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linear_algebra:03_gaussian_elimination [2014/01/20 13:30] marje |
linear_algebra:03_gaussian_elimination [2014/01/20 13:34] marje |
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| decide weather it has unique solution or not. | decide weather it has unique solution or not. | ||
| - | |$\left(\begin{array}{rrrrr|r}1&0&0&5&8&0\\0&0&1&2&7&0\\0&0&0&0&0&1\end{array}\right)$ | $\left(\begin{array}{rrrrr|r}1&0&6&5&8&2\\0&0&2&2&7&3\end{array}\right)$ | $\left(\begin{array}{rrr|r}1&0&0&3\\0&1&0&-2\\0&0&1&10\\0&0&0&0\end{array}\right)$ | $\left(\begin{array}{rrrrr|r}1&0&0&0&0&3\\0&0&9&1&0&-3\\0&1&-2&0&0&2\\0&0&0&0&1&5\end{array}\right)$ | | + | |$\left(\begin{array}{rrrrr|r}1&0&0&5&8&0\\0&0&1&2&7&0\\0&0&0&0&0&1\end{array}\right)$ | $\left(\begin{array}{rrrrr|r}1&0&6&5&8&2\\0&0&2&2&7&3\end{array}\right)$ | |
| - | |++Answer | This one is. However, the system has no solution. ++ | ++Answer | This one is not. The entry $a_{23}=2$ but in case of general Gaussian elimination algorithm there should be $1$. By multiplying the second row by $\frac{1}{2}$ we obtain the desired final form of Gaussian elimination. The system has more then one solution. ++ | ++Answer | No, since the last row consists of only zeros. If we drop the last row, then the augmented matrix is in the needed form. The system has precisely one solution. ++ | ++Answer | No. The 3rd and 2nd row should be switched and instead of $9$ there should be $1$ and instead of $-2$ there should be $0$. ++ | | + | |++Answer | This one is. However, the system has no solution. ++ | ++Answer | This one is not. The entry $a_{23}=2$ but in case of general Gaussian elimination algorithm there should be $1$. By multiplying the second row by $\frac{1}{2}$ we obtain the desired final form of Gaussian elimination. The system has more then one solution. ++ | |
| + | |$\left(\begin{array}{rrr|r}1&0&0&3\\0&1&0&-2\\0&0&1&10\\0&0&0&0\end{array}\right)$ | $\left(\begin{array}{rrrrr|r}1&0&0&0&0&3\\0&0&9&1&0&-3\\0&1&-2&0&0&2\\0&0&0&0&1&5\end{array}\right)$ | | ||
| + | | ++Answer | No, since the last row consists of only zeros. If we drop the last row, then the augmented matrix is in the needed form. The system has precisely one solution. ++ | ++Answer | No. The 3rd and 2nd row should be switched and instead of $9$ there should be $1$ and instead of $-2$ there should be $0$. ++ | | ||
| </WRAP> | </WRAP> | ||