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Using Properties of Determinants Show that ⎡ ⎢ ⎣ 1 1 1 + X 1 1 + Y 1 1 + Z 1 1 ⎤ ⎥ ⎦ = X Y Z + Y Z + Z X + X Y . - Mathematics

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Question

Using properties of determinants show that

`[[1,1,1+x],[1,1+y,1],[1+z,1,1]] = xyz+ yz +zx+xy.`

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Solution

To prove:  `[[1,1,1+x],[1,1+y,1],[1+z,1,1]] = xyz+ yz +zx+xy.`

LHS: Let   `Δ = [[1,1,1+x],[1,1+y,1],[1+z,1,1]] = xyz+ yz +zx+xy.`

Take x, y and z common from C3, Cand C1 respectively.

Therefore, Δ = xyz `[[1/z,1/y,1/x+1],[1/z,1/y+1,1/x],[1/z+1,1/y,1/x]]`

`C_3 → C_3+C+C_1`

`Δ = xyz [[1/z,1/y,1+1/x+1/y+1/z],[1/z,1/y+1,1+1/x+1/y+1/x],[1/x+1,1/y,1+1/x+1/y+1/z]]`

Taking `1+1/x+1/y+1/z` common

`Δ = xyz (1+1/x+1/y+1/z) [[1/z,1/y,1],[1/z,1/y+1,1],[1/z+1,1/y,1]]`

Applying `R_2 → R_2-R_1,R_3 →R_3-R_1`

`Δ = xyz (1+1/x+1/y+1/z)[[1/z,1/y,1],[0,1,0],[1,0,0]]`

On expanding we get` Δ = xyz (1+1/x+1/y+1/z) = xyz+ yz +zx+xy` 

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2016-2017 (March) Foreign Set 3

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