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:

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

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$\left\{\begin{array}{c}3\mathrm{x}+\mathrm{y}+3\mathrm{z}++\mathrm{w}=0\\ 4\mathrm{x}-7\mathrm{y}-3\mathrm{z}+5\mathrm{W}=0\\ \mathrm{x}+3++4\mathrm{z}-3\mathrm{w}=0\\ 3\mathrm{x}+2++\mathrm{w}=0\end{array}\right.$

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

Convert the given equations into the matrix form.

$\left[\begin{array}{ccccc}3& 1& 3& 6& 0\\ 4& -7& -3& 5& 0\\ 1& 3& 4& -3& 0\\ 3& 0& 2& 7& 0\end{array}\right]$

Divide row 1 by 3:${\mathrm{R}}_1=\frac{{\mathrm{R}}_1}{3}$.

$\left[\begin{array}{ccccc}1& \frac{1}{3}& 1& 2& 0\\ 4& -7& -3& 5& 0\\ 1& 3& 4& -3& 0\\ 3& 0& 2& 7& 0\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2:${\mathrm{R}}_2={\mathrm{R}}_2-4{\mathrm{R}}_1$.

localid="1659025081392" $\left[\begin{array}{ccccc}1& \frac{1}{3}& 1& 2& 0\\ 0& -\frac{25}{3}& -7& -3& 0\\ 1& 3& 4& -3& 0\\ 3& 0& 2& 7& 0\end{array}\right]$

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

localid="1659025174183" $\left[\begin{array}{ccccc}1& \frac{1}{3}& 1& 2& 0\\ 0& -\frac{25}{3}& -7& -3& 0\\ 0& \frac{8}{3}& 3& -5& 0\\ 3& 0& 2& 7& 0\end{array}\right]$

Subtract row 1 multiplied by 3 from row 4:${\mathrm{R}}_4={\mathrm{R}}_4-3{\mathrm{R}}_1.$

$\left[\begin{array}{ccccc}1& \frac{1}{3}& 1& 2& 0\\ 0& -\frac{25}{3}& -7& -3& 0\\ 0& \frac{8}{3}& 3& -5& 0\\ 3& -1& -1& 1& 0\end{array}\right]$

Multiply row 2 by $-\frac{3}{25}$:${\mathrm{R}}_2=-\frac{3{\mathrm{R}}_2}{25}$.

$\left[\begin{array}{ccccc}1& 0& \frac{18}{25}& \frac{47}{25}& 0\\ 0& 1& \frac{21}{25}& \frac{9}{25}& 0\\ 0& \frac{8}{3}& 3& -5& 0\\ 3& -1& -1& 1& 0\end{array}\right]$

Subtract row 2 multiplied by $\frac{1}{3}$from row 1:${\mathrm{R}}_1={\mathrm{R}}_1-\frac{{\mathrm{R}}_2}{3}$.

$\left[\begin{array}{ccccc}1& 0& \frac{18}{25}& \frac{47}{25}& 0\\ 0& 1& \frac{21}{25}& \frac{9}{25}& 0\\ 0& \frac{8}{3}& 3& -5& 0\\ 0& -1& -1& 1& 0\end{array}\right]$

Subtract row 2 multiplied by $\frac{8}{3}$from row 3:${\mathrm{R}}_3={\mathrm{R}}_3-\frac{8{\mathrm{R}}_2}{3}$

$\left[\begin{array}{ccccc}1& 0& \frac{18}{25}& \frac{47}{25}& 0\\ 0& 1& \frac{21}{25}& \frac{9}{25}& 0\\ 0& 0& \frac{19}{25}& -\frac{149}{25}& 0\\ 0& -1& -1& 1& 0\end{array}\right]$

Add row 2 to row 4:${\mathrm{R}}_4={\mathrm{R}}_4+{\mathrm{R}}_2$.

$\left[\begin{array}{ccccc}1& 0& \frac{18}{25}& \frac{47}{25}& 0\\ 0& 1& \frac{21}{25}& \frac{9}{25}& 0\\ 0& 0& \frac{19}{25}& -\frac{149}{25}& 0\\ 0& 0& -\frac{4}{25}& \frac{34}{25}& 0\end{array}\right]$

Multiply row 3 by $\frac{25}{19}$:${\mathrm{R}}_3=\frac{25{\mathrm{R}}_3}{19}$.

$\left[\begin{array}{ccccc}1& 0& \frac{18}{25}& \frac{47}{25}& 0\\ 0& 1& \frac{21}{25}& \frac{9}{25}& 0\\ 0& 0& 1& -\frac{149}{25}& 0\\ 0& 0& -\frac{4}{25}& \frac{34}{25}& 0\end{array}\right]$

Subtract row 3 multiplied by $\frac{18}{25}$from row 1:${\mathrm{R}}_1={\mathrm{R}}_1-\frac{18{\mathrm{R}}_3}{25}$.

$\left[\begin{array}{ccccc}1& 0& 0& \frac{143}{19}& 0\\ 0& 1& \frac{21}{25}& \frac{9}{25}& 0\\ 0& 0& 1& -\frac{149}{19}& 0\\ 0& 0& -\frac{4}{25}& \frac{34}{25}& 0\end{array}\right]$

Subtract row 3 multiplied by $\frac{21}{25}$from row 2:${\mathbf{R}}_{\mathbf{2}}\mathbf{=}{\mathbf{R}}_{\mathbf{2}}\mathbf{-}\frac{\mathbf{21}{\mathbf{R}}_{\mathbf{3}}}{\mathbf{25}}$.

$\left[\begin{array}{ccccc}1& 0& 0& \frac{143}{19}& 0\\ 0& 1& 0& \frac{132}{19}& 0\\ 0& 0& 1& -\frac{149}{25}& 0\\ 0& 0& -\frac{4}{25}& \frac{34}{25}& 0\end{array}\right]$

Add row 3 multiplied by $\frac{4}{25}$to row 4:${\mathrm{R}}_4={\mathrm{R}}_4+\frac{4{\mathrm{R}}_3}{25}$.

$\left[\begin{array}{ccccc}1& 0& 0& \frac{143}{19}& 0\\ 0& 1& 0& \frac{132}{19}& 0\\ 0& 0& 1& -\frac{149}{19}& 0\\ 0& 0& 0& \frac{2}{19}& 0\end{array}\right]$

Multiply row 4 by $\frac{19}{2}$: ${\mathrm{R}}_4=\frac{19{\mathrm{R}}_4}{2}$.

$\left[\begin{array}{ccccc}1& 0& 0& \frac{143}{19}& 0\\ 0& 1& 0& \frac{132}{19}& 0\\ 0& 0& 1& \frac{149}{19}& 0\\ 0& 0& 0& 1& 0\end{array}\right]$

Subtract row 4 multiplied by $\frac{143}{19}$from row 1:${\mathrm{R}}_1={\mathrm{R}}_1-\frac{143{\mathrm{R}}_4}{19}$.

$\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 1& 0& \frac{132}{19}& 0\\ 0& 0& 1& -\frac{149}{19}& 0\\ 0& 0& 0& 1& 0\end{array}\right]$

Subtract row 4 multiplied by $\frac{132}{19}$from row 2:${\mathrm{R}}_2={\mathrm{R}}_2-\frac{132{\mathrm{R}}_4}{19}$.

$\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 0& 1& -\frac{149}{19}& 0\\ 0& 0& 0& 1& 0\end{array}\right]$

Add row 4 multiplied by $\frac{149}{19}$to row 3: ${\mathrm{R}}_3={\mathrm{R}}_3+\frac{149{\mathrm{R}}_4}{19}$.

$\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\end{array}\right]$

Therefore,$x\mathit{=}\mathit{0}\mathit{,}y\mathit{=}\mathit{0}\mathit{,}z\mathit{=}\mathit{0}\mathit{,}\mathit{ }and\mathit{ }w\mathit{=}\mathit{0}$are the sets of given homogeneous equations.