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Question

Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant

-2aa+ba+cb+a-2bb+cc+ac+b-2c

The other factor in the value of the determinant is
(a) 4
(b) 2
(c) a + b + c
(d) none of these

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Solution

(a) 4

Δ=-2a a + b a + cb + a -2b b + cc + a c + b -2c Let a + b = 2C, b + c = 2A and c + a = 2B.a + b + b + c + c + a = 2A + 2B + 2C2a + b + c = 2A + B + Ca + b + c = A + B + CAlso,a = a + b + c - b + c = A + B + C - 2A = B + C - ASimilarly,b = C + A - B, c = A + B - CΔ = 2A - 2B - 2C 2C 2B 2C 2B-2C-2A 2A 2B 2A 2C-2A-2B = 8 × A - B - C C B C B - C - A A B A C - A - B taking out 2 common from R1 R2 R3=8 × A - B C + B B B - A B - C A B + A C - B C-A-B Applying C1C1+C2 , C2C2 + C3=8 × A - B C + B B 0 2B A + B 2B 0 C - B Applying R2 R1+ R2, R3 R2 +R3=8 × A - B 2B A + B 0 C - B + 2B × C + B B 2B A + B Expanding along C1=16 BA - BC - B + C + BA + B - 2B2= 32 ABC=32b + c2c + a2a + b2=4a + bb + cc + aHence, 4 is the other factor of the determinant.

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