell end table close square brackets

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