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Question
Using properties of determinants, prove that
`|[b+c , a ,a ] ,[ b , a+c, b ] ,[c , c, a+b ]|` = 4abc
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 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
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.
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