Question

Find the least square${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}}^{\mathbf{*}}$ of the system$\mathit{A}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}\mathbf{=}\mathit{b}$ where$\mathit{A}\mathbf{=}\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$ and$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{b}}\mathbf{=}\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$ .

Short answer

The solution $\stackrel{\rightharpoonup }{x}=\left[\begin{array}{c}-\frac{7}{6}\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]$.

Step-by-step solution

Explanation of the solution

Consider the system as follows.

$A\stackrel{\rightharpoonup }{x}=b$where $A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$and $\stackrel{\rightharpoonup }{b}=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$.

Consider the matrix A as follows.

$A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$

Now, find $A^{T}A$as follows.

$$\begin{gathered} A^{T}A={\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]}^T\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right] \\ =\left[\begin{array}{ccc}1& 4& 7\\ 2& 5& 8\\ 3& 6& 9\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right] \\ =\left[\begin{array}{ccc}66& 78& 90\\ 78& 93& 108\\ 90& 108& 126\end{array}\right] \\ {A}^{T}A=\left[\begin{array}{ccc}66& 78& 90\\ 78& 93& 108\\ 90& 108& 126\end{array}\right] \end{gathered}$$

The exact solution of the system is as follows.

$$\begin{gathered} \left(A^{T}A\right)\stackrel{\rightharpoonup }{x}=A^T\stackrel{\rightharpoonup }{b} \\ \left[\begin{array}{ccc}66& 78& 90\\ 78& 93& 108\\ 90& 108& 126\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right] \\ \left[\begin{array}{ccc}66& 78& 90\\ 78& 93& 108\\ 90& 108& 126\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right] \\ \left[\begin{array}{ccc}66& 78& 90\\ 78& 93& 108\\ 90& 108& 126\end{array}\right]\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right] \end{gathered}$$

After solving the matrix calculation as follows.

$\left[\overline{)\begin{array}{ccc}1& 0& -1\\ 0& 1& 2\\ 0& 0& 0\end{array}}\begin{array}{c}-\frac{7}{6}\\ 1\\ 0\end{array}\right]$

Then write the equation as follows.

$$\begin{gathered} x_1-x_3=-\frac{7}{6} \\ {x}_2+2x_3=1 \\ {x}_2+2x_3=1 \\ {x}_3=t \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}-\frac{7}{6}+t\\ 1-2t\\ t\end{array}\right] \\ \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]\left[\begin{array}{c}-\frac{7}{6}+t\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}1\\ -2\\ 3\end{array}\right] \end{gathered}$$

Hence, the least square for the system$\left(A^{T}A\right)\overrightarrow{x}=A^T\stackrel{\rightharpoonup }{b}$ is $\stackrel{\rightharpoonup }{x}=\left[\begin{array}{c}-\frac{7}{6}\\ 1\\ 0\end{array}\right]+t\left[\begin{array}{c}1\\ -2\\ 1\end{array}\right]$.