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Consider the matrix M$M=\left[\begin{array}{ccc}2& -2& 18\\ 2& 1& 0\\ 1& 2& 0\end{array}\right]$and role="math" localid="1659936948232" ${\stackrel{鈫�}{v}}_1=\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right],{\stackrel{鈫�}{v}}_2=\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$, and ${\stackrel{鈫�}{v}}_3=\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$.

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

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

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

$$\begin{gathered} r_{11}=\left|\right|{\stackrel{鈫�}{v}}_1\left|\right| \\ {r}_{11}=\left|\left|\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]\right|\right| \\ {r}_{11}=\sqrt{2^2+2^2+1^2} \\ {r}_{11}=3 \end{gathered}$$

Substitute the values 3 for $r_{11}$and $\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$for ${\stackrel{鈫�}{v}}_1$in the equation ${\stackrel{鈫�}{u}}_1=\frac{1}{r_{11}}{\stackrel{鈫�}{v}}_1$as follows:

$$\begin{gathered} {\stackrel{鈫�}{u}}_1=\frac{1}{r_{11}}{\stackrel{鈫�}{v}}_1 \\ {\stackrel{鈫�}{u}}_1=\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \\ {\stackrel{鈫�}{u}}_1=\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right] \end{gathered}$$

Therefore,${\stackrel{鈫�}{u}}_1=\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]$ and$r_{11}=3$ .

As $r_{12}={\stackrel{鈫�}{u}}_1.{\stackrel{鈫�}{v}}_2$, substitute the values $\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$for ${\stackrel{鈫�}{v}}_2$and${\left[\begin{array}{ccc}\frac{2}{3}& \frac{2}{3}& \frac{2}{3}\end{array}\right]}^T$for in the equation

$r_{12}={\stackrel{鈫�}{u}}_1.{\stackrel{鈫�}{v}}_2$as follows:

$r_{12}={\stackrel{鈫�}{u}}_1.{\stackrel{鈫�}{v}}_2$

$$\begin{gathered} r_{12}={\left[\begin{array}{ccc}\frac{2}{3}& \frac{2}{3}& \frac{2}{3}\end{array}\right]}^T.\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right] \\ {r}_{12}=-\begin{array}{ccc}\frac{4}{3}& +\frac{2}{3}& \frac{2}{3}\end{array} \\ {r}_{12}=0 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$for ${\stackrel{鈫�}{v}}_2$,0 for $r_{12}$and$\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]$ for in the equation

${\stackrel{鈫�}{v}}_2^{鈯�}=\stackrel{鈫�}{v_2}-r_{12}{\stackrel{鈫�}{u}}_1$as follows :

${\stackrel{鈫�}{v}}_2^{鈯�}=\stackrel{鈫�}{v_2}-r_{12}{\stackrel{鈫�}{u}}_1$

${\stackrel{鈫�}{v}}_2^{鈯�}=\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]-\left(0\right)\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]$

${\stackrel{鈫�}{v}}_2^{鈯�}=\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$

Therefore ,${\stackrel{鈫�}{v}}_2^{鈯�}=\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]\mathrm{and}{r}_{12}=0$

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

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

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

$$\begin{gathered} r_{22}=||{\stackrel{鈫�}{v}}_2^{鈯�}|| \\ {r}_{22}||\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]|| \\ {r}_{22}=\sqrt{{\left(-2\right)}^2+1^2+2^2} \\ {r}_{22}=3 \end{gathered}$$

Substitute the values 3 for $r_{22}$and $\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$for ${\stackrel{鈫�}{v}}_2^{鈯�}$in the equation role="math" localid="1659938646006" ${\stackrel{鈫�}{u}}_2=\frac{1}{r_{22}}{\stackrel{鈫�}{v}}_2^{鈯�}$as follows:

$$\begin{gathered} {\stackrel{鈫�}{u}}_2=\frac{1}{r_{22}}{\stackrel{鈫�}{v}}_2^{鈯�} \\ {\stackrel{鈫�}{u}}_2=\frac{1}{3}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right] \\ {\stackrel{鈫�}{u}}_2=\left[\begin{array}{c}\frac{-2}{3}\\ \frac{1}{3}\\ \frac{2}{3}\end{array}\right] \end{gathered}$$

Therefore ,${\stackrel{鈫�}{u}}_2=\left[\begin{array}{c}\frac{-2}{3}\\ \frac{1}{3}\\ \frac{2}{3}\end{array}\right]\mathrm{and}{\mathrm{r}}_{22}=3$

As $r_{13}={\stackrel{鈫�}{u}}_1.{\stackrel{鈫�}{v}}_3$, substitute the values $\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$for ${\stackrel{鈫�}{v}}_3$and ${\left[\frac{2}{3}\frac{2}{3}\frac{1}{3}\right]}^T$for ${\stackrel{鈫�}{u}}_1$in the equation

$r_{13}={\stackrel{鈫�}{u}}_1.{\stackrel{鈫�}{v}}_3$as follows:

$r_{13}={\stackrel{鈫�}{u}}_1.{\stackrel{鈫�}{v}}_3$

$r_{12}={\left[\frac{2}{3}\frac{2}{3}\frac{1}{3}\right]}^T.\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$

$r_{12}=12$

Substitute the values $\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$for ${\stackrel{鈫�}{v}}_3$and ${\left[-\frac{2}{3}\frac{1}{3}\frac{2}{3}\right]}^T$for ${\stackrel{鈫�}{u}}_1$in the equation $r_{23}={\stackrel{鈫�}{u}}_2.{\stackrel{鈫�}{v}}_3$as follows:

$$\begin{gathered} r_{23}={\stackrel{鈫�}{u}}_2.{\stackrel{鈫�}{v}}_3 \\ {r}_{23}={\left[-\frac{2}{3}\frac{1}{3}\frac{2}{3}\right]}^T.\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right] \\ {r}_{23}=-12 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]$for $\stackrel{鈫�}{v_3}$, 12 for $r_{13}$, -12 for $r_{23}$, $\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]$for ${\stackrel{鈫�}{u}}_1$and $\left[\begin{array}{c}-\frac{2}{3}\\ \frac{1}{3}\\ \frac{2}{3}\end{array}\right]$for ${\stackrel{鈫�}{u}}_2$in the equation ${\stackrel{鈫�}{v}}_3^{鈯�}={\stackrel{鈫�}{v}}_3-r_{13}\stackrel{鈫�}{u}-r_{23}{\stackrel{鈫�}{u}}_2$as follows:

role="math" localid="1659939547497" ${\stackrel{鈫�}{v}}_3^{鈯�}={\stackrel{鈫�}{v}}_3-r_{13}\stackrel{鈫�}{u}-r_{23}{\stackrel{鈫�}{u}}_2\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]$

$$\begin{gathered} {\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]-12\left[\begin{array}{c}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{array}\right]+12\left[\begin{array}{c}-\frac{2}{3}\\ \frac{1}{3}\\ \frac{2}{3}\end{array}\right] \\ {\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}18\\ 0\\ 0\end{array}\right]-\left[\begin{array}{c}8\\ 8\\ 4\end{array}\right]+\left[\begin{array}{c}-8\\ 4\\ 8\end{array}\right] \\ {\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right] \end{gathered}$$

Therefore ,${\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right]$,$r_{12}=12\mathrm{and}{\mathrm{r}}_{23}=-12$

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

$$\begin{gathered} r_{33}=||{\stackrel{鈫�}{v}}_3^{鈯�}|| \\ {\stackrel{鈫�}{u}}_3=\frac{1}{r_{33}}{\stackrel{鈫�}{v}}_3^{鈯�} \end{gathered}$$

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

$$\begin{gathered} r_{33}=||{\stackrel{鈫�}{v}}_3^{鈯�}|| \\ {r}_{33}=||\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right]|| \\ {r}_{33}=\sqrt{{\left(-2\right)}^2+{\left(-4\right)}^2+{\left(4\right)}^2} \\ {r}_{33}=6 \end{gathered}$$

Substitute the values 6 for $r_{33}$and $\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right]$for ${\stackrel{鈫�}{v}}_3^{鈯�}$in the equation ${\stackrel{鈫�}{u}}_3=\frac{1}{r_{33}}{\stackrel{鈫�}{v}}_3^{鈯�}$as follows:

$$\begin{gathered} {\stackrel{鈫�}{u}}_3=\frac{1}{r_{33}}{\stackrel{鈫�}{v}}_3^{鈯�} \\ {\stackrel{鈫�}{u}}_3=\frac{1}{6}\left[\begin{array}{c}2\\ -4\\ 4\end{array}\right] \\ {\stackrel{鈫�}{u}}_3=\left[\begin{array}{c}\frac{1}{3}\\ -\frac{2}{3}\\ \frac{2}{3}\end{array}\right] \end{gathered}$$

Therefore, the matrices $\mathrm{Q}=\left[\begin{array}{ccc}\frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{1}{3}& \frac{2}{3}& \frac{2}{3}\end{array}\right]\mathrm{and}R=\left[\begin{array}{ccc}3& 0& 12\\ 0& 3& -12\\ 0& 0& 6\end{array}\right]$

Hence, the QR factorization of the matrix is$\left[\begin{array}{ccc}2& -2& 18\\ 2& 1& 0\\ 1& 2& 0\end{array}\right]=\left[\begin{array}{ccc}\frac{2}{3}& -\frac{2}{3}& \frac{1}{3}\\ \frac{2}{3}& \frac{1}{3}& -\frac{2}{3}\\ \frac{1}{3}& \frac{2}{3}& \frac{2}{3}\end{array}\right]\left[\begin{array}{ccc}3& 0& 12\\ 0& 3& -12\\ 0& 0& 6\end{array}\right]$ .