We're looking for a line of best fit of the form \(\hat y=c_1x+c_0\). Set up a matrix equation:
\(\begin{bmatrix}1&1\\3&1\\5&1\end{bmatrix}\begin{bmatrix}c_1\\c_0\end{bmatrix}=\begin{bmatrix}5\\13\\21\end{bmatrix}\)
Multiply both sides on the left by the transpose of the coefficient matrix:
\(\begin{bmatrix}1&3&5\\1&1&1\end{bmatrix}\begin{bmatrix}1&1\\3&1\\5&1\end{bmatrix}\begin{bmatrix}c_1\\c_0\end{bmatrix}=\begin{bmatrix}1&3&5\\1&1&1\end{bmatrix}\begin{bmatrix}5\\13\\21\end{bmatrix}\)
\(\begin{bmatrix}35&9\\9&3\end{bmatrix}\begin{bmatrix}c_1\\c_0\end{bmatrix}=\begin{bmatrix}149\\39\end{bmatrix}\)
Multiply both sides by the inverse of the new coefficient matrix:
\(\begin{bmatrix}35&9\\9&3\end{bmatrix}^{-1}\begin{bmatrix}35&9\\9&3\end{bmatrix}\begin{bmatrix}c_1\\c_0\end{bmatrix}=\begin{bmatrix}35&9\\9&3\end{bmatrix}^{-1}\begin{bmatrix}149\\39\end{bmatrix}\)
\(\begin{bmatrix}c_1\\c_0\end{bmatrix}=\begin{bmatrix}4\\1\end{bmatrix}\)
So we end up with a best-fit line of
\(\hat y=4x+1\)
