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Consider the matrix $M=\left[\begin{array}{ccc}1& 0& 1\\ 7& 7& 8\\ 1& 2& 1\\ 7& 7& 6\end{array}\right]$and ${\stackrel{鈫�}{v}}_1=\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right]$, ${\stackrel{鈫�}{v}}_2=\left[\begin{array}{c}0\\ 7\\ 2\\ 7\end{array}\right]$and ${\stackrel{鈫�}{v}}_3=\left[\begin{array}{c}1\\ 8\\ 1\\ 6\end{array}\right]$.

By the theorem of QR method, the value of${\stackrel{鈫�}{u}}_1$ and$r_{11}$ is defined as follows.

$$\begin{gathered} r_{11}=\left|\left|{\stackrel{鈫�}{v}}_1\right|\right| \\ {\stackrel{鈫�}{u}}_1=\frac{1}{r_{11}}{\stackrel{鈫�}{v}}_1 \end{gathered}$$

Simplify the equation$r_{11}=\left|\left|{\stackrel{鈫�}{v}}_1\right|\right|$ as follows.

$$\begin{gathered} r_{11}=\left|\left|{\stackrel{鈫�}{v}}_1\right|\right| \\ {r}_{11}=\left|\left|\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right]\right|\right| \\ {r}_{11}=\sqrt{1^2+7^2+1^2+7^2} \\ {r}_{11}=10 \end{gathered}$$

Substitute the values 10 for$r_{11}$and$\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right]$for in the equation${\stackrel{鈫�}{u}}_1=\frac{1}{r_{11}}{\stackrel{鈫�}{v}}_1$as follows.

$$\begin{gathered} {\stackrel{鈫�}{u}}_1=\frac{1}{r_{11}}{\stackrel{鈫�}{v}}_1 \\ {\stackrel{鈫�}{u}}_1=\frac{1}{10}\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \\ {\stackrel{鈫�}{u}}_1=\left[\begin{array}{c}1/10\\ 7/10\\ 1/10\\ 7/10\end{array}\right] \end{gathered}$$

Therefore, the values${\stackrel{鈫�}{u}}_1=\left[\begin{array}{c}1/10\\ 7/10\\ 1/10\\ 7/10\end{array}\right]$and $r_{11}=10$.

As $r_{12}={\stackrel{鈫�}{u}}_1路{\stackrel{鈫�}{v}}_2$, substitute the values$\left[\begin{array}{c}0\\ 7\\ 2\\ 7\end{array}\right]$ for${\stackrel{鈫�}{v}}_2$ and$\left[\frac{1}{10}\frac{7}{10}\frac{1}{10}\frac{7}{10}\right]$ for${\stackrel{鈫�}{u}}_1$ in the equation as follows.

$$\begin{gathered} r_{12}={\stackrel{鈫�}{u}}_1路{\stackrel{鈫�}{v}}_2 \\ {r}_{12}=\left[\frac{1}{10}\frac{7}{10}\frac{1}{10}\frac{7}{10}\right].\left[\begin{array}{c}0\\ 7\\ 2\\ 7\end{array}\right] \\ {r}_{12}=\frac{49}{10}+\frac{1}{5}+\frac{49}{10} \\ {r}_{12}=10 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}0\\ 7\\ 2\\ 7\end{array}\right]$for ${\stackrel{鈫�}{v}}_2$, 10 for $r_{12}$and $\left[\begin{array}{c}1/10\\ 7/10\\ 1/10\\ 7/10\end{array}\right]$for${\stackrel{鈫�}{u}}_1$ in the equation${\stackrel{鈫�}{v}}_2^{鈯�}={\stackrel{鈫�}{v}}_2-r_{12}{\stackrel{鈫�}{u}}_1$ as follows.

$$\begin{gathered} {\stackrel{鈫�}{v}}_2^{鈯�}={\stackrel{鈫�}{v}}_2-r_{12}{\stackrel{鈫�}{u}}_1 \\ {\stackrel{鈫�}{v}}_2^{鈯�}=\left[\begin{array}{c}0\\ 7\\ 2\\ 7\end{array}\right]-10\left[\begin{array}{c}1/10\\ 7/10\\ 1/10\\ 7/10\end{array}\right] \\ {\stackrel{鈫�}{v}}_2^{鈯�}=\left[\begin{array}{c}0\\ 7\\ 2\\ 7\end{array}\right]-\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \\ {\stackrel{鈫�}{v}}_2^{鈯�}=\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right] \end{gathered}$$

Therefore, the values${\stackrel{鈫�}{v}}_2^{鈯�}=\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right]$ and $r_{12}=10$.

The value of${\stackrel{鈫�}{u}}_2$and$r_{22}$is defined as follows.

$$\begin{gathered} r_{22}=\left|\left|{\stackrel{鈫�}{v}}_2^{鈯�}\right|\right| \\ {\stackrel{鈫�}{u}}_2=\frac{1}{r_{22}}{\stackrel{鈫�}{v}}_2^{鈯�} \end{gathered}$$

Simplify the equation$r_{22}=\left|\left|{\stackrel{鈫�}{v}}_2^{鈯�}\right|\right|$as follows.

$$\begin{gathered} r_{22}=\left|\left|{\stackrel{鈫�}{v}}_2^{鈯�}\right|\right| \\ {r}_{22}=\left|\left|\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right]\right|\right| \\ {r}_{22}=\sqrt{{\left(-1\right)}^2+{\left(0\right)}^2+{\left(1\right)}^2+{\left(0\right)}^2} \\ {r}_{22}=\sqrt{2} \end{gathered}$$

Substitute the values$\sqrt{2}$for$r_{22}$and$\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right]$for${\stackrel{鈫�}{v}}_2^{鈯�}$in the equation ${\stackrel{鈫�}{u}}_2=\frac{1}{r_{22}}{\stackrel{鈫�}{v}}_2^{鈯�}$as follows.

localid="1660111423835" $$\begin{gathered} {\stackrel{鈫�}{u}}_2=\frac{1}{r_{22}}{\stackrel{鈫�}{v}}_2^{鈯�} \\ {\stackrel{鈫�}{u}}_2=\frac{1}{\sqrt{2}}\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right] \\ {\stackrel{鈫�}{u}}_2=\left[\begin{array}{c}-1/\sqrt{2}\\ 0\\ 1/\sqrt{2}\\ 0\end{array}\right] \end{gathered}$$

Therefore, the values${\stackrel{鈫�}{u}}_2=\left[\begin{array}{c}-1/\sqrt{2}\\ 0\\ 1/\sqrt{2}\\ 0\end{array}\right]$and $r_{22}=\sqrt{2}$.

As $r_{13}={\stackrel{鈫�}{u}}_1路{\stackrel{鈫�}{v}}_3$, substitute the values$\left[\begin{array}{c}1\\ 8\\ 1\\ 6\end{array}\right]$ for${\stackrel{鈫�}{v}}_3$ and$\left[\frac{1}{10}\frac{7}{10}\frac{1}{10}\frac{7}{10}\right]$ for${\stackrel{鈫�}{u}}_1$ in the equation as follows.

role="math" localid="1660112927183" $$\begin{gathered} r_{13}={\stackrel{鈫�}{u}}_1路{\stackrel{鈫�}{v}}_3 \\ {r}_{13}=\left[\frac{1}{10}\frac{7}{10}\frac{1}{10}\frac{7}{10}\right].\left[\begin{array}{c}1\\ 8\\ 1\\ 6\end{array}\right] \\ {r}_{23}=\frac{1}{10}+\frac{7}{10}+\frac{1}{10}+\frac{7}{10} \\ {r}_{13}=10 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}1\\ 8\\ 1\\ 6\end{array}\right]$for ${\stackrel{鈫�}{v}}_3$and$\left[-\frac{1}{\sqrt{2}}0\frac{1}{\sqrt{2}}0\right]$ for in the equation $r_{23}={\stackrel{鈫�}{u}}_2路{\stackrel{鈫�}{v}}_3$as follows.

$$\begin{gathered} r_{23}={\stackrel{鈫�}{u}}_2路{\stackrel{鈫�}{v}}_3 \\ {r}_{23}=\left[-\frac{1}{\sqrt{2}}0\frac{1}{\sqrt{2}}0\right].\left[\begin{array}{c}1\\ 8\\ 1\\ 6\end{array}\right] \\ {r}_{23}=-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \\ {r}_{23}=0 \end{gathered}$$

Substitute the values$\left[\begin{array}{c}1\\ 8\\ 1\\ 6\end{array}\right]$ for ${\stackrel{鈫�}{v}}_3$, 10 for $r_{13}$, 0 for $r_{23}$,$\left[\begin{array}{c}1/10\\ 7/10\\ 1/10\\ 7/10\end{array}\right]$ for${\stackrel{鈫�}{u}}_1$ and$\left[\begin{array}{c}-1/\sqrt{2}\\ 0\\ 1\sqrt{2}\\ 0\end{array}\right]$ for${\stackrel{鈫�}{u}}_2$ in the equation${\stackrel{鈫�}{v}}_3^{鈯�}={\stackrel{鈫�}{v}}_3-r_{13}{\stackrel{鈫�}{u}}_1-r_{23}{\stackrel{鈫�}{u}}_2$ as follows.

$$\begin{gathered} {\stackrel{鈫�}{v}}_3^{鈯�}={\stackrel{鈫�}{v}}_3-r_{13}{\stackrel{鈫�}{u}}_1-r_{23}{\stackrel{鈫�}{u}}_2 \\ {\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}1\\ 8\\ 1\\ 6\end{array}\right]-10\left[\begin{array}{c}1/10\\ 7/10\\ 1/10\\ 7/10\end{array}\right]-0\left[\begin{array}{c}-1/\sqrt{2}\\ 0\\ 1/\sqrt{2}\\ 0\end{array}\right] \\ {\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}1\\ 8\\ 1\\ 6\end{array}\right]-\left[\begin{array}{c}1\\ 7\\ 1\\ 7\end{array}\right] \\ {\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}0\\ 1\\ 0\\ 1\end{array}\right] \end{gathered}$$

Therefore, the values role="math" localid="1660111878414" ${\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}0\\ 1\\ 0\\ 1\end{array}\right]$,$r_{13}=10$ and role="math" localid="1660111919189" $r_{23}=0$.

The value of${\stackrel{鈫�}{u}}_3$and$r_{33}$is defined as follows.

$\begin{array}{l}{r}_{33}=\left|\left|{\stackrel{鈫�}{v}}_3^{鈯�}\right|\right|\\ {\stackrel{鈫�}{u}}_3=\frac{1}{r_{33}}{\stackrel{鈫�}{v}}_3^{鈯�}\end{array}$

Simplify the equation$r_{33}=\left|\left|{\stackrel{鈫�}{v}}_3^{鈯�}\right|\right|$as follows.

$$\begin{gathered} r_{33}=\left|\left|{\stackrel{鈫�}{v}}_3^{鈯�}\right|\right| \\ {r}_{33}=\left|\left|\left[\begin{array}{c}0\\ 1\\ 0\\ 1\end{array}\right]\right|\right| \\ {r}_{33}=\sqrt{{\left(0\right)}^2+{\left(1\right)}^2+{\left(0\right)}^2+{\left(1\right)}^2} \\ {r}_{33}=\sqrt{2} \end{gathered}$$

Substitute the values$\sqrt{2}$forlocalid="1660113946949" $r_{33}$and$\left[\begin{array}{c}0\\ 1\\ 0\\ 1\end{array}\right]$forlocalid="1660113959100" ${\stackrel{鈫�}{v}}_3^{鈯�}$in the equationlocalid="1660113976831" ${\stackrel{鈫�}{u}}_3=\frac{1}{r_{33}}{\stackrel{鈫�}{v}}_3^{鈯�}$as follows.

localid="1660114003302" $$\begin{gathered} {\stackrel{鈫�}{u}}_3=\frac{1}{r_{33}}{\stackrel{鈫�}{v}}_3^{鈯�} \\ {\stackrel{鈫�}{u}}_3=\frac{1}{\sqrt{2}}\left[\begin{array}{c}-1\\ 0\\ 1\\ 0\end{array}\right] \\ {\stackrel{鈫�}{u}}_3=\left[\begin{array}{c}-1/\sqrt{2}\\ 0\\ 1/\sqrt{2}\\ 0\end{array}\right] \end{gathered}$$

Therefore, the matriceslocalid="1660114249217" $Q=\left[\begin{array}{ccc}1/10& -1/\sqrt{2}& 0\\ 7/10& 0& 1/\sqrt{2}\\ 1/10& 1/\sqrt{2}& 0\\ 7/10& 0& 1/\sqrt{2}\end{array}\right]$and $R=\left[\begin{array}{ccc}10& 10& 10\\ 0& \sqrt{2}& 0\\ 0& 0& \sqrt{2}\end{array}\right]$.

Hence, the QR factorization of the matrix is $\left[\begin{array}{ccc}1& 0& 1\\ 7& 7& 8\\ 1& 2& 1\\ 7& 7& 6\end{array}\right]=\left[\begin{array}{ccc}1/10& -1/\sqrt{2}& 0\\ 7/10& 0& 1/\sqrt{2}\\ 1/10& 1/\sqrt{2}& 0\\ 7/10& 0& 1/\sqrt{2}\end{array}\right]\left[\begin{array}{ccc}10& 10& 10\\ 0& \sqrt{2}& 0\\ 0& 0& \sqrt{2}\end{array}\right]$.