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Properties of Determinants

Prove that: ...

Question

Prove that: $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 x & y & z y z & z x & x y \end{matrix} ∣ ∣ ∣ ∣ = (x - y) (y - z) (z - x)$ .

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Solution

$∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 x & y & z y z & z x & x y \end{matrix} ∣ ∣ ∣ ∣ C_{1} \to C_{1} - C_{2} ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 1 x - y & y & z y z - z x & z x & x y \end{matrix} ∣ ∣ ∣ ∣ (y - x) ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 1 - 1 & y & z - z & z x & x y \end{matrix} ∣ ∣ ∣ ∣ C_{2} \to C_{2} - C_{1} (y - x) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 1 - 1 & y - z & z + z & x (z - y) & x y \end{matrix} ∣ ∣ ∣ ∣ ∣ (y - x) (z - y) ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 - 1 & 1 & z + z & x & x y \end{matrix} ∣ ∣ ∣ ∣ (y - x) (z - y) ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 - 1 & 1 & z z & x & x y \end{matrix} ∣ ∣ ∣ ∣ (y - x) (z - y) [1 - (- x - z)] = (x - y) (y - z) (z - x)$

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∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 x & y & z x^{2} & y^{2} & z^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = [(x - y) (y - z) (z - x)]

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & x & x^{3} 1 & y & y^{3} 1 & z & z^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = (x + y + z) (x - y) (y - z) (z - x)

Q. Without expanding, prove that

∣ ∣ ∣ ∣ \begin{matrix} x + y & y + z & z + x z & x & y 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 x^{2} & y^{2} & z^{2} x^{3} & y^{3} & z^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣

(x - y) (y - z) (z - x) (x y + y z + z x)

(x + y) (y + z) (z + x) \geq 8 x y z

x, y, z > 0

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Properties of Determinants

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