91Ó°ÊÓ

A linear system of equations with no constant terms is called a homogeneous system of linear equations. A homogeneous linear system, in other words, has the following form:

$\begin{array}{l}{\mathbf{a}}_{\mathbf{11}}\mathbf{,}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{a}}_{\mathbf{12}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{a}}_{\mathbf{1}\mathbf{n}}{\mathbf{x}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\\ {\mathbf{a}}_{\mathbf{21}}\mathbf{,}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{a}}_{\mathbf{22}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{a}}_{\mathbf{2}\mathbf{n}}{\mathbf{x}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\\ \mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\\ {\mathbf{a}}_{\mathbf{m}\mathbf{1}}\mathbf{,}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{a}}_{\mathbf{m}\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{a}}_{\mathbf{mn}}{\mathbf{x}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\end{array}$

For examples:

$\{\begin{array}{l}2x-5y=0\\ x-2y=0\end{array}$ is a homogeneous system in two variables.

$\{\begin{array}{l}x+y+z=0\\ y-Z=0\\ x+2y=0\end{array}$ is a homogeneous system in three variables.

The given Homogeneous equations are $\left\{\begin{array}{c}x-2y+3z=0\\ x+4y-6z=0\\ 2x+2y-3z=0\end{array}\right.$.

Find the sets of homogeneous equations with the help of the row reduction method.

Convert the given equations into the matrix form.

$\left[\begin{array}{cccc}1& -2& 3& 0\\ 1& 4& -6& 0\\ 2& 2& -3& 0\end{array}\right]$

Subtract row 1 from row 2:$R_2=R_2-R_1$.

localid="1659009605589" $\left[\begin{array}{cccc}1& -2& 3& 0\\ 1& 6& -9& 0\\ 2& 2& -3& 0\end{array}\right]$

Subtract row 1 multiplied by 2 from row 3:${\mathrm{R}}_3={\mathrm{R}}_3-2{\mathrm{R}}_1$.

$\left[\begin{array}{cccc}1& -2& 3& 0\\ 0& 6& -9& 0\\ 0& 6& -9& 0\end{array}\right]$

Divide row 2 by 6: ${\mathrm{R}}_2=\frac{{\mathrm{R}}_2}{6}$.

$\left[\begin{array}{cccc}1& -2& 3& 0\\ 0& 1& -\frac{3}{2}& 0\\ 0& 6& -9& 0\end{array}\right]$

Add row 2 multiplied by 2 to row 1:${\mathrm{R}}_1={\mathrm{R}}_1+2{\mathrm{R}}_2$.

$\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& -\frac{3}{2}& 0\\ 0& 6& -9& 0\end{array}\right]$

Subtract row 2 multiplied by 6 from row 3:${\mathrm{R}}_3={\mathrm{R}}_3-6{\mathrm{R}}_2$.

$$\begin{gathered} \left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& -\frac{3}{2}& 0\\ 0& 0& 0& 0\end{array}\right] \\ \mathrm{Therefore},\mathrm{x}=0,\mathrm{y}=\frac{3}{2}\mathrm{z}\mathrm{are}\mathrm{the}\mathrm{sets}\mathrm{of}\mathrm{given}\mathrm{homogeneous}\mathrm{equations}. \end{gathered}$$