Find $x,y$ and $z$, if $A = \left[ {\begin{array}{cccccccccccccccccccc}0&{2y}&z\\x&y&{ - z}\\x&{ - y}&z\end{array}} \right]$ satisfies ${A^\prime } = {A^{ - 1}}$.

We have, $A = \left[ {\begin{array}{cccccccccccccccccccc}0&{2y}&z\\x&y&{ - z}\\x&{ - y}&z\end{array}} \right]$ and ${A^\prime } = \left[ {\begin{array}{cccccccccccccccccccc}0&x&x\\{2y}&y&{ - y}\\z&{ - z}&z\end{array}} \right]$

By using elementary row transformations,

we get
$A = IA$
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}0&{2y}&z\\x&y&{ - z}\\x&{ - y}&z\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\0&1&0\\0&0&1\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}0&{2y}&z\\x&y&{ - z}\\0&{ - 2y}&{2z}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\0&1&0\\0&{ - 1}&1\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}0&{2y}&z\\x&{3y}&0\\0&0&{3z}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\1&1&0\\1&{ - 1}&1\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{ - x}&{ - y}&z\\x&{3y}&0\\0&0&z\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}0&{ - 1}&0\\1&1&0\\{\frac{1}{3}}&{\frac{{ - 1}}{3}}&{\frac{1}{3}}\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{ - x}&{ - y}&0\\x&{3y}&0\\0&0&z\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{\frac{{ - 1}}{3}}&{\frac{{ - 2}}{3}}&{\frac{{ - 1}}{3}}\\1&1&0\\{\frac{1}{3}}&{\frac{{ - 1}}{3}}&{\frac{1}{3}}\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{ - x}&{ - y}&0\\0&{2y}&0\\0&0&z\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}{\frac{{ - 1}}{3}}&{\frac{{ - 2}}{3}}&{\frac{{ - 1}}{3}}\\{\frac{2}{3}}&{\frac{1}{3}}&{\frac{{ - 1}}{3}}\\{\frac{1}{3}}&{\frac{{ - 1}}{3}}&{\frac{1}{3}}\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{ - x}&0&0\\0&{2y}&0\\0&0&z\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}0&{\frac{{ - 1}}{2}}&{\frac{{ - 1}}{2}}\\{\frac{2}{3}}&{\frac{1}{3}}&{\frac{{ - 1}}{3}}\\{\frac{1}{3}}&{\frac{{ - 1}}{3}}&{\frac{1}{3}}\end{array}} \right]A$

$\Rightarrow$ $\left[ {\begin{array}{llllllllllllllllllll}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}0&{\frac{1}{{2x}}}&{\frac{1}{{2x}}}\\{\frac{1}{{3y}}}&{\frac{1}{{6y}}}&{\frac{{ - 1}}{{6y}}}\\{\frac{1}{{3z}}}&{\frac{{ - 1}}{{3z}}}&{\frac{1}{{3z}}}\end{array}} \right]A$

$\therefore$ ${A^{ - 1}} = \left[ {\begin{array}{cccccccccccccccccccc}0&{\frac{1}{{2x}}}&{\frac{1}{{2x}}}\\{\frac{1}{{3y}}}&{\frac{1}{{6y}}}&{\frac{{ - 1}}{{6y}}}\\{\frac{1}{{3z}}}&{\frac{{ - 1}}{{3z}}}&{\frac{1}{{3z}}}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}0&x&x\\{2y}&y&{ - y}\\z&{ - z}&z\end{array}} \right]$

$\Rightarrow$ $\frac{1}{{2x}} = x \Rightarrow = \pm \frac{1}{{\sqrt 2 }}$

$\Rightarrow$ $\frac{1}{{6y}} = y \Rightarrow y = \pm \frac{1}{{\sqrt 6 }}$

and $\frac{1}{{3z}} = z \Rightarrow z = \pm \frac{1}{{\sqrt 3 }}$

Alternate Method

We have,
$A = \left[ {\begin{array}{cccccccccccccccccccc}0&{2y}&z\\x&y&{ - z}\\x&{ - y}&z\end{array}} \right]$ and ${A^\prime } = \left[ {\begin{array}{cccccccccccccccccccc}0&x&x\\{2y}&y&{ - y}\\z&{ - z}&z\end{array}} \right]$

Also, ${A^\prime } = {A^{ - 1}}$
$\Rightarrow$ $A{A^\prime } = A{A^{ - 1}}$

$\Rightarrow$ $A{A^\prime } = I$
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}0&{2y}&z\\x&y&{ - z}\\x&{ - y}&z\end{array}} \right]\left[ {\begin{array}{cccccccccccccccccccc}0&x&x\\{2y}&y&{ - y}\\z&{ - z}&z\end{array}} \right]$

$= \left[ {\begin{array}{llllllllllllllllllll}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}{4{y^2} + {z^2}}&{2{y^2} - {z^2}}&{ - 2{y^2} + {z^2}}\\{2{y^2} - {z^2}}&{{x^2} + {y^2} + {z^2}}&{{x^2} - {y^2} - {z^2}}\\{ - 2{y^2} + {z^2}}&{{x^2} - {y^2} - {z^2}}&{{x^2} + {y^2} + {z^2}}\end{array}} \right]$

$= \left[ {\begin{array}{cccccccccccccccccccc}1&0&0\\0&1&0\\0&0&1\end{array}} \right]$

$\Rightarrow$ $2{y^2} - {z^2} = 0 \Rightarrow 2{y^2} = {z^2}$

$\Rightarrow$ $4{y^2} + {z^2} = 1$
$\Rightarrow$ $2 \cdot {z^2} + {z^2} = 1$

$z = \pm \frac{1}{{\sqrt 3 }}$
$\therefore$ ${y^2} = \frac{{{z^2}}}{2} \Rightarrow y = \pm \frac{1}{{\sqrt 6 }}$

${x^2} + {y^2} + {z^2} = 1$
$\Rightarrow$ ${x^2} = 1 - {y^2} - {z^2} = 1 - \frac{1}{6} - \frac{1}{3}$

$= 1 - \frac{3}{6} = \frac{1}{2}$
$\Rightarrow$ $x = \pm \frac{1}{{\sqrt 2 }}$

$\therefore$ $x = \pm ,\frac{1}{{\sqrt 2 }},y = \pm \frac{1}{{\sqrt 6 }}$

and $z = \pm \frac{1}{{\sqrt 3 }}$