Q. 74.0( 2 Votes )

# Mark the correct

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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