Answer
The equation of the least squares line associated with the given set of data points is $y=\frac{3}{8}x+\frac{3}{4}$
Work Step by Step
The matrices can be formed as:
$A=\begin{bmatrix}
6& 1 \\
-2 & 1 \\\end{bmatrix} \rightarrow A^T=\begin{bmatrix}
6& -2 \\
1 & 1 \\\end{bmatrix}$
$x=\begin{bmatrix}
a \\
b \\
\end{bmatrix}$
$b=\begin{bmatrix}
3 \\
0
\end{bmatrix}$
Apply matrices to the least square solution:
$x_0=(A^TA)^{-1}A^Tb$
$=(\begin{bmatrix}
6& -2 \\
1 & 1 \\\end{bmatrix} \begin{bmatrix}
6& 1 \\
-2 & 1 \\\end{bmatrix})^{-1} \begin{bmatrix}
6& -2 \\
1 & 1 \\\end{bmatrix} \begin{bmatrix}
3 \\
0
\end{bmatrix}$
$=\begin{bmatrix}
40 &4 \\
4&2
\end{bmatrix}^{-1}\begin{bmatrix}
6& -2 \\
1 & 1 \\\end{bmatrix}\begin{bmatrix}
3 \\
0
\end{bmatrix}$
$=\frac{1}{32}\begin{bmatrix}
1 & -2 \\
-2& 20
\end{bmatrix}\begin{bmatrix}
6& -2 \\
1 & 1 \\\end{bmatrix}\begin{bmatrix}
3 \\
0
\end{bmatrix}$
$=\frac{1}{8}\begin{bmatrix}
1 &-1 \\
2& 6
\end{bmatrix}\begin{bmatrix}
3 \\
0
\end{bmatrix}$
$=\frac{1}{8}\begin{bmatrix}
3 \\
6
\end{bmatrix}$
$=\begin{bmatrix}
\frac{3}{8} \\
\frac{3}{4}
\end{bmatrix}$
The equation of the least squares line associated with the given set of data points is $y=\frac{3}{8}x+\frac{3}{4}$
