Answer
$y=\frac{11}{10} x-\frac{9}{5} $
Work Step by Step
The matrices can be formed as:
$A=\begin{bmatrix}
0 & 1 \\
1& 1 \\
2 & 1\\
3 &1 \\
4 & 1
\end{bmatrix}$
$x=\begin{bmatrix}
a \\
b
\end{bmatrix}$
$b=\begin{bmatrix}
-2\\
-1\\
1 \\
2 \\
2
\end{bmatrix}$
Apply matrices to the least square solution:
$x_0=(A^TA)^{-1}A^Tb$
$=(\begin{bmatrix}
0& 1& 2 & 3 & 4\\
1 & 1 & 1 & 1 & 1\\
\end{bmatrix} \begin{bmatrix}
0 & 1 \\
1& 1 \\
2 & 1\\
3 &1 \\
4 & 1
\end{bmatrix})^{-1} \begin{bmatrix}
0& 1& 2 & 3 & 4\\
1 & 1 & 1 & 1 & 1\\
\end{bmatrix}\begin{bmatrix}
-2\\
-1\\
1 \\
2 \\
2
\end{bmatrix}$
$=\begin{bmatrix}
30 & 10\\
10 & 5
\end{bmatrix}^{-1} \begin{bmatrix}
0& 1& 2 & 3 & 4\\
1 & 1 & 1 & 1 & 1\\
\end{bmatrix} \begin{bmatrix}
-2\\
-1\\
1 \\
2 \\
2
\end{bmatrix}$
$=\frac{1}{10}\begin{bmatrix}
1 & -2\\
-2 & 6
\end{bmatrix} \begin{bmatrix}
0& 1& 2 & 3 & 4\\
1 & 1 & 1 & 1 & 1\\
\end{bmatrix} \begin{bmatrix}
-2\\
-1\\
1 \\
2 \\
2
\end{bmatrix}$
$=\frac{1}{10}\begin{bmatrix}
-2& -1& 0 & 1 & 2\\
6 & 4 & 2 & 0 & -2\\
\end{bmatrix}\begin{bmatrix}
-2\\
-1\\
1 \\
2\\
2
\end{bmatrix}$
$=\frac{1}{10}\begin{bmatrix}
11 \\
-18 \end{bmatrix}$
$=\begin{bmatrix}
\frac{11}{10} \\
\frac{-9}{5}
\end{bmatrix}$
The equation of the least squares line associated with the given set of data points is $y=\frac{11}{10} x-\frac{9}{5} $
