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at the moment I am a little bit confused. Here is the matrix I am trying to solve

$$ \left( \begin{array}{cc|c} 5 & -1& 12 \\ -1 & 2& 12 \end{array} \right) $$

I tried to solve it by multiplying the rows with several numbers but it does not show up a possible solution...

Pls give me a hint...

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1 Answer 1

up vote 8 down vote accepted

We use elementary row operations.

It's very much like solving a system of two equations in two unknowns:

$$ \begin{eqnarray*} 5x - \,\;y &= 12\\ \\ -x + 2y &= 12\\ \end{eqnarray*} $$ But instead, we use the *augmented coefficient matrix (remove the variables).

$$ \left( \begin{array}{cc|c} 5 & -1& 12 \\ -1 & 2& 12 \end{array} \right) $$ First multiply the second row by 5, and then add row 1 to row 2:

$$ \left( \begin{array}{cc|c} 5 & -1& 12 \\ -1 & 2& 12 \end{array} \right) = \left( \begin{array}{cc|c} 5 & -1& 12 \\ -5 & 10& 60 \end{array} \right)= \left( \begin{array}{cc|c} 5 & -1& 12 \\ 0 & 9& 72 \end{array} \right) $$

Can you take it from here?

We want to end with something like

$$ \left( \begin{array}{cc|c} a & 0& x \\ 0 & b& y \end{array} \right) $$

That is, we end with $x = a, y = b$.

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ok. Thx for your hint. However, I do not get it;( Which row would you multiply with which value? –  Le Chifre Jan 2 '13 at 14:12
    
lol;P Ok got it! THX!!!! –  Le Chifre Jan 2 '13 at 14:14
    
See the expanded answer above for the first reduction... –  amWhy Jan 2 '13 at 14:14
    
It is somewhat easier to first multiply the top row by 2 and add it to the second row. Then one does not have to care about fractions –  N3buchadnezzar Jan 2 '13 at 14:20
    
Or multiply the second row by 5, then add the first row to the second row. –  amWhy Jan 2 '13 at 14:24

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