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

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

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

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

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

Substitute the values 2 for $r_{11}$and $\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]$for ${\stackrel{鈫�}{v}}_1$in the equation role="math" localid="1660107365367" ${\stackrel{鈫�}{u}}_1=\frac{1}{r_{11}}{\stackrel{鈫�}{v}}_1$as follows.

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

Therefore, the values ${\stackrel{鈫�}{u}}_1=\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]\mathrm{and}{r}_{11}=2$.

As $r_{12}={\stackrel{鈫�}{u}}_1.{\stackrel{鈫�}{v}}_2$, substitute the values $\left[\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right]$for ${\stackrel{鈫�}{v}}_2$and $\left[\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right]$for ${\stackrel{鈫�}{u}}_1$in the equation role="math" localid="1660108053453" $r_{11}={\stackrel{鈫�}{u}}_1.{\stackrel{鈫�}{v}}_2$as follows.

role="math" localid="1660108155180" $$\begin{gathered} r_{12}={\stackrel{鈫�}{u}}_1.{\stackrel{鈫�}{v}}_2 \\ {r}_{12}=\left[\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right].\left[\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right] \\ {r}_{12}=\frac{1}{2}+\frac{1}{2} \\ {r}_{12}=1 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right]\mathrm{for}{\stackrel{鈫�}{\mathrm{v}}}_2,1\mathrm{for}{\mathrm{r}}_{12}\mathrm{and}\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]\mathrm{for}{\stackrel{鈫�}{\mathrm{u}}}_1$in the equation role="math" localid="1660108359838" ${\stackrel{鈫�}{v}}_2^{鈯�}={\stackrel{鈫�}{v}}_2-r_{12}{\stackrel{鈫�}{u}}_1$as follows.

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

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

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

role="math" localid="1660108863808" $$\begin{gathered} r_{22}=||{\stackrel{鈫�}{v}}_2^{鈯�}|| \\ {\stackrel{鈫�}{u}}_2=\frac{1}{r_{22}}{\stackrel{鈫�}{v}}_2^{鈯�} \end{gathered}$$

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

role="math" localid="1660111638709" $$\begin{gathered} {\stackrel{鈫�}{u}}_2=\frac{1}{r_{22}}{\stackrel{鈫�}{v}}_2^{鈯�} \\ {\stackrel{鈫�}{u}}_2=\left[\begin{array}{c}1/2\\ -1/2\\ -1/2\\ 1/2\end{array}\right] \\ {\stackrel{鈫�}{u}}_2=\left[\begin{array}{c}1/2\\ -1/2\\ -1/2\\ 1/2\end{array}\right] \end{gathered}$$

Substitute the values 7 for role="math" localid="1660111513038" $r_{22}$and role="math" localid="1660111523533" $\left[\begin{array}{c}1/2\\ 1/2\\ -1/2\\ -1/2\end{array}\right]\mathrm{for}{\stackrel{鈫�}{v}}_2^{鈯�}$in the equation role="math" localid="1660111540204" ${\stackrel{鈫�}{u}}_2=\frac{1}{r_{22}}{\stackrel{鈫�}{v}}_2^{鈯�}$as follows.

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

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

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

$$\begin{gathered} r_{13}={\stackrel{鈫�}{u}}_1.{\stackrel{鈫�}{v}}_3 \\ {r}_{13}=\left[\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right]\left[\begin{array}{c}0\\ 2\\ 1\\ -1\end{array}\right] \\ {r}_{13}=+1+\frac{1}{2}-\frac{1}{2} \\ {r}_{13}=1 \end{gathered}$$

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

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

Substitute the values $\left[\begin{array}{c}0\\ 2\\ 1\\ -1\end{array}\right]\mathrm{for}{\stackrel{鈫�}{v}}_3,1\mathrm{for}{r}_{13},-2\mathrm{for}{r}_{23},\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]\mathrm{for}{\stackrel{鈫�}{u}}_1\mathrm{and}\left[\begin{array}{c}1/2\\ -1/2\\ -1/2\\ 1/2\end{array}\right]\mathrm{for}{\stackrel{鈫�}{u}}_2$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.

$$\begin{gathered} {\stackrel{鈫�}{v}}_3^{鈯�}={\stackrel{鈫�}{v}}_3-r_{13}{\stackrel{鈫�}{u}}_1-r_{23}{\stackrel{鈫�}{u}}_2 \\ {\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}0\\ 2\\ 1\\ -1\end{array}\right]-1\left[\begin{array}{c}1/2\\ 1/2\\ 1/2\\ 1/2\end{array}\right]+2\left[\begin{array}{c}1/2\\ -1/2\\ -1/2\\ 1/2\end{array}\right] \\ {\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}-1/2\\ 3/2\\ 1/2\\ -3/2\end{array}\right]+\left[\begin{array}{c}1\\ -1\\ -1\\ 1\end{array}\right] \\ {\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}1/2\\ 1/2\\ -1/2\\ -1/2\end{array}\right] \end{gathered}$$

Therefore, the values ${\stackrel{鈫�}{v}}_3^{鈯�}=\left[\begin{array}{c}1/2\\ 1/2\\ -1/2\\ -1/2\end{array}\right],r_{13}=1\mathrm{and}{r}_{23}=2$.

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

localid="1660113997835" $$\begin{gathered} r_{33}=||{\stackrel{鈫�}{v}}_3^{鈯�}|| \\ {\stackrel{鈫�}{u}}_3=\frac{1}{r_{33}}{\stackrel{鈫�}{v}}_3^{鈯�} \end{gathered}$$

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

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

Substitute the values 7 for $r_{33}$and $\left[\begin{array}{c}1/2\\ 1/2\\ -1/2\\ -1/2\end{array}\right]$for ${\stackrel{鈫�}{v}}_3^{鈯�}$in the equation localid="1660113500881" ${\stackrel{鈫�}{u}}_3=\frac{1}{r_{33}}{\stackrel{鈫�}{v}}_3^{鈯�}$as follows.

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

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

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