91Ó°ÊÓ

Consider the solution of the equation matrix $A\stackrel{→}{x}=\stackrel{→}{b}$where$A=\left[\begin{array}{cc}3& 2\\ 5& 3\\ 4& 5\end{array}\right]$and $\stackrel{→}{b}=\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right]$.

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 $\left[\begin{array}{cc}3& 2\\ 5& 3\\ 4& 5\end{array}\right]$for A and $\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right]$for $\stackrel{→}{b}$in the equation $\stackrel{→}{x}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{→}{b}$.

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

$$\begin{gathered} \stackrel{→*}{x}={\left({\left[\begin{array}{cc}3& 2\\ 5& 3\\ 4& 5\end{array}\right]}^T\left[\begin{array}{cc}3& 2\\ 5& 3\\ 4& 5\end{array}\right]\right)}^{-1}{\left[\begin{array}{cc}3& 2\\ 5& 3\\ 4& 5\end{array}\right]}^T\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right] \\ \stackrel{→*}{x}={\left(\left[\begin{array}{ccc}3& 5& 4\\ 2& 3& 5\end{array}\right]\left[\begin{array}{cc}3& 2\\ 5& 3\\ 4& 5\end{array}\right]\right)}^{-1}\left[\begin{array}{ccc}3& 5& 4\\ 2& 3& 5\end{array}\right]\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right] \\ \stackrel{→*}{x}={\left(\left[\begin{array}{cc}50& 41\\ 41& 38\end{array}\right]\right)}^{-1}\left[\begin{array}{c}68\\ 47\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} \stackrel{→*}{x}={\left(\left[\begin{array}{cc}50& 41\\ 41& 38\end{array}\right]\right)}^{-1}\left[\begin{array}{c}68\\ 47\end{array}\right] \\ \stackrel{→*}{x}=\left[\begin{array}{cc}38/219& 41/219\\ -41/219& 50/219\end{array}\right]\left[\begin{array}{c}147\\ 392\end{array}\right] \\ \stackrel{→*}{x}=\left[\begin{array}{c}3\\ -2\end{array}\right] \end{gathered}$$

Therefore, the least square solution of$\stackrel{→}{x}$ is$\left[\begin{array}{c}3\\ -2\end{array}\right]$ .

Substitute the values $\left[\begin{array}{c}3\\ -2\end{array}\right]$for$\stackrel{→}{x}$, $\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right]$for $\stackrel{→}{b}$and $\left[\begin{array}{cc}3& 2\\ 5& 3\\ 4& 5\end{array}\right]$for A in $||\stackrel{→}{b}-A\stackrel{→*}{x}||$as follows.

$$\begin{gathered} ||\stackrel{→}{b}-A\stackrel{→*}{x}||=||\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right]-\left[\begin{array}{cc}3& 2\\ 5& 3\\ 4& 5\end{array}\right]\left[\begin{array}{c}3\\ -2\end{array}\right]|| \\ =||\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right]-\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right]|| \\ =||\begin{array}{c}0\\ 0\\ 0\end{array}|| \\ ||\stackrel{→}{b}-A\stackrel{→*}{x}||=0 \end{gathered}$$

Hence, the least square solution of$\stackrel{→*}{x}$ is$\left[\begin{array}{c}3\\ -2\end{array}\right]$ and the error$||\stackrel{→}{b}-A\stackrel{→*}{x}||$ is 0.