91Ó°ÊÓ

Two vectors are perpendicular to each other if their dot product is zero,

Let us suppose that$\stackrel{\mathbf{→}}{\mathbf{x}}$and $\stackrel{\mathbf{→}}{\mathbf{y}}$are two perpendicular then these two vectors are said to be perpendicular, if

$\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{y}}\mathbf{=}\mathbf{0}$

$\stackrel{→}{\mathrm{b}}={\mathrm{x}}_1\stackrel{→}{{\mathrm{v}}_1}+{\mathrm{x}}_2\stackrel{→}{{\mathrm{v}}_2}+{\mathrm{x}}_3\stackrel{→}{{\mathrm{v}}_3}$,

$\stackrel{→}{\mathrm{b}}=\left[\begin{array}{c}-8\\ -1\\ 2\\ 15\end{array}\right]$, $\stackrel{→}{{\mathrm{v}}_1}=\left[\begin{array}{c}1\\ 4\\ 7\\ 5\end{array}\right]$, $\stackrel{→}{{\mathrm{v}}_2}=\left[\begin{array}{c}2\\ 5\\ 8\\ 3\end{array}\right]$,$\stackrel{→}{{\mathrm{v}}_3}=\left[\begin{array}{c}4\\ 6\\ 9\\ 1\end{array}\right]$

Consider the equation,

$\stackrel{→}{\mathrm{b}}={\mathrm{x}}_1\stackrel{→}{{\mathrm{v}}_1}+{\mathrm{x}}_2\stackrel{→}{{\mathrm{v}}_2}+{\mathrm{x}}_3\stackrel{→}{{\mathrm{v}}_3}$

Substitute the respective given values,

$\left[\begin{array}{c}-8\\ -1\\ 2\\ 15\end{array}\right]={\mathrm{x}}_1\left[\begin{array}{c}1\\ 4\\ 7\\ 5\end{array}\right]+{\mathrm{x}}_2\left[\begin{array}{c}2\\ 5\\ 8\\ 3\end{array}\right]+{\mathrm{x}}_3\left[\begin{array}{c}4\\ 6\\ 9\\ 1\end{array}\right]$

Now the matrix form of the above equation is,

$\left[\begin{array}{ccc}1& 2& 4\\ 4& 5& 6\\ 7& 8& 9\\ 5& 3& 1\end{array}\right]\left[\begin{array}{c}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ {\mathrm{x}}_3\end{array}\right]=\left[\begin{array}{c}-8\\ -1\\ 2\\ 15\end{array}\right]$

Augmented form of the above matrix is,

$\left[\begin{array}{cccc}1& 2& 4& |-8\\ 4& 5& 6& |-1\\ 7& 8& 9& |2\\ 5& 3& 1& |15\end{array}\right]$

Now use Gauss-Jordan elimination method to perform the elimination on the above matrix,

Use operation, ${\mathrm{R}}_2→{\mathrm{R}}_2-4{\mathrm{R}}_1$,${\mathrm{R}}_3→{\mathrm{R}}_3-7{\mathrm{R}}_1$and${\mathrm{R}}_4→{\mathrm{R}}_4-5{\mathrm{R}}_1$

$\left[\begin{array}{cccc}1& 2& 4& -8\\ 0& -3& -10& 31\\ 0& -6& -19& 58\\ 0& -7& -19& 55\end{array}\right]$

Use operation,${\mathrm{R}}_2→-\frac{{\mathrm{R}}_2}{3}$

$\left[\begin{array}{cccc}1& 2& 4& -8\\ 0& 1& \frac{10}{3}& \frac{-31}{3}\\ 0& -6& -19& 58\\ 0& -7& -19& 55\end{array}\right]$

Use operation, ${\mathrm{R}}_1→{\mathrm{R}}_1-2{\mathrm{R}}_2$,${\mathrm{R}}_3→{\mathrm{R}}_3+6{\mathrm{R}}_2$and${\mathrm{R}}_4→{\mathrm{R}}_4+7{\mathrm{R}}_2$

$\left[\begin{array}{cccc}1& 0& \frac{-8}{3}& \frac{38}{3}\\ 0& 1& \frac{10}{3}& \frac{-31}{3}\\ 0& 0& 1& -4\\ 0& 0& \frac{13}{3}& \frac{-52}{3}\end{array}\right]$

Use operation, ${\mathrm{R}}_1→{\mathrm{R}}_1+\frac{8}{3}{\mathrm{R}}_3$,${\mathrm{R}}_3→{\mathrm{R}}_2-\frac{10}{3}{\mathrm{R}}_3$and${\mathrm{R}}_4→{\mathrm{R}}_4-\frac{13}{3}{\mathrm{R}}_3$

$\left[\begin{array}{cccc}1& 0& 0& 2\\ 0& 1& 0& 3\\ 0& 0& 1& -4\\ 0& 0& 0& 0\end{array}\right]$

So, the system of equation is,

$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}2\\ 3\\ -4\\ 0\end{array}\right]$

Solution of the above system is,

$$\begin{gathered} {\mathrm{x}}_1=2 \\ {\mathrm{x}}_2=3 \\ {\mathrm{x}}_3=-4 \end{gathered}$$

Thus the solution of the system of equation $\stackrel{→}{\mathrm{b}}={\mathrm{x}}_1\stackrel{→}{{\mathrm{v}}_1}+{\mathrm{x}}_2\stackrel{→}{{\mathrm{v}}_2}+{\mathrm{x}}_3\stackrel{→}{{\mathrm{v}}_3}$is, ${\mathrm{x}}_1=2$,${\mathrm{x}}_2=3$ and ${\mathrm{x}}_3=-4$.