Question

Find the least-squares solution ${\overrightarrow{\mathbf{x}}}^{\mathbf{*}}$of the system $\mathbf{A}\overrightarrow{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{b}}$, where $\mathbf{A}\mathbf{=}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$and$\overrightarrow{\mathbf{b}}\mathbf{=}\left[\begin{array}{c}3\\ 2\\ 7\end{array}\right]$. Draw a sketch showing the vector $\overrightarrow{\mathbf{b}}$, the image of A, the vector$\mathbf{A}{\overrightarrow{\mathbf{x}}}^{\mathbf{*}}$, and the vector$\overrightarrow{\mathbf{b}}\mathbf{-}\mathit{A}\overrightarrow{\mathbf{x}}$.

Short answer

The least square solution of ${\overrightarrow{x}}^{*}$is $\left[2\right]$and the graph is

Step-by-step solution

Determine the least squares solution.

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

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

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

$$\begin{gathered} {\overrightarrow{x}}^{\ast }={\left(A^{T}A\right)}^{-1}{A}^T\overrightarrow{b} \\ {\overrightarrow{x}}^{\ast }={\left({\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]}^{\mathrm{\top }}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\right)}^{-1}{\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]}^T\left[\begin{array}{c}3\\ 2\\ 7\end{array}\right] \\ {\overrightarrow{x}}^{\ast }={\left(\left[\begin{array}{lll}1& 2& 3\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\right)}^{-1}\left[\begin{array}{lll}1& 2& 3\end{array}\right]\left[\begin{array}{c}3\\ 2\\ 7\end{array}\right] \\ {\overrightarrow{x}}^{\ast }=\left(\right[14\left]{)}^{-1}\right[28] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} {\overrightarrow{x}}^{\ast }=\left(\right[14\left]{)}^{-1}\right[28] \\ {\overrightarrow{x}}^{\ast }=\left[2\right] \end{gathered}$$

Draw the graph of the vector b→, image of A, the vector Ax→*and the vector b→=Ax→.

The image of the matrix A is defined as follows.

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

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

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

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

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

Draw the graph that contains the vector$\overrightarrow{b}$, image of A, the vector$A{\overrightarrow{x}}^{*}$and the vector $\overrightarrow{b}-A{\overrightarrow{x}}^{*}$.

Hence, the least square solution of${\overrightarrow{x}}^{*}$is [2] and the graph is sketched.