By using properties of determinants,show that:I\(\begin{vmatrix} x & x^2 & yz\\ y & y^2 & zx\\z&z^2&xy \end{vmatrix}\)=(x-y)(y-z)(z-x)(xy+yz+zx)
Solution and Explanation
Let △=\(\begin{vmatrix} x & x^2 & yz\\ y & y^2 & zx\\z&z^2&xy \end{vmatrix}\)
Applying R2 → R2 − R1 and R3 → R3 − R1, we have:
△=\(\begin{vmatrix} x & x^2 & yz\\ y-x & y^2-x^3 & zx-yz\\z-x&z^2-x^2&xy-yz \end{vmatrix}\)
=\(\begin{vmatrix} x & x^2 & yz\\- (x-y) & -(x-y)(x+y)& zx-yz\\z-x&z^2-x^2&-y(z-x) \end{vmatrix}\)
=(x-y)(z-x)\(\begin{vmatrix} x & x^2 & yz\\ -1 &-x-y & z\\1&z+x&-y \end{vmatrix}\)
Applying R3 → R3 + R2, we have:
△=(x-y)(z-x)\(\begin{vmatrix} x & x^2 & yz\\ -1 &-x-y & z\\0&z-y&z-y \end{vmatrix}\)
=(x-y)(z-x)(z-y)\(\begin{vmatrix} x & x^2 & yz\\ -1 &-x-y & z\\0&1&1 \end{vmatrix}\)
Expanding along R3, we have:
△=[(x-y)(z-x)(z-y)]\(\bigg[(-1)\)\(\begin{vmatrix} x &yz \\ -1 & z \end{vmatrix}\)+1\(\begin{vmatrix} x &x^2 \\ -1 & -x-y \end{vmatrix}\) \(\bigg]\)
=[(x-y)(z-x)(z-y)][(-xz-yz)+(-x2-xy+x2)]
=(x-y)(y-z)(z-x)(xy+yz+zx)
Hence, the given result is proved.
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Concepts Used:
Properties of Determinants
Properties of Determinants:
- Reflection Property: The value of the determinant remains unchanged if its rows and columns are interchanged.
- Switching Property: The interchanging of any two rows (or columns) of the Determinant changes its signs.
- All- Zero Property: The Determinants will be equivalent to zero if each term of rows and columns is zero.
- Proportionality (Repetition) Property: If all elements of a row (or column) are proportional or identical to the elements of some other row (or column), then the determinant is zero.
- Scalar Multiple Property: If all the elements of a row (or column) of a determinant are multiplied by a non-zero constant, then the determinant gets multiplied by the same constant.
- Sum Property: If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants.
- Property of Invariance: If to each element of any row or column of a determinant, the equimultiples of corresponding elements of other row (or column) are added, then value of determinant remains the same, i.e., the value of determinant remain same if we apply the operation Ri → Ri + kRj or Ci → Ci + kCj
- Factor Property: If a Determinant Δ becomes 0 while considering the value of x = α, then (x - α) is considered as a factor of Δ
- Triangle Property: If all the elements of a determinant above or below the main diagonal consist of zeros, then the determinant is equal to the product of diagonal elements.
Read More: Properties of Determinants