0
$\begingroup$

I have something like this:

$2x + y = 160$

$x + 2y = 170$

$x + z = 95$

$z = 45$

$2z + y = 150$

I'm trying to use gauss elimination method, but exacly I've matrix form $5$ x $3$, where for example this algorihm using square matrix:

https://martin-thoma.com/solving-linear-equations-with-gaussian-elimination/

  • 0
    This is a system of five equations for three unknowns. That means that either two of the equations can be obtained through linear combinations of the other three and are thus redundant - leaving you with three equations for three unknowns and thus a square matrix again - or the system has no solution. Can you figure out which of these cases holds here? – 2017-02-14
  • 0
    $z=45$ and $x+z=95$, so $x = 50$, and $2z+y=150$, so $y=60$. – 2017-02-14
  • 0
    yeah I know it. but what if I have 64x 30? – 2017-02-14
  • 0
    Possible duplicate of [Solving non square matrix equations](http://math.stackexchange.com/questions/22616/solving-non-square-matrix-equations) – 2017-02-14

1 Answers 1

0

Use the system

$\begin{cases}x+z = 95 \\ z=45 \\ 2z+y = 150 \end{cases}$

Then in order to solve it represent it as $Ax=b:$

$Ax = b \Rightarrow\begin{bmatrix}1&0&1 \\ 0&0&1 \\ 0&1&2\end{bmatrix}\cdot x = \begin{bmatrix}95\\45\\150\end{bmatrix}$

Then $x=50,y=60,z=45$.

As you have more equations than unknowns ($5$ equations vs $3$ unknowns) then you have to get rid of $2$ equations. This is called an overdetermined system of equations.

But first, you have to prove that there exists a lineal combination involving the coefficients of the $5x3$ matrix , this is done measuring the rank of the coefficient matrix and comparing it to the rank of the augmented matrix, which in your case both have rank $3$, satisfying Rouché–Capelli theorem

  • 0
    Yes, here it is possible to use a square matrix. The OP also asks, however, what to do in general for a non-square matrix. – 2017-02-14
  • 0
    @DietrichBurde: It is said that the system is Overdetermined, since it has more equations than unknowns. If the coefficients of the system are linearly independent (no linear combination), then the system doesn't have a solution. This can be proven with the rank. – 2017-02-14
  • 0
    Yes, exactly. This is what I meant with "Solving non-square matrix equations", which is what the OP has asked. – 2017-02-14
  • 0
    @DietrichBurde: I will update my answer to introduce the concept to the user as said in my comment. – 2017-02-14