Question

Find the least-squares solution$\overrightarrow{\mathbf{x}}$ of the system$\mathit{A}\overrightarrow{\mathbf{x}}\mathbf{=}\overrightarrow{\mathbf{b}}$, where$\mathit{A}\mathbf{=}\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]$ and$\overrightarrow{\mathbf{b}}\mathbf{=}\left[\begin{array}{c}5\\ 6\end{array}\right]$. Use only paper and pencil. Draw a sketch.

Short answer

The least square solution of${\overrightarrow{x}}^{*}$is$\overrightarrow{x_1}=1-3\overrightarrow{x_2}$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$A=\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]$and $\overrightarrow{b}=\left[\begin{array}{c}5\\ 0\end{array}\right]$.

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

Substitute the value $\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]$for A and $\left[\begin{array}{c}3\\ 2\\ 7\end{array}\right]$for $\overrightarrow{b}$in the equation $\left(A^{T}A\right){\overrightarrow{x}}^{*}=A^T\overrightarrow{b}$.

$$\begin{gathered} \left(A^{T}A\right){\overrightarrow{x}}^{*}{A}^T\overrightarrow{b} \\ \left({\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]}^T\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]\right){\overrightarrow{x}}^{*}={\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]}^T\left[\begin{array}{c}5\\ 0\end{array}\right] \\ \left(\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right]\right){\overrightarrow{x}}^{*}={\left[\begin{array}{cc}1& 2\\ 3& 6\end{array}\right]}^T\left[\begin{array}{c}5\\ 0\end{array}\right] \\ \left(\left[\begin{array}{cc}5& 15\\ 15& 45\end{array}\right]\right){\overrightarrow{x}}^{*}=\left[\begin{array}{c}5\\ 15\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} \left(\left[\begin{array}{cc}5& 15\\ 15& 45\end{array}\right]\right)\left[\begin{array}{c}{\overrightarrow{x}}_1\\ {\overrightarrow{x}}_2\end{array}\right]=\left[\begin{array}{c}5\\ 15\end{array}\right] \\ 5{\overrightarrow{x}}_1=15{\overrightarrow{x}}_2=5 \\ {\overrightarrow{x}}_1=1-3{\overrightarrow{x}}_2 \end{gathered}$$

Therefore, the least square solution of ${\overrightarrow{x}}^{*}$is in the form of${\overrightarrow{x}}_1=1-3{\overrightarrow{x}}_2$.

Draw the graph of.

Draw the graph of the least square solution of the equation$\overrightarrow{x_1}=1-3\overrightarrow{x_2}$as follows.

Hence, the least square solution of ${\overrightarrow{x}}^{*}$is ${\overrightarrow{x}}_1=1-3{\overrightarrow{x}}_2$and the graph is sketched.