91Ó°ÊÓ

Consider the solution of the equation matrix$A\stackrel{→}{x}=\stackrel{→}{b}$where$A=\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]$and $\stackrel{→}{b}=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$.

If is the solution of the equation$A\stackrel{→}{x}=\stackrel{→}{b}$then the least-square solution$\stackrel{→*}{x}$is defined as$\stackrel{→}{x}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{→}{b}$.

Substitute the value $\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]$for A and $\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right]$for in the equation $\stackrel{→*}{x}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{→}{b}$.

$\stackrel{→*}{x}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{→}{b}$

role="math" localid="1660126240637" $$\begin{gathered} \stackrel{→*}{x}={\left({\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]}^T\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]\right)}^{-1}{\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]}^T\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right] \\ \stackrel{→*}{x}={\left(\left[\begin{array}{ccc}1& 2& 1\\ 1& 8& 5\end{array}\right]\left[\begin{array}{cc}1& 1\\ 2& 8\\ 1& 5\end{array}\right]\right)}^{-1}\left[\begin{array}{ccc}3& 5& 4\\ 2& 3& 5\end{array}\right]\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right] \\ \stackrel{→*}{x}={\left(\left[\begin{array}{cc}6& 22\\ 22& 90\end{array}\right]\right)}^{-1}\left[\begin{array}{c}68\\ 47\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

role="math" localid="1660126334311" $$\begin{gathered} \stackrel{→*}{x}={\left(\left[\begin{array}{cc}6& 22\\ 22& 90\end{array}\right]\right)}^{-1}\left[\begin{array}{c}68\\ 47\end{array}\right] \\ \stackrel{→*}{x}=\left[\begin{array}{cc}90/28& -22/28\\ -22/28& 6/28\end{array}\right]\left[\begin{array}{c}68\\ 47\end{array}\right] \\ \stackrel{→*}{x}=\left[\begin{array}{c}0\\ 0\end{array}\right] \end{gathered}$$

Therefore, the least square solution of$\stackrel{→*}{x}$ is $\left[\begin{array}{c}0\\ 0\end{array}\right]$.

If$\stackrel{→*}{x}=0$implies $A\stackrel{→*}{x}=\stackrel{→}{0}$.

As $A\stackrel{→*}{x}$is orthogonal projection of $\stackrel{→}{b}$onto $Im\left(A\right)$and$A\stackrel{→*}{x}=\stackrel{→}{0}$ means the vector$\stackrel{→}{b}$ is perpendicular to the column of A.

Hence, the least square solution of $\stackrel{→*}{x}$is $\left[\begin{array}{c}0\\ 0\end{array}\right]$and the vector $\stackrel{→}{b}$is perpendicular to the column of A .