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$\mathit{A}\mathbf{=}\left[\begin{array}{cc}6& 9\\ 3& 8\\ 2& 10\end{array}\right]$ and$\overrightarrow{\mathbf{b}}\mathbf{=}\left[\begin{array}{c}0\\ 49\\ 0\end{array}\right]$. Determine the error$||\overrightarrow{b}-A\stackrel{\to *}{x}||$.

Short answer

The least square solution of $\stackrel{\to *}{x}$is$\left[\begin{array}{c}-1\\ 2\end{array}\right]$ and the error is 42.

Step-by-step solution

Determine the least squares solution.

Consider the solution of the equation matrix $A\overrightarrow{x}=\overrightarrow{b}$= where $A=\left[\begin{array}{cc}6& 9\\ 3& 8\\ 2& 10\end{array}\right]$and $\overrightarrow{b}=\left[\begin{array}{c}0\\ 49\\ 0\end{array}\right]$.

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 $\left[\begin{array}{cc}6& 9\\ 3& 8\\ 2& 10\end{array}\right]$for A and $\left[\begin{array}{c}0\\ 49\\ 0\end{array}\right]$for $\overrightarrow{b}$in the equation $\overrightarrow{x}={\left(A^{T}A\right)}^{-1}{A}^T\overrightarrow{b}$.

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

$$\begin{gathered} \stackrel{\to *}{x}={\left({\left[\begin{array}{cc}6& 9\\ 3& 8\\ 2& 10\end{array}\right]}^T\left[\begin{array}{cc}6& 9\\ 3& 8\\ 2& 10\end{array}\right]\right)}^{-1}{\left[\begin{array}{cc}6& 9\\ 3& 8\\ 2& 10\end{array}\right]}^T\left[\begin{array}{c}0\\ 49\\ 0\end{array}\right] \\ \stackrel{\to *}{x}={\left(\left[\begin{array}{ccc}6& 3& 2\\ 9& 8& 10\end{array}\right]\left[\begin{array}{cc}6& 9\\ 3& 8\\ 2& 10\end{array}\right]\right)}^{-1}\left[\begin{array}{ccc}6& 3& 2\\ 9& 8& 10\end{array}\right]\left[\begin{array}{c}0\\ 49\\ 0\end{array}\right] \\ \stackrel{\to *}{x}={\left(\left[\begin{array}{cc}45& 98\\ 98& 245\end{array}\right]\right)}^{-1}\left[\begin{array}{c}147\\ 392\end{array}\right] \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} \stackrel{\to *}{x}={\left(\left[\begin{array}{cc}45& 98\\ 98& 245\end{array}\right]\right)}^{-1}\left[\begin{array}{c}147\\ 392\end{array}\right] \\ \stackrel{\to *}{x}=\left[\begin{array}{cc}5/49& -2/49\\ -2/49& 1/49\end{array}\right]\left[\begin{array}{c}147\\ 392\end{array}\right] \\ \stackrel{\to *}{x}=\left[\begin{array}{c}-1\\ 2\end{array}\right] \end{gathered}$$

Therefore, the least square solution of is$\stackrel{\to *}{x}=\left[\begin{array}{c}-1\\ 2\end{array}\right]$ .

Draw the error ||b→-Ax|| .

Substitute the values $\left[\begin{array}{c}-1\\ 2\end{array}\right]$for $\stackrel{\to *}{x}$, $\left[\begin{array}{c}0\\ 49\\ 0\end{array}\right]$for $\overrightarrow{b}$and $\left[\begin{array}{cc}6& 9\\ 3& 8\\ 2& 10\end{array}\right]$for A in $||\overrightarrow{b}-A\stackrel{\to *}{x}||$ as follows.

$$\begin{gathered} ||\overrightarrow{b}-A\stackrel{\to *}{x}||=||\left[\begin{array}{c}0\\ 49\\ 0\end{array}\right]-\left[\begin{array}{cc}6& 9\\ 3& 8\\ 2& 10\end{array}\right]\left[\begin{array}{c}-1\\ 2\end{array}\right]|| \\ =||\left[\begin{array}{c}0\\ 49\\ 0\end{array}\right]-\left[\begin{array}{c}12\\ 13\\ 18\end{array}\right]|| \\ =||\begin{array}{c}-12\\ 36\\ -18\end{array}|| \\ ||\overrightarrow{b}-A\stackrel{\to *}{x}||=\sqrt{{\left(-12\right)}^2+{\left(36\right)}^2{\left(-18\right)}^2} \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} ||\overrightarrow{b}-A\stackrel{\to *}{x}||=\sqrt{{\left(-12\right)}^2+{\left(36\right)}^2{\left(-18\right)}^2} \\ =\sqrt{1764} \\ ||\overrightarrow{b}-A\stackrel{\to *}{x}||=42 \end{gathered}$$

Hence, the least square solution of$\overrightarrow{x}$ is $\left[\begin{array}{c}-1\\ 2\end{array}\right]$and the error $||\overrightarrow{b}-A\stackrel{\to *}{x}||$is 42.