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Text Solution

Solution :

`|(x,y,z),(x^2,y^2,z^2),(x^3 , y^3,z^3)|` <br>
`= xyz|(1,1,1),(x,y,z),(x^2,y^2,z^2)|` <br>
`c_2 -> c_2 - c_1` <br>
`c_3 -> c_3 - c_1` <br>
`=> |(1,0,0),(x, y-x, z-x),(x^2, y^2-x^2, z^2-x^2)|` <br>
`= 1((y-x)(z^2-x^2) - (z-x)(y^2 - x^2))` <br>
`= xyz(y-x)(z-x)(z+x-y-x)` <br>
`= xyz(y-x)(z-x)(z-y)`= RHS <br>
hence proved
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**Definition and example**

**Definition**

**Let A be a square matrix of order n; then the sum of the product of elements of any row (column) with their cofactors is always equal to `|A|`**

**Let A be a square matrix of order n; then the sum of the product of elements of any row (column) with the cofactors of the corresponding elements of some other row (column) is 0.**

**Let A be a square matrix of order n then `|A| = |A^T|`**

**Let `A=[a_(ij)]` be a square matrix of order `n>=2` and let B be a matrix obtained from A by interchanging any two rows (columns) of A then `|B|=-|A|`**