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Using Properties of Determinants, Prove that ∣ ∣ ∣ ∣ B + C a A B C + a B C C a + B ∣ ∣ ∣ ∣ = 4abc - Mathematics

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Sum

Using properties of determinants, prove that

`|[b+c , a ,a  ] ,[ b , a+c, b ] ,[c , c, a+b ]|` = 4abc 

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Solution 1

`|[b+c , a ,a  ] ,[ b , a+c , b ] ,[c , c, a+b ]|` = 4abc 

Applying R1 ⇒ R1 + R2 + R3

`Delta = |[2(b+c) , 2(a +c) ,2(a + b)],[b , a +c ,b ],[ c ,c , a+b]|`

`Delta =2 |[b+c , a +c , a + b],[b , a +c ,b ],[ c ,c , a+b]|` 

Applying R⇒ R2 - R1 and  R3 ⇒ R3 - R

`Delta =2 |[b+c , c +a , a + b],[-c, 0  ,-a ],[ -b , -a , 0]|` 

Applying R1 ⇒ R1 + R2 + R3

⇒  `Delta =2 |[0 , c  ,  b],[-c, 0  ,-a ],[ -b , -a , 0]|` 

Δ =  [ -c ( 0 - ab ) + b ( ac - 0)]

Δ = 2 ( abc + abc) 

Δ = 4 abc

Solution 2

Let Δ = `|(b+c , a , a), (b , c+a, b), (c, c, a+b)|`

Applying R1 → R1 - R2 - R3 to Δ, we get

Δ = `|(0 ,-2c ,-2b), (b , c+a, b), (c, c, a+b)|`

Expending along R1 we obtain

Δ=`0|(c+a, b),(c, a+b)| -(-2c)|(b, b),(c, a+b)| +(-2b)|(b, c+a),(c, c)|`

= 2c( ab + b2 - bc ) - 2b( bc - c2 - ac )

= 2abc + 2cb2 - 2bc2 - 2b2c + 2bc2 + 2abc

= 4abc
Hence proved.

Concept: Properties of Determinants
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