Find `A=[(0,2y,z),(x,y,-z),(x,-y,z)]` satisfies `A^(T) = A^(-1)`

Explanation

We have `A=[{:(0,2y,z),(x,-y,z):}]` and `A=[{:(0,x,x),(2y,y,-y),(z,-z,z):}]`
By using elementary row transformation, we get
`rArr [{:(0,2y,z),(x,y,-z),(x,-y,z):}]=[{:(1,0,0),(0,1,0),(0,0,1):}]A`
`rArr [{:(0,2y,z),(x,y,-z),(0,2y,2z):}]=[{:(1,0,0),(0,1,0),(0,-1,1):}]A [because R_(3)rArrR_(3)-R_(2)]`
`rArr [{:(0,2y,z),(x,3y,0),(0,0,3z):}]=[{:(1,0,0),(1,1,0),(1,-1,1):}]A [{:(becauseR_(3)rArrR_(3),+R_(1)),("and"R_(2)rArrR_(2),+R_(1)):}]`
`rArr [{:(-x,-y-z),(x,3y,0),(0,0,z):}]=[{:(0,-1,0),(1,1,0),((1)/(3),(-1)/(3),(1)/(3)):}]A[because R_(1)rArrR_(1)-R_(2)` and `R_(3)rArr(1)/(3)R_(3)`]
`rArr[{:(-x,-y,0),(x,3y,0),(0,0,z):}]=[{:((-1)/(3),(-2)/(3),(-1)/(3)),(1,1,0),((1)/(3),(-1)/(3),(1)/(3)):}]A [because R_(10rArrR_(1)-R_(3)]`
`rArr [{:(-x,-y,0),(0,2y,0),(0,0,z):}]=[{:((-1)/(3),(-2)/(3),(-1)/(3)),((2)/(3),(1)/(3),(-1)/(3)),((1)/(3),(-1)/(3),(1)/(3)):}]A[because R_(2)rArrR_(2)+R_(1)]`
`rArr[{:(-x,0,0),(0,2y,0),(0,0z):}]=[{:(0,(-1)/(2),(-1)/(2)),((2)/(3),(1)/(3),(-1)/(3)),((1)/(3),(-1)/(3),(1)/(3)):}]A[because R_(1)rArrR_(1)+(1)/(2)R_(2)]`
`rArr [{:(1,0,0),(0,1,0),(0,0,1):}]=[{:(0,(1)/(2x),(1)/(2x)),((1)/(3y),(1)/(6y),(-1)/(6y)),((1)/(3z),(-1)/(3z),(1)/(3z)):}]A[because R_(1)rArr(-1)/(x)R_(1)R_(2)rArr(1)/(2y)R_(2) "and" R_(3)rArr(1)/(2)R_(3)]`
`[{:(0,(1)/(2x),(1)/(2x)),((1)/(3y),(1)/(6 y)),(-1)/(6y)),((1)/(3z),(-1)/(3z),(1)/(3z)):}]=[{:(0,x,x),(2y,y,-y),(z,-z,z):}]`
`rArr (1)/(2x)=xrArr=+-(1)/sqrt(2)`
`rArr (1)/(6y)=yrArry=+-(1)/sqrt(6)`
and `(1)/(3z)=zrArrz=+-(1)/sqrt(3)`
Alternate method
We have, `A=[{:(0,2y,z),(x,y,-z),(x,-y,-z),(x,-y,z):}]` and `A=[{:(0,x,x),(2y,y, -y),(z,-z,z):}]`
Also, `A'=A^(-1)`'
`rArr A A'=A A^(-1) [because A A^(-1)=I]`
`rArr A A=I`
`rArr [{:(0,2y,z),(x,y,-2),(x,y,-z),(x,-y,z):}][{:(0,x,x),(2y,y,-y),(z,-z,z):}]=[{:(1,0,0),(0,1,0),(0,0,1):}]`
`rArr[{:(4y^(2)+z^(2),2y^(2)-z^(2),-2y^(2)+z^(2)),(2y^(2)-z^(2),x^(2)+y^(2)+z^(2),x^(2)-y^(2)-z^(2)),(-2y^(2)+z^(2),x^(2)-y^(2)-z^(2),x^(2)+y^(2)+z^(3)):}]=[{:(1,0,0),(0,1,0),(0,0,1):}]`
`rArr 2y^(2)-z^(2)=0rArr2y^(2)=z^(2)`
`rArr 4y^(2)+z^(2)=1`
`rArr 2.z^(2)+z^(2)=1`
`z=+-(1)/sqrt(3)`
`therefore y^(2)=(z^(2))/(2)rArry=+-(1)/sqrt(6)`
Also, `x^(2)+y^(2)+z^(2)=1`
`rArr x^(2)=1-y^(2)-z^(2)=1-(1)/(6)-(1)/(3)`
`=1-(3)/(6)=(1)/(2)`
`rArr x=+-(1)/sqrt(2)`
`therefore x=+-(1)/sqrt(2).y=+-(1)/sqrt(6)`
and `z=+-(1)/sqrt(3)`