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Question
Using properties of determinants, prove that:
`|[a^2 + 1, ab, ac], [ba, b^2 + 1, bc ], [ca, cb, c^2+1]| = a^2 + b^2 + c^2 + 1`
Solution
L.H.S. Δ = `|[a^2 + 1, ab, ac], [ba, b^2 + 1, bc ], [ca, cb, c^2+1]| `
Operating `R_1 → (1)/(a) R_1,R_2 → (1)/(b) R_2 and R_3 → 1/c R_3 "we have"`
Δ = abc `|[a+ (1)/(a), b, c], [a , b +(1)/(b), c], [a , b , c + (1)/(c)]|`
Multiplying C1 by a, C2 by b and C3 by c, we have
Δ = `|[ a^2+1, -b^2 , c^2], [ a^2, b^2+1, c^2], [a^2, b^2, c^2+1]|`
Operating C1 → C1 + C2+C3, we have
Δ = `|[ 1+a^2 +b^2+c^2, b^2, c^2],[1+a^2 +b^2+c^2 , b^2+1, c^2],[1+a^2 +b^2+c^2, b^2, c^2+1]|`
Δ = `(1 + a^2 + b^2+c^2) |[ 1, b^2, c^2],[1, b^2+1, c^2], [1, b^2, c^2+1]|`
Operating R2 → R2 → R1 and R3 → R3 → R1, we have
Δ = `(1 + a^2 + b^2+c^2) |(1,b^2,c^2),(0,1,0),(0,0,1)|`
Expanding along C1 , we have
Δ = `(1 + a^2 + b^2+c^2) |[1, 0],[0,1]|`
= `a^2 +b^2+c^2 +1 = R.H.S.`
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