Mo logo [home] [lexicon] [problems] [tests] [courses] [auxiliaries] [notes] [staff] german flag
Mathematics-Online course: Linear Algebra - Linear Systems of Equations - Classification and General Structure

Solvability of a linear system of equations

[previous page] [next page] [table of contents][page overview]
The solution set of a homogeneous LSE

$\displaystyle Ax = 0 $

with $ m\times n$-coefficient matrix $ A$ forms a subspace $ U$ of $ K^n$.

If the inhomogeneous LSE

$\displaystyle Ax = b $

has a solution $ v$, then for the general solution we have

$\displaystyle x \in v + U\, , $

that is, the solution set is an affine subspace of $ K^n$. In particular, an inhomogeneous LSE can have either no solution, or exactly one solution ($ U=\{0\}$), or an infinite number of solutions ( $ \operatorname{dim} U>0$).
(Authors: App/Burkhardt/Höllig/Kimmerle)
Let $ x_1$, $ x_2$ be solutions of the homogeneous LSE and let $ \lambda\in K$, then

$\displaystyle A(x_1-x_2)$ $\displaystyle = Ax_1 -Ax_2 = 0 -0 =0$    

and


$\displaystyle A(\lambda x_1)$ $\displaystyle = \lambda Ax_1 = \lambda 0 = 0\,,$    

that is, the set of solutions of the homogeneous LSE forms a subspace $ U$.

If $ v$ is a solution of the inhomogeneous LSE and if $ u\in U$, then $ x=v+u$ is also a solution of the inhomogeneous LSE since

$\displaystyle Ax = A(v+u) = Av +Au = b+0 =b\,. $

Conversely, if $ v$ and $ w$ are solutions of the inhomogeneous LSE, then their difference $ v-w$ is a solution of the homogeneous LSE since

$\displaystyle A(v-w) = Av -Aw =b -b =0\,. $

Thus, we obtain all solutions of a (solvable) inhomogeneous LSE by taking an arbitrary solution of the inhomogeneous LSE and then adding to it all solutions of the homogeneous LSE.

(Authors: App/Burkhardt/Höllig)
  1. The LSE

    \begin{displaymath} \left( \begin{array}{cc} 3 & 5 \\ 7 & 4 \end{array} \... ... = \left( \begin{array}{c} 29\\ 60 \end{array} \right) \end{displaymath}

    has the unique solution $ x_1=8$, $ x_2=1$.
  2. The LSE

    \begin{displaymath} \left( \begin{array}{ccc} 1 & -2 & 4 \\ 3 & -2 & 4 \en... ...) = \left( \begin{array}{c} 2\\ 6 \end{array} \right) \end{displaymath}

    has no unique solution. Choosing an arbitrary $ x_3=t$ we obtain the solution $ x_2=2t$, $ x_1=2$.
  3. The LSE

    \begin{displaymath} \left( \begin{array}{cc} 1 & 1 \\ 3 & 1 \\ 0 & 2 \en... ...\left( \begin{array}{c} 6\\ 7\\ 8 \end{array} \right) \end{displaymath}

    has no solution. The last row yields $ x_2=4$. Inserting this into the other rows we obtain from the first row $ x_1=2$. From the second row, however, we obtain $ x_1=1$.
(Authors: App/Burkhardt/Höllig)
  automatically generated 4/21/2005