 Solution

For a real linear change of coordinates $x^{\prime }=Ax$, one has $d{x}^{\prime }=Adx$. The Euclidean metric therefore transforms toIt agrees with $d{x}^{T}dx$ for every displacement precisely when $A^{T}A=I$. For example, equality of the associated quadratic forms determines their symmetric coefficient matrices by polarization identity. The orthogonal group is defined by this condition, soTaking determinants gives $(\det A{)}^2=1$. Thus these matrices are invertible automatically; both signs of the determinant are allowed, since the metric alone does not specify an orientation.