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linear_algebra:03_gaussian_elimination [2014/01/20 12:50] marje |
linear_algebra:03_gaussian_elimination [2014/01/20 13:24] marje |
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| Analyse the result: what should be changed in order to obtain the final augmented matrix from above.\\ | Analyse the result: what should be changed in order to obtain the final augmented matrix from above.\\ | ||
| - | </WRAP> | ||
| <hidden **Solution**> | <hidden **Solution**> | ||
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| Do you get the correct answer? ++Answer | The final augmented matrix is $\left(\begin{array}{rrrr|r}1&-\frac{3}{2}&-\frac{1}{2}&\frac{5}{2}&\frac{1}{2}\\0&0&0&1&0\\0&0&0&0&0\end{array}\right)$ ++ | Do you get the correct answer? ++Answer | The final augmented matrix is $\left(\begin{array}{rrrr|r}1&-\frac{3}{2}&-\frac{1}{2}&\frac{5}{2}&\frac{1}{2}\\0&0&0&1&0\\0&0&0&0&0\end{array}\right)$ ++ | ||
| - | </WRAP> | + | </WRAP>\\ |
| As you can see, the simple Gaussian elimination algorithm does not work in such a case. Next we will generalize | As you can see, the simple Gaussian elimination algorithm does not work in such a case. Next we will generalize | ||
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| if $r=n$ and $a_{ii}=1$ for every $i=1,\dots,n$, then **the system has a unique solution**. The solution is | if $r=n$ and $a_{ii}=1$ for every $i=1,\dots,n$, then **the system has a unique solution**. The solution is | ||
| the last column of the augmented matrix, | the last column of the augmented matrix, | ||
| - | see [[gaussian_elimination#simple_gaussian|simple Gaussian elimination algorithm]]. | + | see [[03_gaussian_elimination#simple_gaussian|simple Gaussian elimination algorithm]]. |
| If $r<n$, then **the system has more than one solution**. In the case of solving such a system over $\mathbb{R}$ | If $r<n$, then **the system has more than one solution**. In the case of solving such a system over $\mathbb{R}$ | ||