Using properties of determinants- prove that following- x2 - 1 xy xz xy y2 - 1 yz xz yz z2 - 1 - 1 - x2 - y2 - z2-

LHS = x^2+1 amp; xy amp;xz xy #160; amp;y^2+1 #160; amp;yz xz amp;yz #160; amp;z^2+1 Taking out common factors x, y ...

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Properties of Determinants

Using propert...

Question

Using properties of determinants, prove that following:
$∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} + 1 & x y & x z x y & y^{2} + 1 & y z x z & y z & z^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 1 + x^{2} + y^{2} + z^{2} .$

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∣ ∣ ∣ ∣ ∣ \begin{matrix} (x + y)^{2} & z x & z y z x & (z + y)^{2} & x y z y & x y & (z + x)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = 2 x y z (x + y + z)^{3}

⎛ ⎜ ⎝ \begin{matrix} 1 & x & x + 1 2 x & x (x - 1) & x (x + 1) 3 x (1 - x) & x (x - 1) (x - 2) & x (x + 1) (x - 1) \end{matrix} ⎞ ⎟ ⎠ = 6 x^{2} (1 - x^{2})

x + y + z = x y z

\frac{2 x}{1 - x^{2}} + \frac{2 y}{1 - y^{2}} + \frac{2 z}{1 - z^{2}} = \frac{2 x}{1 - x^{2}} \cdot \frac{2 y}{1 - y^{2}} \cdot \frac{2 z}{1 - z^{2}}

u = l o g \sqrt{(x^{2} + y^{2} + z^{2})}

(x^{2} + y^{2} + z^{2}) (\frac{d^{2} u}{d x^{2}} + \frac{d^{2} u}{d y^{2}} + \frac{d^{2} u}{d z^{2}}) = 1

I f x y + y z + z x = 1,

\frac{x}{1 - x^{2}} + \frac{y}{1 - y^{2}} + \frac{z}{1 - z^{2}} = \frac{4 x y z}{(1 - x^{2}) (1 - y^{2}) (1 - z^{2})} .

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