# The value of determinant |(a-b-c,2a,2a),(2b,b-c-a,2b),(2c,2c,c-a-b)| is?

Aug 1, 2018

$D = | \left(a - b - c , 2 a , 2 a\right) , \left(2 b , b - c - a , 2 b\right) , \left(2 c , 2 c , c - a - b\right) | = {\left(a + b + c\right)}^{3}$

#### Explanation:

Here .

$D = | \left(a - b - c , 2 a , 2 a\right) , \left(2 b , b - c - a , 2 b\right) , \left(2 c , 2 c , c - a - b\right) |$

${R}_{1} + {R}_{2} + {R}_{3}$
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color(red)(a-b-c+2b+2c),color(blue)(2a+b-c-a+2c),color(green)(2a+2b+c-a-b
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$\therefore D = | \left(a + b + c , a + b + c , a + b + c\right) , \left(2 b , b - c - a , 2 b\right) , \left(2 c , 2 c , c - a - b\right) |$

${R}_{1} \left(\frac{1}{a + b + c}\right)$

$\therefore D = \left(a + b + c\right) | \left(1 , 1 , 1\right) , \left(2 b , b - c - a , 2 b\right) , \left(2 c , 2 c , c - a - b\right) |$

${C}_{2} - {C}_{1} \mathmr{and} {C}_{3} - {C}_{1}$

$\therefore D = \left(a + b + c\right) | \left(1 , 1 - 1 , 1 - 1\right) , \left(2 b , b - c - a - 2 b , 2 b - 2 b\right) , \left(2 c , 2 c - 2 c , c - a - b - 2 c\right) |$

$\therefore D = \left(a + b + c\right) | \left(1 , 0 , 0\right) , \left(2 b , - b - c - a , 0\right) , \left(2 c , 0 , - c - a - b\right) |$

Expansion of determinant:

$\therefore D = \left(a + b + c\right) \left\{1 \left(- b - c - a\right) \left(- c - a - b\right) - 0 - 0\right\}$

$\therefore D = \left(a + b + c\right) \left(b + c + a\right) \left(c + a + b\right)$

$\therefore D = {\left(a + b + c\right)}^{3}$