Linear Algebra/Describing The Solution Set - Wikibooks

This system has many solutions because in echelon form

2 x + z = 3 x − y − z = 1 3 x − y = 4 → − ( 3 2 ) ρ 1 + ρ 3 − ( 1 2 ) ρ 1 + ρ 2 2 x + z = 3 − y − ( 3 2 ) z = − 1 2 − y − ( 3 2 ) z = − 1 2 → − ρ 2 + ρ 3 2 x + z = 3 − y − ( 3 2 ) z = − 1 2 0 = 0 {\displaystyle {\begin{array}{rcl}{\begin{array}{*{3}{rc}r}2x&&&+&z&=&3\\x&-&y&-&z&=&1\\3x&-&y&&&=&4\end{array}}&{\xrightarrow[{-\left({\frac {3}{2}}\right)\rho _{1}+\rho _{3}}]{-\left({\frac {1}{2}}\right)\rho _{1}+\rho _{2}}}&{\begin{array}{*{3}{rc}r}2x&&&+&z&=&3\\&&-y&-&\left({\frac {3}{2}}\right)z&=&-{\frac {1}{2}}\\&&-y&-&\left({\frac {3}{2}}\right)z&=&-{\frac {1}{2}}\end{array}}\\[3em]&{\xrightarrow[{}]{-\rho _{2}+\rho _{3}}}&{\begin{array}{*{3}{rc}r}2x&&&+&z&=&3\\&&-y&-&\left({\frac {3}{2}}\right)z&=&-{\frac {1}{2}}\\&&&&0&=&0\end{array}}\end{array}}}

not all of the variables are leading variables. The Gauss' method theorem showed that a triple satisfies the first system if and only if it satisfies the third. Thus, the solution set { ( x , y , z ) | 2 x + z = 3  and  x − y − z = 1  and  3 x − y = 4 } {\displaystyle {\Big \{}(x,y,z){\Big |}2x+z=3{\text{ and }}x-y-z=1{\text{ and }}3x-y=4{\Big \}}} can also be described as { ( x , y , z ) | 2 x + z = 3  and  − y − 3 z 2 = − 1 2 } {\displaystyle \left\{(x,y,z){\Big |}2x+z=3{\text{ and }}-y-{\frac {3z}{2}}=-{\frac {1}{2}}\right\}} . However, this second description is not much of an improvement. It has two equations instead of three, but it still involves some hard-to-understand interaction among the variables.

To get a description that is free of any such interaction, we take the variable that does not lead any equation, z {\displaystyle z} , and use it to describe the variables that do lead, x {\displaystyle x} and y {\displaystyle y} . The second equation gives y = 1 2 − 3 2 z {\displaystyle y={\frac {1}{2}}-{\frac {3}{2}}z} and the first equation gives x = 3 2 − 1 2 z {\displaystyle x={\frac {3}{2}}-{\frac {1}{2}}z} . Thus, the solution set can be described as { ( x , y , z ) = ( 3 2 − 1 2 z , 1 2 − 3 2 z , z ) | z ∈ R } {\displaystyle \left\{(x,y,z)=\left({\frac {3}{2}}-{\frac {1}{2}}z,{\frac {1}{2}}-{\frac {3}{2}}z,z\right){\Big |}z\in \mathbb {R} \right\}} . For instance, ( 1 2 , − 5 2 , 2 ) {\displaystyle \left({\frac {1}{2}},-{\frac {5}{2}},2\right)} is a solution because taking z = 2 {\displaystyle z=2} gives a first component of 1 2 {\displaystyle {\frac {1}{2}}} and a second component of − 5 2 {\displaystyle -{\frac {5}{2}}} .

The advantage of this description over the ones above is that the only variable appearing, z {\displaystyle z} , is unrestricted — it can be any real number.