effect on first fundamental form of a surface by isometry Show that applying an isometry of $\mathbb{R}^3$ to a surface does not change its first fundamental form. What is the effect of a dilation ?

This is a problem from Presley book, and it has a solution, but I could not understand the solution at all. Can someone please explain it in detail, please?
Thanks.
 A: We will go over each of the three types of isometry: Translation, reflection, and rotation. Then we will cover dilation separtately.
Suppose $\sigma(u,v)=(x,y,z)$ is a surface patch for some surface. To achieve a translation of our surface we define a new surface patch $\phi$ and associate to it a displacement vector, say $\hat{b}$. So $\phi(u,v) = \sigma(u,v) +\hat{b}=(x,y,z)+\hat{b}$. Now we take the partial derivatives of $\sigma$ and $\phi$. Since we have done nothing but add a constant vector, the partials will be equal (i.e. $\sigma_u=\phi_u$, $\sigma_v=\phi_v$). Thus, the first fundamental forms must also be equal (since these partials define the same tangent space and the first fundamental form is merely the dot product of vectors in the tangent space of a surface at some point).
Now we consider a rotation or a reflection of $\mathbb{R}^3$. Recall, from linear algebra, that every rotation or reflection from $\mathbb{R}^3$ to $\mathbb{R}^3$ may be defined by a unitary linear map. This means we can represent a rotation or reflection of $\mathbb{R}^3$ by matrix multiplication of some matrix, $T$, and the vectors defining our surface (what we call a surface patch) $\sigma$. In other words, we need only consider $T(\sigma)$. But, by differentiating this, we see $(T\sigma)_u=T\sigma_u$ and $(T\sigma)_v=T\sigma_v$.
Now, consider two arbitray vectors in the tangent space of $\sigma$ at some point $p$. We will call the vectors $x$ and $y$. Then the first fundamental form of these vectors is simply the dot product, $x\cdot y$. From the above, we may represent the same vectors after rotation or reflection by $$Tx\cdot Ty=(Tx)^T(Ty)=x^TT^TTy=x^T(T^TT)y$$
Since T is a unitary matrix, we get that $T^TT=I$ where $I$ is the identity matrix. Thus $Tx\cdot Ty= x^Ty=x\cdot y$. So the first fundamental form is preserved.
Lastly, we consider a dilation of $\mathbb{R}^3$ by some scalar $c\neq 0$. Note that this is the same as taking a scalar matrix $C=cI$ in replace of $T$ in the last explanation. Since $C$ is a scalar matrix we get $$Cx\cdot Cy= x^TC^TCy=x^TCCy=x^TC^2y=c^2x^TIy=c^2x^Ty=c^2(x\cdot y)$$
So we can see that this is simply the square of $c$ times the first fundamental form.
