91Ó°ÊÓ

Consider the solution of the equation matrix$A\stackrel{→}{x}=\stackrel{→}{b}$ where$A=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$ and $\stackrel{→}{b}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$.

If$\stackrel{→}{x}$ 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$\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$ for A and$\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ for$\stackrel{→}{b}$ in the equation ${\stackrel{→}{x}}^{*}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{→}{b}$.

$$\begin{gathered} {\stackrel{→}{x}}^{*}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{→}{b} \\ {\stackrel{→}{x}}^{*}={\left({\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]}^T\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]\right)}^{-1}{\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]}^T\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right] \\ {\stackrel{→}{x}}^{*}={\left(\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]\right)}^{-1}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right] \\ {\stackrel{→}{x}}^{*}={\left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right)}^{-1}\left[\begin{array}{c}1\\ 1\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\begin{array}{l}{\stackrel{→}{x}}^{*}=\left(\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right)\end{array}}^{-1}\left[\begin{array}{c}1\\ 1\end{array}\right] \\ {\stackrel{→}{x}}^{*}=\left[\begin{array}{c}1\\ 1\end{array}\right] \end{gathered}$$

Therefore, the least square solution of${\stackrel{→}{x}}^{*}$ is $\left[\begin{array}{c}1\\ 1\end{array}\right]$.

Substitute the values$\left[\begin{array}{c}1\\ 1\end{array}\right]$ for${\stackrel{→}{x}}^{*}$ and$\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$ for A in$A{\stackrel{→}{x}}^{*}$ as follows.

$$\begin{gathered} A{\stackrel{→}{x}}^{*}=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right] \\ A{\stackrel{→}{x}}^{*}=\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right] \end{gathered}$$

Substitute the values$\left[\begin{array}{c}1\\ 1\end{array}\right]$ for ${\stackrel{→}{x}}^{*}$,$\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$ for$\stackrel{→}{b}$and $\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$for A in$\stackrel{→}{b}-A{\stackrel{→}{x}}^{*}$ as follows.

$$\begin{gathered} \stackrel{→}{b}-A{\stackrel{→}{x}}^{*}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right] \\ =\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]-\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right] \\ \stackrel{→}{b}-A{\stackrel{→}{x}}^{*}=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$

Draw the graph that containsthe vector$\stackrel{→}{b}$, image of A, the vector$A{\stackrel{→}{x}}^{*}$ and the vector$\stackrel{→}{b}=A{\stackrel{→}{x}}^{*}$.

Hence, the least square solution of${\stackrel{→}{x}}^{*}$ is$\left[\begin{array}{c}1\\ 1\end{array}\right]$ and the graph is sketched.