In Exercises 1-4, find the equation \(y = {\beta _0} + {\beta _1}x\) of the least-square line that best fits the given data points.

4. \(\left( {2,3} \right),\left( {3,2} \right),\left( {5,1} \right),\left( {6,0} \right)\)

Use the x and y coordinates to find the \(X\) and \(y\) matrices.

\(X = \left[ {\begin{aligned}1&2\\1&3\\1&5\\1&6\end{aligned}} \right]\) and \(y = \left[ {\begin{aligned}3\\2\\1\\0\end{aligned}} \right]\)

The normal equation of \(X\beta = y\) can be obtained using \({X^T}X\beta = {X^T}y\), which is equivalent to \(\beta = {\left( {{X^T}X} \right)^{ - 1}}{X^T}y\).

Find \({X^T}X\) as follows:

\(\begin{aligned}{X^T}X &= \left[ {\begin{aligned}1&1&1&1\\2&3&5&6\end{aligned}} \right]\left[ {\begin{aligned}1&2\\1&3\\1&5\\1&6\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{1 + 1 + 1 + 1}&{2 + 3 + 5 + 6}\\{2 + 3 + 5 + 6}&{4 + 9 + 25 + 36}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}4&{16}\\{16}&{74}\end{aligned}} \right]\end{aligned}\)

Find the inverse of \({X^T}X\) as follows:

\(\begin{aligned}{\left( {{X^T}X} \right)^{ - 1}} &= {\left[ {\begin{aligned}4&{16}\\{16}&{74}\end{aligned}} \right]^{ - 1}}\\ &= \frac{1}{{296 - 256}}\left[ {\begin{aligned}{74}&{ - 16}\\{ - 16}&4\end{aligned}} \right]\\ &= \frac{1}{{40}}\left[ {\begin{aligned}{74}&{ - 16}\\{ - 16}&4\end{aligned}} \right]\end{aligned}\)

Find \({X^T}y\) as follows:

\(\begin{aligned}{X^T}y &= \left[ {\begin{aligned}1&1&1&1\\2&3&5&6\end{aligned}} \right]\left[ {\begin{aligned}3\\2\\1\\0\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{3 + 2 + 1 + 0}\\{6 + 6 + 5 + 0}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}6\\{17}\end{aligned}} \right]\end{aligned}\)

Substitute the calculated values in \(\beta = {\left( {{X^T}X} \right)^{ - 1}}{X^T}y\) and solve it as follows:

\(\begin{aligned}\beta &= {\left( {{X^T}X} \right)^{ - 1}}{X^T}y\\\beta &= \frac{1}{{40}}\left[ {\begin{aligned}{74}&{ - 16}\\{ - 16}&4\end{aligned}} \right]\left[ {\begin{aligned}6\\{17}\end{aligned}} \right]\\\beta &= \frac{1}{{40}}\left[ {\begin{aligned}{444 - 272}\\{ - 96 + 68}\end{aligned}} \right]\\\beta &= \frac{1}{{40}}\left[ {\begin{aligned}{172}\\{ - 28}\end{aligned}} \right]\\\left[ {\begin{aligned}{{\beta _0}}\\{{\beta _1}}\end{aligned}} \right] &= \left[ {\begin{aligned}{4.3}\\{ - 0.7}\end{aligned}} \right]\end{aligned}\)

Hence, the equation of the least-square line that best fits is \(y = 4.3 - 0.7x\).