By using the method of least squares, find the best line through the points:
$(-1, 0)$,   $(1, -2)$,   $(-2, 2)$.

Step 1. The general equation of a line is $c_0 + c_1x = y$. Plugging the data points into this formula gives a matrix equation $\mathrm{A}\mathbf{c} = \mathbf{y}$.
$\left[\Rule{0pt}{3.6em}{0pt}\right.$$\left]\Rule{0pt}{3.6em}{0pt}\right.$ $\left[\begin{matrix}c_0\\c_1\end{matrix}\right]$ $=$ $\left[\Rule{0pt}{3.6em}{0pt}\right.$$\left]\Rule{0pt}{3.6em}{0pt}\right.$

Step 2. The matrix equation $\mathrm{A}\mathbf{c} = \mathbf{y}$ has no solution, so instead we use the normal equation $\mathrm{A^T} \mathrm{A}\, \hat{\mathbf{c}} = \mathrm{A^T} \mathbf{y}$
$\mathrm{A^T} \mathrm{A} =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$

$\mathrm{A^T} \mathbf{y} =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$

Step 3. Solving the normal equation gives the answer
$\hat{\mathbf{c}} =$ $\left[\Rule{0pt}{2.4em}{0pt}\right.$$\left]\Rule{0pt}{2.4em}{0pt}\right.$

which corresponds to the formula
$y =$

Analysis. Compute the predicted $y$ values: $\hat{\mathbf{y}} = \mathrm{A}\hat{\mathbf{c}}$.
$\hat{\mathbf{y}} =$ $\left[\Rule{0pt}{3.6em}{0pt}\right.$$\left]\Rule{0pt}{3.6em}{0pt}\right.$

Compute the error vector: $\mathbf{e} = \mathbf{y} - \hat{\mathbf{y}}$.
$\mathbf{e} =$ $\left[\Rule{0pt}{3.6em}{0pt}\right.$$\left]\Rule{0pt}{3.6em}{0pt}\right.$

Compute the total error: $\mathrm{SSE} = e_1^2+e_2^2+e_3^2$.
$\mathrm{SSE} =$

You can earn partial credit on this problem.