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The Inner Productfor any two arbitrary functions is given by:

\(\left\langle {f,g} \right\rangle = \int_0^{2\pi } {f\left( t \right)g\left( t \right)dt} \)

As per the question, we have the weighting matrix\(W\), design matrix\(X\), parametric matrix\(\beta \),and the observation matrix\(y\)as:

\[\begin{array}{l}W = \left[ {\begin{array}{*{20}{c}}1&0&0&0&0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&2&0\\0&0&0&0&1\end{array}} \right]\\X = \left[ {\begin{array}{*{20}{c}}1&{ - 2}\\1&{ - 1}\\1&0\\1&1\\1&2\end{array}} \right]\\\beta = \left[ {\begin{array}{*{20}{c}}{{\beta _1}}\\{{\beta _2}}\end{array}} \right]\\y = \left[ {\begin{array}{*{20}{c}}0\\0\\2\\4\\4\end{array}} \right]\end{array}\]

Usingscaling parameters, we have:

\[\begin{array}{l}WX = \left[ {\begin{array}{*{20}{c}}1&{ - 2}\\2&{ - 2}\\2&0\\2&2\\1&2\end{array}} \right]\\Wy = \left[ {\begin{array}{*{20}{c}}0\\0\\4\\8\\4\end{array}} \right]\end{array}\]

Now, the parameter vector will be:

\[\begin{array}{c}\hat \beta = {\left\{ {{{\left( {WX} \right)}^T}WX} \right\}^{ - 1}}{\left( {WX} \right)^T}Wy\\ = \left[ {\begin{array}{*{20}{c}}{\frac{1}{{14}}}&0\\0&{\frac{1}{{16}}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{28}\\{24}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}2\\{\frac{3}{2}}\end{array}} \right]\end{array}\]

Hence, the least square line is \(y = 2 + \left( {\frac{3}{2}} \right)x\).