Chapter 5 - Inner Product Spaces - 5.4 Least Squares Approximation - Problems - Page 375: 9

$y=2x^2-5x$

The matrices can be formed as: $A=\begin{bmatrix} 2^2& 2 & 1 \\ 1 & 1& 1 \\ 0^2 & 0 & 1\\ \end{bmatrix}=\begin{bmatrix} 4& 2 &1 \\ 1 & 1 &1 \\ 0 & 0 & 1\end{bmatrix}$ $x=\begin{bmatrix} a \\ b \\ c \end{bmatrix}$ $b=\begin{bmatrix} -2\\ -3\\ 0 \\ \end{bmatrix}$ Apply matrices to the least square solution: $x_0=(A^TA)^{-1}A^Tb$ $=(\begin{bmatrix} 4& 1& 0\\ 2 & 1& 0 \\ 1 & 1 & 1\end{bmatrix}\begin{bmatrix} 4& 2 & 1\\ 1& 1& 1 \\ 0 & 0 & 1\\ \end{bmatrix} )^{-1}\begin{bmatrix} 4 & 1 & 0 \\ 2 & 1 & 0\\ 1 & 1 & 1\end{bmatrix} \begin{bmatrix} -2\\ -3\\ 0 \\ \end{bmatrix}$ $=\begin{bmatrix} 17 & 9 & 5\\ 9& 5 & 3 \\ 5 & 3 & 3 \end{bmatrix}^{-1}\begin{bmatrix} 4 & 1 & 0\\ 2 & 1& 0 \\ 1 & 1&1 \\\end{bmatrix}\begin{bmatrix} -2\\ -3\\ 0 \\ \end{bmatrix}$ $=\frac{1}{2}\begin{bmatrix} 3 & -6 & 1\\ -6 & 13 & -3 \\ 1 & -3 & 2 \end{bmatrix}\begin{bmatrix} 4 & 1 & 0\\ 2& 1&0 \\ 1 & 1&1 \\\end{bmatrix}\begin{bmatrix} -2\\ -3\\ 0 \\ \end{bmatrix}$ $=\frac{1}{2}\begin{bmatrix} 1 & -2 &1 \\ -1 & 4 & -3 \\ 0& 0 & 2 \end{bmatrix}\begin{bmatrix} -2\\ -3\\ 0 \\ \end{bmatrix}$ $=\frac{1}{2}\begin{bmatrix} 4 \\ -10 \\ 0\end{bmatrix}$ $=\begin{bmatrix} 2 \\ -5 \\ 0 \end{bmatrix}$ The equation of the least squares parabola is $y=2x^2-5x$