Linear inhomogeneous system/Elimination/2x+5y+2z-v is 3, 3x-4y+u+2v is 1, 4x -2z+2u is 7/Example

We want to solve the inhomogeneous linear system

${\begin{matrix}2x&+5y&+2z&&-v&=&3\\3x&-4y&&+u&+2v&=&1\\4x&&-2z&+2u&&=&7\,\end{matrix}}$

over ${}\mathbb {R}$ (or over ${}\mathbb {Q}$). Firstly, we eliminate ${}x$ by keeping the first row ${}I$, replacing the second row ${}II$ by ${}II-{\frac {3}{2}}I$, and replacing the third row ${}III$ by ${}III-2I$. This yields

${\begin{matrix}2x&+5y&+2z&&-v&=&3\\&-{\frac {23}{2}}y&-3z&+u&+{\frac {7}{2}}v&=&{\frac {-7}{2}}\\&-10y&-6z&+2u&+2v&=&1\,.\end{matrix}}$

Now, we can eliminate ${}y$ from the (new) third row, with the help of the second row. Because of the fractions, we rather eliminate ${}z$ (which eliminates also ${}u$). We leave the first and the second row as they are, and we replace the third row ${}III$ by ${}III-2II$. This yields the system, in a new ordering of the variables,

${\begin{matrix}2x&+2z&&+5y&-v&=&3\\&-3z&+u&-{\frac {23}{2}}y&+{\frac {7}{2}}v&=&{\frac {-7}{2}}\\&&&13y&-5v&=&8\,.\end{matrix}}$

Now we can choose an arbitrary (free) value for ${}v$. The third row determines then ${}y$ uniquely, we must have

${}y={\frac {8}{13}}+{\frac {5}{13}}v\,.$

In the second equation, we can choose ${}u$ arbitrarily, this determines ${}z$ via

${}{\begin{aligned}z&=-{\frac {1}{3}}{\left(-{\frac {7}{2}}-u-{\frac {7}{2}}v+{\frac {23}{2}}{\left({\frac {8}{13}}+{\frac {5}{13}}v\right)}\right)}\\&=-{\frac {1}{3}}{\left(-{\frac {7}{2}}-u-{\frac {7}{2}}v+{\frac {92}{13}}+{\frac {115}{26}}v\right)}\\&=-{\frac {1}{3}}{\left({\frac {93}{26}}-u+{\frac {12}{13}}v\right)}\\&=-{\frac {31}{26}}+{\frac {1}{3}}u-{\frac {4}{13}}v.\end{aligned}}$

The first row determines ${}x$, namely

${}{\begin{aligned}x&={\frac {1}{2}}{\left(3-2z-5y+v\right)}\\&={\frac {1}{2}}{\left(3-2{\left(-{\frac {31}{26}}+{\frac {1}{3}}u-{\frac {4}{13}}v\right)}-5{\left({\frac {8}{13}}+{\frac {5}{13}}v\right)}+v\right)}\\&={\frac {1}{2}}{\left({\frac {30}{13}}-{\frac {2}{3}}u-{\frac {4}{13}}v\right)}\\&={\frac {15}{13}}-{\frac {1}{3}}u-{\frac {2}{13}}v.\end{aligned}}$

Hence, the solution set is

${\left\{{\left({\frac {15}{13}}-{\frac {1}{3}}u-{\frac {2}{13}}v,{\frac {8}{13}}+{\frac {5}{13}}v,-{\frac {31}{26}}+{\frac {1}{3}}u-{\frac {4}{13}}v,u,v\right)}\mid u,v\in \mathbb {R} \right\}}.$

A particularly simple solution is obtained by equating the free variables ${}u$ and ${}v$ with ${}0$. This yields the special solution

${}(x,y,z,u,v)=\left({\frac {15}{13}},\,{\frac {8}{13}},\,-{\frac {31}{26}},\,0,\,0\right)\,.$

The general solution set can also be written as

${\left\{{\left({\frac {15}{13}},{\frac {8}{13}},-{\frac {31}{26}},0,0\right)}+u{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}+v{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}\mid u,v\in \mathbb {R} \right\}}.$

Here,

${\left\{u{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}+v{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}\mid u,v\in \mathbb {R} \right\}}$

is a description of the general solution of the corresponding homogeneous linear system.