We are given that,
Δ1 = \(\begin{vmatrix}
1 &1 & 1 \\[0.3em]
a & b &c \\[0.3em]
a^2 & b^2 & c^2
\end{vmatrix}\) and Δ2 = \(\begin{vmatrix}
1 &bc & a \\[0.3em]
1 & ca &b \\[0.3em]
1 & ab & c
\end{vmatrix}\)
Let us find the determinants Δ1 and Δ2.
We know that,
Determinant of 3 × 3 matrix is given as,
⇒ ∆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.