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Question
Using properties of determinants, prove that:
`|(a,b,b+c),(c,a,c+a),(b,c,a+b)|` = (a+b+c)(a-c)2
Solution
Let Δ = `|(a,b,b+c),(c,a,c+a),(b,c,a+b)|`
Operating C3 → C3 - C2 , we obtain
Δ =`|(a,b,c),(c,a,c),(b,c,a+b-c)|`
Operating R1 → R1 + R2 + R3 , we obtain
Δ =`|(a+b+c,a+b+c,a+b+c),(c,a,c),(b,c,a+b-c)|`
Taking (a + b + c) common from R1 , we obtain
= (a+b+c) `|(1,1,1),(c,a,c),(b,c,a+b-c)|`
Operating C1 → C1 - C3 , C2 → C2 - C3 , we obtain
= (a+b+c) `|(0,0,0),(0,a-c,c),(c-a,2c-a-b,a+b-c)|`
Expanding along R1 , we obtain
= (a+b+c){-(a-c)(a-c)}
= (a+b+c)(a-c)2
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