Find the values of x, y, z if the matrix $A =\begin{bmatrix}0&2y&z\\x&y&-z\\x&-y&z\end{bmatrix}$ satisfies the equation $A^TA = I$. 

$±\frac{1}{\sqrt{2}},±\frac{1}{\sqrt{6}},±\frac{1}{\sqrt{3}}$

It is given that $A =\begin{bmatrix}0&2y&z\\x&y&-z\\x&-y&z\end{bmatrix}$

$∴A^T=\begin{bmatrix}0&x&x\\2y&y&-y\\z&-z&z\end{bmatrix}$

Now, $A^TA=I$

$⇒\begin{bmatrix}0&x&x\\2y&y&-y\\z&-z&z\end{bmatrix}\begin{bmatrix}0&2y&z\\x&y&-z\\x&-y&z\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

or $\begin{bmatrix}0+x^2+x^2&0+ xy-xy&0-xz + zx\\0+ xy-xy&4y^2+ y^2+ y^2&2yz-yz-yz\\0-xz + zx&2yz-yz-yz&z^2+z^2+z^2\end{bmatrix}$

or $\begin{bmatrix}2x^2&0&0\\0&6y^2&0\\0&0&3z^2\end{bmatrix}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

On comparing the corresponding elements, we have

$2x^2 = 1$ or $x=±\frac{1}{\sqrt{2}}$

$6y^2=1$ or $y=±\frac{1}{\sqrt{6}}$

$3z^2=1$ or $z=±\frac{1}{\sqrt{3}}$