Answer
See explanation
Work Step by Step
(a) From Theorem $6(\mathrm{c}),$ we know that if $A$ is invertible, then so is $A^{T}$ The purple box below from Theorem 6 that if two matrices $M$ and $N$ are invertible, then so is their product $M N$. So, since $A$ and $A^{T}$ are invertible in this case, so is $A^{T} A$.
(b) To prove that the inverse of $A$ is $\left(A^{T} A\right)^{-1} A^{T},$ it suffices to show that
\[
A\left[\left(A^{T} A\right)^{-1} A^{T}\right]=I
\]
So let's show that:
\[
\begin{aligned}
A\left[\left(A^{T} A\right)^{-1} A^{T}\right] & \stackrel{*}{=} A\left[A^{-1}\left(A^{T}\right)^{-1} A^{T}\right] \\
&=A\left[A^{-1} I\right] \\
&=A A^{-1} \\
&=I
\end{aligned}
\]
where $*$ uses the fact that $(M N)^{-1}=N^{-1} M^{-1}$
