Answer
$y=-\frac{15}{37} x-\frac{12}{7}$
Work Step by Step
The matrices can be formed as:
$A=\begin{bmatrix}
-3 & 1 \\
-2& 1 \\
-1 & 1\\
0 &1 \\
2 & 1
\end{bmatrix}$
$x=\begin{bmatrix}
a \\
b
\end{bmatrix}$
$b=\begin{bmatrix}
1\\
0\\
1\\
-1 \\
-1
\end{bmatrix}$
Apply matrices to the least square solution:
$x_0=(A^TA)^{-1}A^Tb$
$=(\begin{bmatrix}
-3& -2& -1 & 0 &2 \\
1 & 1 & 1 & 1 & 1
\end{bmatrix} \begin{bmatrix}
-3 & 1 \\
-2& 1 \\
-1& 1\\
0&1 \\
2 & 1
\end{bmatrix})^{-1} \begin{bmatrix}
-3& -2& -1 & 0 &2 \\
1 & 1 & 1 & 1 & 1
\end{bmatrix} \begin{bmatrix}
1\\
0 \\
1\\
-1\\
-1
\end{bmatrix}$
$=\begin{bmatrix}
18 & -4\\
-4 & 5
\end{bmatrix}^{-1} \begin{bmatrix}
-3& -2& -1 & 0 &2 \\
1 & 1 & 1 & 1 & 1
\end{bmatrix} \begin{bmatrix}
1\\
0 \\
1\\
-1 \\
-1
\end{bmatrix}$
$=\frac{1}{74}\begin{bmatrix}
5 & 4\\
4 & 18
\end{bmatrix} \begin{bmatrix}
-3& -2& -1 & 0 &2 \\
1 & 1 & 1 & 1 & 1
\end{bmatrix} \begin{bmatrix}
1\\
0 \\
1 \\
-1 \\
-1
\end{bmatrix}$
$=\frac{1}{74} \begin{bmatrix}
-11& -6& -1 & 4 & 14 \\
6 & 10 & 14 & 18 & 26
\end{bmatrix} \begin{bmatrix}
1\\
0 \\
1\\
-1 \\
-1
\end{bmatrix}$
$=\frac{1}{74}\begin{bmatrix}
-30\\
-24 \end{bmatrix}$
$=\begin{bmatrix}
-\frac{15}{37} \\
-\frac{12}{7}
\end{bmatrix}$
The equation of the least squares line associated with the given set of data points is $y=-\frac{15}{37} x-\frac{12}{7}$
