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Addition of Vectors

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Question

Shows that $∣ ∣ ∣ ∣ \begin{matrix} y + z & x & x y & z + x & y z & z & x + y \end{matrix} ∣ ∣ ∣ ∣ = 4 x y z$

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Solution

The determinant $d e t (A) = l e t (a_{1}, a_{2}, a_{3})$ is an alternatively mulk linear form that means
$d e t (a_{1} + α a_{2}, a_{2}, a_{3}) < d e t (a_{1}, a_{2}, a_{3}) + d e t (a_{2}, a_{2}, a_{3})$
Because det is linear is first argument and mulk linear form
Because it is in alternating form exchanging from two argument change the sign
$d e t (a_{1}, a_{2}, a_{3}) = - d e t (a_{2}, a_{2}, a_{3})$
Which means $d e t (a_{2}, a_{2}, a_{3})$ vanishes
The leads to
$d e t (a_{1} + a_{2}, a_{2}, a_{3}) = d e t (a_{1}, a_{2}, a_{3})$
Similar for another arguments
Using two property we simplify determinant
$d = ⎡ ⎢ ⎣ \begin{matrix} y + z & x & x y & z + x & y z & z & x + y \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 0 & - 2 z & - 2 y y & z + x & y z & z & x + y \end{matrix} ⎤ ⎥ ⎦ = 2 ⎡ ⎢ ⎣ \begin{matrix} 0 & - z & - y y & z + x & y z & z & x + y \end{matrix} ⎤ ⎥ ⎦ = 2 ⎡ ⎢ ⎣ \begin{matrix} 0 & - z & - y y & x & 0 z & 0 & x \end{matrix} ⎤ ⎥ ⎦ = - 2 ⎡ ⎢ ⎣ \begin{matrix} 0 & z & y y & x & 0 z & 0 & x \end{matrix} ⎤ ⎥ ⎦ = 2 ⎡ ⎢ ⎣ \begin{matrix} 0 & y & z y & z & z z & x & 0 \end{matrix} ⎤ ⎥ ⎦$
If $x = y, x = 0$ then $d = 0$ and formula holds. If two variables $x, y$ and $z$ is zero then one of the columns we have
$d e t (0, a_{2}, a_{3}) = d e t (0 + a_{2}, a_{2}, a_{3}) = d e t (a_{2}, a_{2}, a_{3}) = 0$
If $x = 0$
$= 2 ⎡ ⎢ ⎣ \begin{matrix} 0 & y & z y & 0 & 0 z & 0 & 0 \end{matrix} ⎤ ⎥ ⎦ = 2 y z ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 1 y & 0 & 0 z & 0 & 0 \end{matrix} ⎤ ⎥ ⎦ = 2 y z ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 1 y & 0 & 0 z & 0 & 0 \end{matrix} ⎤ ⎥ ⎦ = - 2 y z ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 1 0 & y & 0 0 & z & 0 \end{matrix} ⎤ ⎥ ⎦ = 0 d = 2 z ⎡ ⎢ ⎣ \begin{matrix} 0 & y & z y & 0 & x z & x & 0 \end{matrix} ⎤ ⎥ ⎦ = \frac{2}{x y z} ⎡ ⎢ ⎣ \begin{matrix} 0 & x y & x z y z & 0 & x z y z & x y & 0 \end{matrix} ⎤ ⎥ ⎦ = \frac{2}{x y z} ⎡ ⎢ ⎣ \begin{matrix} 0 & x y & x z y z & 0 & 2 x z y z & 0 & - 2 x z \end{matrix} ⎤ ⎥ ⎦ = \frac{4}{y} ⎡ ⎢ ⎣ \begin{matrix} 0 & x y & x z 0 & 0 & 1 y z & 0 & - x z \end{matrix} ⎤ ⎥ ⎦ = \frac{4}{y} ⎡ ⎢ ⎣ \begin{matrix} 0 & x y & 0 0 & 0 & 0 y z & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = 4 x y z ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 0 0 & 0 & 1 1 & 0 & 0 \end{matrix} ⎤ ⎥ ⎦ = - 4 x y ⎡ ⎢ ⎣ \begin{matrix} 0 & 1 & 0 1 & 0 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = 4 x y z ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦ = 4 x y z$

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

The value of $∣ ∣ ∣ ∣ \begin{matrix} y + z & x & x y & z + x & y z & z & x + y \end{matrix} ∣ ∣ ∣ ∣$ is equal to

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})} .

Q. Prove the following :

∣ ∣ ∣ ∣ \begin{matrix} y + z & z & y z & z + x & x y & x & x + y \end{matrix} ∣ ∣ ∣ ∣ = 4 x y z

Prove that:
$∣ ∣ ∣ ∣ \begin{matrix} y + z & z & y z & z + x & x y & x & x + y \end{matrix} ∣ ∣ ∣ ∣ = 4 x y z$

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