91Ó°ÊÓ

First of all we will make the augmented matrix to the corresponding system of equations.

$\left|\begin{array}{c}{\mathrm{x}}_2+2{\mathrm{x}}_4+3{\mathrm{x}}_5=0\\ 4{\mathrm{x}}_4+8{\mathrm{x}}_5=0\end{array}\right|$

Corresponding Augmented matrix is written as:

role="math" localid="1659348859008" $\left[\begin{array}{ccccc}0& 1& 0& 2& 3\\ 0& 0& 0& 4& 8\end{array}\overline{)\begin{array}{c}0\\ 0\end{array}}\right]$

There are two linear equation in the augmented matric $\left[\begin{array}{ccccc}0& 1& 0& 2& 3\\ 0& 0& 0& 4& 8\end{array}\overline{)\begin{array}{c}0\\ 0\end{array}}\right]$which means there are three free variables. Apply the row operation role="math" localid="1659349072147" $R_2→R_2-4R_1\mathrm{in}\left[\begin{array}{ccccc}0& 1& 0& 2& 3\\ 0& 0& 0& 4& 8\end{array}\overline{)\begin{array}{c}0\\ 0\end{array}}\right]$as follows:

$\left[\begin{array}{ccccc}0& 1& 0& 2& 3\\ 0& 0& 0& 4& 8\end{array}\overline{)\begin{array}{c}0\\ 0\end{array}}\right]$

Apply the row operation in $R_1→R_1-2R_2\mathrm{in}\left[\begin{array}{ccccc}0& 1& 0& 2& 3\\ 0& 0& 0& 4& 8\end{array}\overline{)\begin{array}{c}0\\ 0\end{array}}\right]$as follows:

$\left[\begin{array}{ccccc}0& 1& 0& 0& -1\\ 0& 0& 0& 1& 2\end{array}\overline{)\begin{array}{c}0\\ 0\end{array}}\right]$

System of equation to the corresponding above matrix will be

$\left|\begin{array}{c}\begin{array}{c}{\mathrm{x}}_2-{\mathrm{x}}_5=0\\ {\mathrm{x}}_4+2{\mathrm{x}}_5=0\end{array}\end{array}\right|$

Simplify further as:

role="math" localid="1659349380833" $\left|\begin{array}{c}{\mathrm{x}}_2={\mathrm{x}}_5\\ x_4=-2x_5\end{array}\right|$

According to system $\left|\begin{array}{c}{\mathrm{x}}_2={\mathrm{x}}_5\\ x_4=-2x_5\end{array}\right|$, the variables ${\mathrm{x}}_{2,}{\mathrm{x}}_4$depend on ${\mathrm{x}}_3,{\mathrm{x}}_5,{\mathrm{x}}_1.\mathrm{So},{\mathrm{x}}_3,{\mathrm{x}}_5,{\mathrm{x}}_1$. So, are free variables.

For the given system of equation${\mathrm{x}}_{1,}{\mathrm{x}}_3,{\mathrm{x}}_5$are free variables and ${\mathrm{x}}_{2,}={\mathrm{x}}_5,{\mathrm{x}}_4=-2{\mathrm{x}}_5$.