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

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

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

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

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

Further, simplify the equation as follows.

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

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

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

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

Further, simplify the equation as follows.

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

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