Prove the following

Question:

Let $A=\left[\begin{array}{lll}x & y & z \\ y & z & x \\ z & x & y\end{array}\right]$, where $x, y$ and $z$ are real

numbers such that $x+y+z>0$ and $x y z=2$.

If $\mathrm{A}^{2}=\mathrm{I}_{3}$, then the value of $\mathrm{x}^{3}+\mathrm{y}^{3}+\mathrm{z}^{3}$ is

Solution:

$\mathrm{A}^{2}=\mathrm{I}$

$\Rightarrow \mathrm{AA}^{\prime}=\mathrm{I} \quad\left(\right.$ as $\left.\mathrm{A}^{\prime}=\mathrm{A}\right)$

$\Rightarrow \mathrm{A}$ is orthogonal

So, $x^{2}+y^{2}+z^{2}=1$ and $x y+y z+z x=0$

$\Rightarrow(x+y+z)^{2}=1+2 \times 0$

$\Rightarrow x+y+z=1$

Thus,

$x^{3}+y^{3}+z^{3}=3 \times 2+1 \times(1-0)$

$=7$

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