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in Determinants by (27.3k points)
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Using the factor theorem it is found that a + b, b + c and c + a are three factors of the determinant. 
\(\begin{vmatrix} -2a &a+b & a+c \\[0.3em] b+a & -2b &b+c \\[0.3em] c+a &c+b &-2c \end{vmatrix}\)

The other factor in the value of the determinant is

A. 4 

B. 2 

C. a + b + c 

D. none of these

1 Answer

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by (26.9k points)
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Best answer

A. 4

\(\begin{vmatrix} -2a &a+b & a+c \\[0.3em] b+a & -2b &b+c \\[0.3em] c+a &c+b &-2c \end{vmatrix}\) 

= k (a + b)(b + c) (c + a)

Let assume a = 0, b = 1, c = 2

Now expending around colume, 

0-1(-4- 6) + 2(3 + 4) = k(1)(3)(2) 

6k = 24 

K = 4

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