Question

Find the least-squares solution$\stackrel{\mathbf{\to }\mathbf{*}}{\mathbf{x}}$ of the system$\mathit{A}\overrightarrow{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{b}}$, where$\mathit{A}\mathbf{=}\left[\begin{array}{cc}3& 2\\ 5& 3\\ 4& 5\end{array}\right]$ androle="math" localid="1660124334357" $\overrightarrow{\mathbf{b}}\mathbf{=}\left[\begin{array}{c}5\\ 9\\ 2\end{array}\right]$. Determine the error$\mathbf{||}\overrightarrow{\mathbf{b}}\mathbf{-}\mathbf{A}\stackrel{\mathbf{\to }\mathbf{*}}{\mathbf{x}}\mathbf{||}$.

Short answer

The least square solution of$\overrightarrow{x}$ is$\left[\begin{array}{c}3\\ -2\end{array}\right]$ and the error is 0 .

Step-by-step solution

Determine the least squares solution.

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

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

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 $\overrightarrow{b}$in the equation $\overrightarrow{x}={\left(A^{T}A\right)}^{-1}{A}^T\overrightarrow{b}$.

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

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

Further, simplify the equation as follows.

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

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

Draw the error ||b→-Ax→*||.

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

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

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