Maths-
General
Easy

Question

Consider open vertical bar table row cell a subscript 1 end subscript end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript end cell end table close vertical bar= – 1, where ai. aj + bi. bj + ci.cj = open square brackets table row cell 0 semicolon end cell cell i not equal to j end cell row cell 1 semicolon end cell cell i equals j end cell end table close and i, j = 1,2,3
Assertion(A) : The value of open vertical bar table row cell a subscript 1 end subscript plus 1 end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript plus 1 end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript plus 1 end cell end table close vertical bar is equal to zero
Reason(R) : If A be square matrix of odd order such that AAT = I, then | A + I | = 0

  1. If both (A) and (R) are true, and (R) is the correct explanation of (A).    
  2. If both (A) and (R) are true but (R) is not the correct explanation of (A).    
  3. If (A) is true but (R) is false.    
  4. If (A) is false but (R) is true.    

The correct answer is: If (A) is true but (R) is false.


    AAT = I
    |A| |AT| = 1 not stretchy rightwards double arrow|A|2 = 1 not stretchy rightwards double arrow |A| = ± 1
    |A + I| = |A + AAT| = |A|3 |I + AT|
    = A3 | I + A |
    when |A| = – 1 not stretchy rightwards double arrow |A + I| = – |A + I|
    not stretchy rightwards double arrow 2|A + I| = 0 not stretchy rightwards double arrow |A + I| = 0
    when |A| = 1 not stretchy rightwards double arrow |A + I| = |A + I|
    therefore Reason is false
    Let A = open square brackets table row cell a subscript 1 end subscript end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript end cell end table close square brackets
    A + I =open square brackets table row cell a subscript 1 end subscript plus 1 end cell cell b subscript 1 end subscript end cell cell c subscript 1 end subscript end cell row cell a subscript 2 end subscript end cell cell b subscript 2 end subscript plus 1 end cell cell c subscript 2 end subscript end cell row cell a subscript 3 end subscript end cell cell b subscript 3 end subscript end cell cell c subscript 3 end subscript plus 1 end cell end table close square brackets
    AAT =open square brackets table row cell a subscript 1 end subscript end cell cell a subscript 2 end subscript end cell cell a subscript 3 end subscript end cell row cell b subscript 1 end subscript end cell cell b subscript 2 end subscript end cell cell b subscript 3 end subscript end cell row cell c subscript 1 end subscript end cell cell c subscript 2 end subscript end cell cell c subscript 3 end subscript end cell end table close square brackets equals open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets= 1 |AAT| = 1 not stretchy rightwards double arrow |A| = ± 1 but |A| = – 1
    therefore|A + I| = |A + AAT| = |A|3 |I + AT| = – |A + I|
    therefore |A + I| = 0
    So assertion is true.

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