Answer:
[tex]\boxed{\bf\:x =\pm \: \dfrac{1}{ \sqrt{2} } ,y =\pm \: \dfrac{1}{ \sqrt{6} } , z = \pm \: \dfrac{1}{ \sqrt{3} } \: } \\ [/tex]
Step-by-step explanation:
Given that,
[tex]\sf\: A = \left[\begin{array}{ccc}0&2y&z\\x&y& - z\\x&-y&z\end{array}\right] \\ [/tex]
Further given that,
[tex]\sf\:A^T = A^{-1}\\ [/tex]
On post multiply by A on both sides, we get
[tex]\sf\:A^T A= A^{-1}A\\ [/tex]
[tex]\sf\:A^TA = I\\ [/tex]
On substituting the values, we get
[tex]\sf\: \left[\begin{array}{ccc}0&x&x\\2y&y& - y\\z&-z&z\end{array}\right]\left[\begin{array}{ccc}0&2y&z\\x&y& - z\\x&-y&z\end{array}\right] = \left[\begin{array}{ccc}1&0&0\\0&1& 0\\0&0&1\end{array}\right]\\ [/tex]
[tex]\sf\: \left[\begin{array}{ccc} {2x}^{2} &0&0\\0& {6y}^{2} & 0\\0&0& {3z}^{2} \end{array}\right] = \left[\begin{array}{ccc}1&0&0\\0&1& 0\\0&0&1\end{array}\right]\\ [/tex]
On comparing,
[tex]\sf\: {2x}^{2} = 1\implies\sf\: {x}^{2} = \dfrac{1}{2} \implies\sf\:x = \pm \: \dfrac{1}{ \sqrt{2} } \\ [/tex]
[tex]\sf\: {6y}^{2} = 1\implies\sf\: {y}^{2} = \dfrac{1}{6} \implies\sf\:y= \pm \: \dfrac{1}{ \sqrt{6} } \\ [/tex]
[tex]\sf\: {3z}^{2} = 1\implies\sf\: {z}^{2} = \dfrac{1}{3} \implies\sf\:z= \pm \: \dfrac{1}{ \sqrt{3} } \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:x =\pm \: \dfrac{1}{ \sqrt{2} } ,y =\pm \: \dfrac{1}{ \sqrt{6} } , z = \pm \: \dfrac{1}{ \sqrt{3} } \: } \\ [/tex]
