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Implicit Differentiation

Prove that:zx...

Question

Prove that :

$|\begin{matrix} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{matrix}| = |\begin{matrix} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \end{matrix}| = |\begin{matrix} x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z \end{matrix}| = x y z (x - y) (y - z) (z - x) (x + y + z) .$

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Solution

$Let Δ_{1} = |z x y z^{2} x^{2} y^{2} z^{4} x^{4} y^{4}|, Δ_{2} = |x y z x^{2} y^{2} z^{2} x^{4} y^{4} z^{4}|, Δ_{3} = |x^{2} y^{2} z^{2} x^{4} y^{4} z^{4} x y z| and Δ_{4} = x y z (x - y) (y - z) (z - x) (x + y + z) Now, Δ_{1 =} |z x y z^{2} x^{2} y^{2} z^{4} x^{4} y^{4}| Using the property that if two rows (or columns) of a determinant are interchanged, the value of the determinant becomes negetive, we get \Rightarrow Δ_{1} = (- 1) |x z y x^{2} z^{2} y^{2} x^{4} z^{4} y^{4}| [∵ C_{1} \leftrightarrow C_{2}] = (- 1) (- 1) |x y z x^{2} y^{2} z^{2} x^{4} y^{4} z^{4}| [∵ C_{2} \leftrightarrow C_{3}] = |x y z x^{2} y^{2} z^{2} x^{4} y^{4} z^{4}| = Δ_{2} . . . (1) = (- 1) |x^{2} y^{2} z^{2} x y z x^{4} y^{4} z^{4}| [Applying R_{1} \leftrightarrow R_{2}] = (- 1) (- 1) |x^{2} y^{2} z^{2} x^{4} y^{4} z^{4} x y^{} z| [Applying R_{2} \leftrightarrow R_{3}] = |x^{2} y^{2} z^{2} x^{4} y^{4} z^{4} x y^{} z| = Δ_{3} . . . (2) Thus, Δ_{1} = Δ_{2} = Δ_{3} [From eqs . (1) and (2)]$

$∆_{2} = |x y z x^{2} y^{2} z^{2} x^{4} y^{4} z^{4}| = x y z |1 1 1 x y^{} z x^{3} y^{3} z^{3}| [Taking out common facto r x from C_{1,} y from C_{2} and z from C_{3}] = xyz |0 0 1 x - y y - z^{} z x^{3} - y^{3} y^{3} - z^{3} z^{3}| [Applying C \to C_{1} - C_{2} and C_{2} \to C_{2} - C_{3}] = xyz (x - y) (y - z) |0 0 1 1 1 z x^{2} + 2 x y + y^{2} y^{2} + 2 y z + z^{2} z^{3}| [∵ (a^{3} - b^{3}) = (a - b) (a^{2} + ab + b^{2})] [Taking out common factor (x - y) from C_{1} and (y - z) from C_{2}] = xyz (x - y) (y - z) \{1 \times |1 1 x^{2} + x y + y^{2} y^{2} + y z + z^{2}|\} [Expanding along R_{1}] = xyz (x - y) (y - z) \{y^{2} + y z + z^{2} - x^{2} - x y - y^{2}\} = xyz (x - y) (y - z) \{y z - x y + z^{2} - x^{2}\} = xyz (x - y) (y - z) \{y (z - x) + (z - x) (z + x)\} = xyz (x - y) (y - z) (z - x) (y + x + z) = xyz (x - y) (y - z) (z - x) (x + y + z) = ∆_{4} Thus, ∆_{1} = ∆_{2} = ∆_{3} = ∆_{4}$

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