Q. 74.0( 2 Votes )

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Answer :

We are given that,


Let us find the determinants ∆1 and ∆2.


We know that,


Determinant of 3 × 3 matrix is given as,




So,




1 = (b × c2 – c × b2) – (a × c2 – c × a2) + (a × b2 – b × a2)


1 = bc2 – b2c – ac2 + a2c + ab2 – a2b …(i)


Also,




2 = (ca × c – b × ab) – bc(1 × c – b × 1) + a(1 × ab – ca × 1)


2 = ac2 – ab2 – bc(c – b) + a(ab – ac)


2 = ac2 – ab2 – bc2 + b2c + a2b – a2c …(ii)


Checking Option (A).


Adding ∆1 and ∆2 by using values from (i) and (ii),


1 + ∆2 = (bc2 – b2c – ac2 + a2c + ab2 – a2b) + (ac2 – ab2 – bc2 + b2c + a2b – a2c)


1 + ∆2 = bc2 – bc2 – b2c + b2c – ac2 + ac2 + ab2 – ab2 – a2b + a2b


1 + ∆2 = 0


Thus, option (A) is correct.


Checking Option (B).


Multiplying 2 by (ii),


2∆2 = 2(ac2 – ab2 – bc2 + b2c + a2b – a2c)


2∆2 = 2ac2 – 2ab2 – 2bc2 + 2b2c + 2a2b – 2a2c …(iii)


Then, adding 2∆2 with ∆1,


1 + 2∆2 = (bc2 – b2c – ac2 + a2c + ab2 – a2b) + (2ac2 – 2ab2 – 2bc2 + 2b2c + 2a2b – 2a2c)


1 + 2∆2 = bc2 – 2bc2 – b2c + 2b2c – ac2 + 2ac2 + ab2 – 2ab2 – a2b + 2a2b


1 + 2∆2 = -bc2 + b2c + ac2 – ab2 + a2b


1 + 2∆2 ≠ 0


Thus, option (B) is not correct.


Checking option (C).


Obviously, ∆1 ≠ ∆2


Since, by (i) and (ii), we can notice ∆1 and ∆2 have different values.


Thus, option (C) is not correct.

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