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If $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=1$, then what is the value of
$\left|\begin{array}{ccc}1 & z & -y \\ -z & 1 & x \\ y & -x & 1\end{array}\right|=?$
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$\left|\begin{array}{ccc}1 & z & -y \\ -z & 1 & x \\ y & -x & 1\end{array}\right|=?$
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The correct answer is:
2
Let $D=\left|\begin{array}{ccc}1 & z & -y \\ -z & 1 & x \\ y & -x & 1\end{array}\right|$
Expanding along $\mathrm{R}_{1}$ $\Rightarrow \mathrm{D}=1\left|\begin{array}{cc}1 & \mathrm{x} \\ -\mathrm{x} & 1\end{array}\right|-\mathrm{z}\left|\begin{array}{cc}-\mathrm{z} & \mathrm{x} \\ \mathrm{y} & 1\end{array}\right|-\mathrm{y}\left|\begin{array}{cc}-\mathrm{z} & 1 \\ \mathrm{y} & -\mathrm{x}\end{array}\right|$
$=\left(1+x^{2}\right)-z(-z-x y)-y(x z-y)$
$=1+x^{2}+z^{2}+x y z-x y z+y^{2}$
$=1+x^{2}+y^{2}+z^{2}=1+1=2$
Expanding along $\mathrm{R}_{1}$ $\Rightarrow \mathrm{D}=1\left|\begin{array}{cc}1 & \mathrm{x} \\ -\mathrm{x} & 1\end{array}\right|-\mathrm{z}\left|\begin{array}{cc}-\mathrm{z} & \mathrm{x} \\ \mathrm{y} & 1\end{array}\right|-\mathrm{y}\left|\begin{array}{cc}-\mathrm{z} & 1 \\ \mathrm{y} & -\mathrm{x}\end{array}\right|$
$=\left(1+x^{2}\right)-z(-z-x y)-y(x z-y)$
$=1+x^{2}+z^{2}+x y z-x y z+y^{2}$
$=1+x^{2}+y^{2}+z^{2}=1+1=2$
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