Question

Find the least-squares solution ${\overrightarrow{\mathbf{x}}}^{\mathbf{*}}$ of the system$\mathit{A}\overrightarrow{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{b}}$, where$\mathit{A}\mathbf{=}\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$ androle="math" localid="1660133788323" $\overrightarrow{\mathbf{b}}\mathbf{=}\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$. Use paper and pencil. Draw a sketch showing the vector$\overrightarrow{\mathbf{b}}$, the image of A, the vector A${\overrightarrow{\mathbf{x}}}^{\mathbf{*}}$, and the vector$\overrightarrow{\mathbf{b}}\mathbf{-}\mathit{A}{\overrightarrow{\mathbf{x}}}^{\mathbf{*}}$.

Short answer

The least square solution of${\overrightarrow{x}}^{*}$ is$\left[\begin{array}{c}1\\ 1\end{array}\right]$ 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$A=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$ and $\overrightarrow{b}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$.

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

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

Further, simplify the equation as follows.

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

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

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

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

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

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

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

Draw the graph that containsthe 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$\left[\begin{array}{c}1\\ 1\end{array}\right]$ and the graph is sketched.