Question

# Using the properties of determinants, show that:$$\begin{vmatrix} x & { x }^{ 2 } & yz \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)$$

Solution

## Let $$\Delta = \begin{vmatrix} x & { x }^{ 2 } & yz \\ y & { y }^{ 2 }&zx \\ z & { z }^{ 2 } & xy \end{vmatrix}$$Applying $$R_1=R_1-R_2, \text{&} R_2=R_2-R_3$$$$\Delta =\begin{vmatrix}x-y&x^2-y^2&z(y-x)\\y-z&y^2-z^2&x(z-y)\\z&z^2&xy \end{vmatrix}$$Taking $$(x-y) \text{&} (y-z)$$ common from first and second row respectively we get,$$\Delta=(x-y)(y-z)\begin{vmatrix}1&x+y&-z\\1&y+z&-x\\z&z^2&xy \end{vmatrix}$$Now applying $$R_1=R_1-R_2$$ we get,$$\Delta=(x-y)(y-z)\begin{vmatrix}0&x-z&x-z\\1&y+z&-x\\z&z^2&xy \end{vmatrix}$$Taking $$(z-x)$$ common from first row, we get,$$\Delta=(x-y)(y-z)(z-x)\begin{vmatrix}0&-1&-1\\1&y+z&-x\\z&z^2&xy \end{vmatrix}$$Expanding along $$R_1$$ we get,$$\Delta =(x-y)(y-z)(z-x)[xy+xz-1(z^2-yz-z^2)]$$$$\therefore \Delta =(x-y)(y-z)(z-x)(xy+yz+zx)$$Hence proved.MathematicsNCERTStandard XII

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