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Using Propertiesof Determinants Prove That: - Mathematics

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Question

Using propertiesof determinants prove that:
`|(x , x(x^2), x+1), (y, y(y^2 + 1), y+1),( z, z(z^2 + 1) , z+1) | = (x-y) (y - z)(z - x)(x + y+ z)`

Sum

Solution

L.H.S
`R_1=> R_1- R_2. R_2 => R_2 - R_3`

`=|(X - Y ,  X^3 +X-Y-Y^3,  X+1-Y-1), (Y-Z ,   Y^3 +Y-Z-Z^3,   Y+ 1 - Z -1),(Z,  Z(Z^2+1), (Z+1) )|`

`as X^3 - Y^3 +X - Y = (X-Y)(X^2 + XY + Y^2)+X-Y = (X - Y)(X^2 + XY + Y^2 +1) `

`=|((X-Y),(X - Y)(X^2+Y^2+XY+1) , (X-Y)), ((Y-Z), (Y-Z)(Y^2+Z^2+YZ + 1),(Y-Z)), (Z , Z(Z^2 +1) , Z+1)|`


`= (X - Y)(Y-Z)     |(1 , X^2+ Y^2+XY+1 , 1 ), (1 , Y^2+ Z^2+YZ+1 ,1),(Z , Z(Z^2+1), Z+1)|`

`R_1 =>R_1- R_2`

= `(X-Y)(Y-Z) |(0, (X-Z)(X+Y+Z), 0), (1 ,Y^2+Z^2+YZ+1 ,1),(Z ,  Z(Z^2+1) , Z+1)|`

`{(as, (X^2+Y^2+XY+1) - (Y^2 + Z^2 +YZ+1)),
(=,X^2+Y^2+XY+1- Y^2-Z^2-YZ-1),
(=,(X-Z)(X+Z)+Y(X-Z)= (X-Z)(X+Y+Z)):}}`

=`(X-Y)(Y-Z)(X-Z)| (0, X+Y+Z, 0),(1 , Y^2+Z^2+YZ+1, 1), (Z , Z(Z^2 + 1), Z+1) |`

`= (X-Y)(Y-Z)(X-Z)[-(X+Y+Z)(Z+1-Z)]`
`= (X-Y)(Y-Z)(Z-X)(X+Y+Z)`

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2018-2019 (March) Set 1

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