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If Δ1\(\begin{vmatrix} 1 &1 & 1 \\[0.3em] a & b &c \\[0.3em] a^2 & b^2 & c^2 \end{vmatrix}\)Δ2\(\begin{vmatrix} 1 &bc & a \\[0.3em] 1 & ca &b \\[0.3em] 1 & ab & c \end{vmatrix}\), then

A. Δ1 + Δ2 = 0 

B. Δ1 + 2Δ2 = 0 

C. Δ1 = Δ2 

D. none of these

1 Answer

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Best answer

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 + a2

⇒ ∆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 + 2a2

⇒ ∆1 + 2∆2 = -bc2 + b2c + ac2 – ab2 + a2

⇒ ∆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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