Question

Using properties of determinants, prove that

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

Answer 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 R₂ ⇒ R₂ - R₁ and  R₃ ⇒ R₃ - R₁ 

`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

Answer 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.