91Ó°ÊÓ

Consider the matrix $M=\left[\begin{array}{cc}6& 2\\ 3& -6\\ 2& 3\end{array}\right]$where ${\stackrel{→}{v}}_1=\left[\begin{array}{c}6\\ -3\\ 2\end{array}\right]$and ${\stackrel{→}{v}}_2=\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$.

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

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

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

$$\begin{gathered} r_{11}=\left|\left|{\stackrel{→}{v}}_1\right|\right| \\ {r}_{11}=\left|\left|\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right]\right|\right| \\ {r}_{11}=\sqrt{6^2+3^2+2^2} \\ {r}_{11}=7 \end{gathered}$$

Substitute the values 7 for $r_{11}$and $\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right]$for in the equation ${\stackrel{↔}{u}}_1=\frac{1}{r_{11}}{\stackrel{→}{v}}_1$as follows.

localid="1659440988937" $$\begin{gathered} {\stackrel{→}{u}}_1=\frac{1}{r_{11}}{\stackrel{→}{v}}_1 \\ {\stackrel{→}{u}}_1=\frac{1}{7}\left[\begin{array}{c}6\\ 3\\ 2\end{array}\right] \\ {\stackrel{→}{u}}_1=\left[\begin{array}{c}6/7\\ 3/7\\ 2/7\end{array}\right] \end{gathered}$$

Therefore, the values ${\stackrel{→}{u}}_1=\left[\begin{array}{c}6/7\\ 3/7\\ 2/7\end{array}\right]$and $r_{11}=7$.

As $r_{12}={\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2$, substitute the values $\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$for ${\stackrel{→}{v}}_2$and $\left[\begin{array}{ccc}\frac{6}{7}& \frac{3}{7}& \frac{2}{7}\end{array}\right]$for ${\stackrel{→}{u}}_1$in the equation $r_{11}={\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2$as follows:

$$\begin{gathered} r_{12}={\stackrel{→}{u}}_1.{\stackrel{→}{v}}_2 \\ {r}_{12}={\left[\begin{array}{ccc}\frac{6}{7}& \frac{3}{7}& \frac{2}{7}\end{array}\right]}^T-\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right] \\ {r}_{12}=\frac{12}{7}-\frac{18}{7}+\frac{6}{7} \\ {r}_{12}=0 \end{gathered}$$

Substitute the values $\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$for ${\stackrel{→}{v}}_2$, 0 for $r_{12}$and $\left[\begin{array}{c}6/7\\ 3/7\\ 2/7\end{array}\right]$for ${\stackrel{→}{u}}_1$in the equation ${\stackrel{→}{v}}_2^{âŠ\yen }={\stackrel{→}{v}}_2-r_{12}{\stackrel{→}{u}}_1$as follows:

$$\begin{gathered} {\stackrel{→}{v}}_2^{âŠ\yen }={\stackrel{→}{v}}_2-r_{12}{\stackrel{→}{u}}_1 \\ {\stackrel{→}{v}}_2^{âŠ\yen }=\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]-0\left[\begin{array}{c}6/7\\ 3/7\\ 2/7\end{array}\right] \\ {\stackrel{→}{v}}_2^{âŠ\yen }=\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right] \end{gathered}$$

Therefore, the values ${\stackrel{→}{v}}_2^{âŠ\yen }=\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$and $r_{12}=0$.

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

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

Simplify the equation $r_{22}=\left|\left|{\stackrel{→}{v}}_2^{âŠ\yen }\right|\right|$as follows:

$$\begin{gathered} r_{22}=\left|\left|{\stackrel{→}{v}}_2^{âŠ\yen }\right|\right| \\ {r}_{22}=\left|\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]\right| \\ {r}_{22}=\sqrt{{\left(2\right)}^2+{\left(-6\right)}^2+{\left(3\right)}^2} \\ {r}_{22}=7 \end{gathered}$$

Substitute the values 7 for $r_{22}$and $\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right]$for${\stackrel{→}{v}}_2^{âŠ\yen }$ in the equation ${\stackrel{→}{u}}_2=\frac{1}{r_{22}}{\stackrel{→}{v}}_2^{âŠ\yen }$as follows:

$$\begin{gathered} {\stackrel{→}{u}}_2=\frac{1}{r_{22}}{\stackrel{→}{v}}_2^{âŠ\yen } \\ {\stackrel{→}{u}}_2=\frac{1}{7}\left[\begin{array}{c}2\\ -6\\ 3\end{array}\right] \\ {\stackrel{→}{u}}_2=\left[\begin{array}{c}2/7\\ -6/7\\ 3/7\end{array}\right] \end{gathered}$$

The values ${\stackrel{→}{u}}_2=\left[\begin{array}{c}2/7\\ -6/7\\ 3/7\end{array}\right]$and $r_{22}=7$.

Therefore, the matrices $Q=\left[\begin{array}{cc}6/7& 2/7\\ 3/7& -6/7\\ 2/7& 3/7\end{array}\right]$and $R=\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$.

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