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