91Ó°ÊÓ

Consider the matrix $M=\left[\begin{array}{ccc}2& 3& 5\\ 0& 4& 6\\ 0& 0& 7\end{array}\right]$and ${\stackrel{→}{v}}_1=\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right]$, ${\stackrel{→}{v}}_2=\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right]$and ${\stackrel{→}{v}}_3=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]$.

By the theorem of QR method, the value of and is defined as follows:

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

Simplify the equation $r_{11}=||{\stackrel{→}{v}}_1||$as follows:

$$\begin{gathered} r_{11}=||{\stackrel{→}{v}}_1|| \\ {r}_{11}\left|\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right]\right| \\ {r}_{11}=\sqrt{2^2+0^2+0^2} \end{gathered}$$

Substitute the values $2$for $r_{11}$and $\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right]$for${\stackrel{→}{v}}_1$ in the equation ${\stackrel{→}{u}}_1=\frac{1}{r_{11}}{\stackrel{→}{v}}_1$as follows:

$$\begin{gathered} {\stackrel{→}{u}}_1=\frac{1}{r_{11}}{\stackrel{→}{v}}_1 \\ {\stackrel{→}{u}}_1=\frac{1}{2}\left[\begin{array}{c}2\\ 0\\ 0\end{array}\right] \\ {\stackrel{→}{u}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \end{gathered}$$

Therefore, ${\stackrel{→}{u}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$and $r_{11}=2$.

As$r_{12}={\stackrel{→}{u}}_1.\stackrel{→}{V_2}$ , substitute the values$\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right]$ for $\stackrel{→}{V_2}$and ${\left[100\right]}^T$for${}_{{\stackrel{→}{u}}_1}$ in the equation $r_{11}={\stackrel{→}{u}}_1.\stackrel{→}{V_2}$as follows:

$$\begin{gathered} r_{11}={\stackrel{→}{u}}_1.{\stackrel{→}{V}}_2 \\ {r}_{12}={\left[100\right]}^T.\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right] \\ {r}_{12}=3 \end{gathered}$$

Substitute the values$\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right]$ for $\stackrel{→}{V_2}$, $3$for $r_{12}$and $\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$for${\stackrel{→}{u}}_1$ in the equation$\stackrel{→}{v}\stackrel{âŠ\yen }{2}={\stackrel{→}{v}}_2-r_{12}{\stackrel{→}{u}}_1$ as follows:

$$\begin{gathered} \stackrel{→}{v}\stackrel{âŠ\yen }{2}={\stackrel{→}{v}}_2-r_{12}{\stackrel{→}{u}}_1 \\ \stackrel{→}{v}\stackrel{âŠ\yen }{2}=\left[\begin{array}{c}3\\ 4\\ 0\end{array}\right]-3\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] \\ \stackrel{→}{v}\stackrel{âŠ\yen }{2}=\left[\begin{array}{c}0\\ 4\\ 0\end{array}\right] \end{gathered}$$

Therefore, $\stackrel{→}{v}\stackrel{âŠ\yen }{2}=\left[\begin{array}{c}0\\ 4\\ 0\end{array}\right]$and $r_{12}=3$.

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

$$\begin{gathered} r_{22}=||\stackrel{→}{v}\stackrel{âŠ\yen }{2}|| \\ {\stackrel{→}{u}}_2=\frac{1}{r_{22}}\stackrel{→}{v}\stackrel{âŠ\yen }{2} \end{gathered}$$

Simplify the equation $r_{22}=\stackrel{→}{v}\stackrel{âŠ\yen }{2}$as follows:

$$\begin{gathered} r_{22}=\stackrel{→}{v}\stackrel{âŠ\yen }{2} \\ {r}_{22}=\left|\left[\begin{array}{c}0\\ 4\\ 0\end{array}\right]\right| \\ {r}_{22}=\sqrt{{\left(0\right)}^2+{\left(4\right)}^2+{\left(0\right)}^2} \\ {r}_{22}=4 \end{gathered}$$

Substitute the values $4$for $r_{22}$and $\left[\begin{array}{c}0\\ 4\\ 0\end{array}\right]$for$\stackrel{→}{v}\stackrel{âŠ\yen }{2}$in the equation${\stackrel{→}{u}}_2=\frac{1}{r_{22}}\stackrel{→}{v}\stackrel{âŠ\yen }{2}$as follows:

$$\begin{gathered} {\stackrel{→}{u}}_2=\frac{1}{r_{22}}\stackrel{→}{v}\stackrel{âŠ\yen }{2} \\ {\stackrel{→}{u}}_2=\frac{1}{4}\left[\begin{array}{c}0\\ 4\\ 0\end{array}\right] \\ {\stackrel{→}{u}}_2=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right] \end{gathered}$$

Therefore, ${\stackrel{→}{u}}_2=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$and $r_{22}=4$.

As $r_{13}={\stackrel{→}{u}}_1-{\stackrel{→}{V}}_3$, substitute the values $\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]$for${\stackrel{→}{V}}_3$ and ${\left[100\right]}^T$for ${\stackrel{→}{u}}_1$in the equation $r_{13}={\stackrel{→}{u}}_1.{\stackrel{→}{V}}_3$as follows:

$$\begin{gathered} r_{13}={\stackrel{→}{u}}_1.{\stackrel{→}{V}}_3 \\ {r}_{13}={\left[100\right]}^T.\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right] \\ {r}_{13}=5 \end{gathered}$$

substitute the values $\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]$for ${\stackrel{→}{v}}_3$ and role="math" localid="1659444143779" ${\left[100\right]}^T$for ${\stackrel{→}{u}}_2$in the equationrole="math" localid="1659444309087" $r_{23}={\stackrel{→}{u}}_2.{\stackrel{→}{V}}_3$ as follows:

$$\begin{gathered} r_{23}={\stackrel{→}{u}}_2.{\stackrel{→}{V}}_3 \\ {r}_{23}={\left[010\right]}^T.\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right] \\ {r}_{23}=6 \end{gathered}$$

Substitute the values$\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]$ for ${\stackrel{→}{V}}_3$, $5$for $r_{13}$, $6$for $r_{23}$, $\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$for${\stackrel{→}{u}}_1$ and $\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$forrole="math" localid="1659444610118" ${\stackrel{→}{u}}_2$ in the equation $\stackrel{→}{v}\stackrel{âŠ\yen }{3}={\stackrel{→}{v}}_3-r_{13}{\stackrel{→}{u}}_1-{\stackrel{→}{u}}_{23}{\stackrel{→}{u}}_2$as follows:

$$\begin{gathered} \stackrel{→}{v}\stackrel{âŠ\yen }{3}={\stackrel{→}{v}}_3-r_{13}{\stackrel{→}{u}}_1-{\stackrel{→}{u}}_{23}{\stackrel{→}{u}}_2 \\ \stackrel{→}{v}\stackrel{âŠ\yen }{3}=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]-5\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]-6\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right] \\ \stackrel{→}{v}\stackrel{âŠ\yen }{3}=\left[\begin{array}{c}5\\ 6\\ 7\end{array}\right]-\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right] \\ \stackrel{→}{v}\stackrel{âŠ\yen }{3}=\left[\begin{array}{c}0\\ 0\\ 7\end{array}\right] \end{gathered}$$

Therefore, $\stackrel{→}{v}\stackrel{âŠ\yen }{3}=\left[\begin{array}{c}0\\ 0\\ 7\end{array}\right]$, role="math" localid="1659445041679" $r_{13}=5$and $r_{23}=6$.

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

$$\begin{gathered} r_{33}=||\stackrel{→}{v}\stackrel{âŠ\yen }{3}|| \\ {\stackrel{→}{u}}_3=\frac{1}{r_{33}}\stackrel{→}{v}\stackrel{âŠ\yen }{3} \end{gathered}$$

Simplify the equation $r_{33}=||\stackrel{→}{v}\stackrel{âŠ\yen }{3}||$as follows:

$$\begin{gathered} r_{33}=||\stackrel{→}{v}\stackrel{âŠ\yen }{3}|| \\ {r}_{33}=\left|\left[\begin{array}{c}0\\ 0\\ 7\end{array}\right]\right| \\ {r}_{33}=\sqrt{{\left(0\right)}^2+{\left(0\right)}^2+{\left(7\right)}^2} \\ {r}_{33}=7 \end{gathered}$$

Substitute the values $7$for $r_{33}$and $\left[\begin{array}{c}0\\ 0\\ 7\end{array}\right]$for$\stackrel{→}{v}\stackrel{âŠ\yen }{3}$ in the equation ${\stackrel{→}{u}}_3=\frac{1}{r_{33}}\stackrel{→}{v}\stackrel{âŠ\yen }{3}$as follows:

$$\begin{gathered} {\stackrel{→}{u}}_3=\frac{1}{r_{33}}\stackrel{→}{v}\stackrel{âŠ\yen }{3} \\ {\stackrel{→}{u}}_3=\frac{1}{7}\left[\begin{array}{c}0\\ 0\\ 7\end{array}\right] \\ {\stackrel{→}{u}}_3=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right] \end{gathered}$$

Therefore, the matrices $Q=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$and $R=\left[\begin{array}{ccc}2& 3& 5\\ 0& 4& 6\\ 0& 0& 7\end{array}\right]$.

Hence, the QR factorization of the matrix is$\left[\begin{array}{ccc}2& 3& 5\\ 0& 4& 6\\ 0& 0& 7\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}2& 3& 5\\ 0& 4& 6\\ 0& 0& 7\end{array}\right]$ .