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,
$x^{3}+y^{3}+z^{3}=3 \times 2+1 \times(1-0)$
$=7$