normal equations and projections (linear algebra) I have three points  $(-1,7)$, $(1,7)$ and $(2,21)$ and a linear model $b = C + Dt$. It is asking to give and solve the normal equations in $\hat{x}$ to find the projection $p = A\hat{x}$ of $b$ onto the column space of $A$ (i.e. find $\hat{x}$ with least $||A\hat{x} - b||^2$).
I think I've found the normal equations to be $3C + 2D = 35$ and $2C + 6D = 42$ and the $b = 9 + 4t$, but I'm not sure where to go from there. 

Problem statement
Given a sequence of $m=3$ data points of the form $\left\{ x_{k}, y_{k} \right\}$, and a model function
$$
 y(x) = a_{0} + a_{1} x
$$
find the least squares solution
$$
 a_{LS} = \left\{
 a\in\mathbb{C}^{n} \colon
\lVert
 \mathbf{A} a - y
\rVert_{2}^{2}
\text{ is minimized}
\right\}
$$
Linear system
$$
%
\begin{align}
%
 \mathbf{A} a & = y \\
%
\left[
\begin{array}{cr}
 1 & -1  \\
 1 & 1  \\
 1 & 2 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 a_{0} \\
 a_{1}
\end{array}
\right]
%
&=
\left[
\begin{array}{c}
 7 \\
 7 \\
 21 \\
\end{array}
\right]
\end{align}
%
$$
Normal equations
$$
%
\begin{align}
%
 \mathbf{A}^{*} \mathbf{A} a &= \mathbf{A}^{*} y \\
%
\left[
\begin{array}{cc}
 3 & 2 \\
 2 & 6 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 a_{0} \\
 a_{1}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
 35 \\
 42
\end{array}
\right]
%
\end{align}
%
$$
Solution via normal equations
$$
%
\begin{align}
%
 a_{LS} &= \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*}y \\
%
&=\frac{1}{14}
%
\left[
\begin{array}{rr}
 6 & -2 \\
 -2 & 3 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
 35 \\
 42
\end{array}
\right] \\
%
&=
%
\left[
\begin{array}{c}
 9 \\
 4
\end{array}
\right]
%
\end{align}
%
$$
Solution function
$$ y =9 + 4 x $$
Is the best fit a good fit?
The residual error vector is
$$
 r =\mathbf{A} a_{LS} - y =
\left[
\begin{array}{r}
  -2 \\
   6 \\
  -4
\end{array}
\right]
$$
The function which was minimized is the total error
$$
 r^{2} = r \cdot r = 56.
$$
Solution plotted against the data points:


Projections
What is the projection on the column space?
$$
 \mathbf{A} a = a_{0} \mathbf{A}_{1} + a_{1} \mathbf{A}_{2} = 9
\left[
\begin{array}{r}
   1 \\
   1 \\
   1
\end{array}
\right] 
+ 4
\left[
\begin{array}{r}
  -1 \\
   1 \\
   2
\end{array}
\right] =
%
\left[
\begin{array}{r}
  5 \\
  13 \\
  17
\end{array}
\right]
\in
\mathcal{R}\left( \mathbf{A} \right)
$$


Errata
Given an invertible matrix $\mathbf{A}$, the inverse matrix can be computed using
$$
\mathbf{A}^{-1} = \frac{\text{adj }\mathbf{A}} {\det \mathbf{A}}
$$
where the adjugate matrix adj $\mathbf{A}$ is the matrix of cofactors and $\det \mathbf{A}$ is the determinant.
For the product matrix in the example, the adjugate matrix is
$$
\text{adj }\mathbf{A}^{*}\mathbf{A} =
%
\left[
\begin{array}{rr}
 6 & -2 \\
 -2 & 3 \\
\end{array}
\right]
%
$$
and the determinant is
$$
\det \mathbf{A}^{*}\mathbf{A} = 18 - 4 = 14.
$$
The inverse of the product matrix is
$$
\left( \mathbf{A}^{*}\mathbf{A} \right) ^{-1}
= \frac{1}{14}
%
\left[
\begin{array}{rr}
 6 & -2 \\
 -2 & 3 \\
\end{array}
\right].
$$
Thanks to @spacedustpi for clarification.
