91Ó°ÊÓ

Consider the solution of the equation matrix $A\stackrel{→}{x}=\stackrel{→}{b}$where $A=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$and $\stackrel{→}{b}\left[\begin{array}{c}3\\ 2\\ 7\end{array}\right]$.

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 $\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$ for A and $\left[\begin{array}{c}3\\ 2\\ 7\end{array}\right]$for $\stackrel{→}{b}$in the equation ${\stackrel{→}{x}}^{*}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{→}{b}$.

$$\begin{gathered} {\stackrel{→}{x}}^{∗}={\left(A^{T}A\right)}^{−1}{A}^T\stackrel{→}{b} \\ {\stackrel{→}{x}}^{∗}={\left({\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]}^{\mathrm{⊤}}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\right)}^{−1}{\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]}^T\left[\begin{array}{c}3\\ 2\\ 7\end{array}\right] \\ {\stackrel{→}{x}}^{∗}={\left(\left[\begin{array}{lll}1& 2& 3\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\right)}^{−1}\left[\begin{array}{lll}1& 2& 3\end{array}\right]\left[\begin{array}{c}3\\ 2\\ 7\end{array}\right] \\ {\stackrel{→}{x}}^{∗}=\left(\right[14\left]{)}^{−1}\right[28] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\stackrel{→}{x}}^{∗}=\left(\right[14\left]{)}^{−1}\right[28] \\ {\stackrel{→}{x}}^{∗}=\left[2\right] \end{gathered}$$

The image of the matrix A is defined as follows.

$$\begin{gathered} Im\left(A\right)=\left\{Ax|x∈□\right\} \\ =\left\{\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]x|x∈□\right\} \\ Im\left(A\right)=\left\{\left[\begin{array}{c}1x\\ 2x\\ 3x\end{array}\right]|x∈□\right\} \end{gathered}$$

Substitute the values$\left[2\right]$for${\stackrel{→}{x}}^{*}$and$\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$for A in$A{\stackrel{→}{x}}^{*}$as follows.

$$\begin{gathered} A{\stackrel{→}{x}}^{*}=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\left[2\right] \\ A{\stackrel{→}{x}}^{*}=\left[\begin{array}{c}2\\ 4\\ 6\end{array}\right] \end{gathered}$$

Substitute the values [2] for${\stackrel{→}{x}}^{*}$,$\left[\begin{array}{c}3\\ 2\\ 7\end{array}\right]$for$\stackrel{→}{b}$and$\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$for A in$\stackrel{→}{b}-A{\stackrel{→}{x}}^{*}$as follows.

$$\begin{gathered} \stackrel{→}{b}-A{\stackrel{→}{x}}^{*}=\left[\begin{array}{c}3\\ 2\\ 7\end{array}\right]-\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\left[2\right] \\ =\left[\begin{array}{c}3\\ 2\\ 7\end{array}\right]-\left[\begin{array}{c}2\\ 4\\ 6\end{array}\right] \\ \stackrel{→}{b}-A\stackrel{→}{x}=\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right] \end{gathered}$$

Draw the graph that contains the 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 [2] and the graph is sketched.