Rozwiąż układ równań:
\(\displaystyle{ \begin{cases} 2 x_{1}+8 x_{2}+5 x_{3} +6 x_{4}=1\\ 3 x_{1}+6 x_{2}+5 x_{3} - x_{4}=1\\ 2 x_{1}+4 x_{2}+3 x_{3} + x_{4}=2 \\
x_{1}+3 x_{2}+2 x_{3} +2 x_{4}=1
\end{cases}}\)
Rozwiąż układ równań
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Mariusz M
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Rozwiąż układ równań
Możesz zastosować rozkład LU opisałem go w kompendium
a dokładniej jest opisany na ważniaku
\(\displaystyle{ \begin{bmatrix} 2&8&5&6 \\3&6&5&-1\\2&4&3&1\\1&3&2&2 \end{bmatrix} \begin{bmatrix} 1 \\2\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\2&8&5&6\\2&4&3&1\\1&3&2&2 \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ \frac{2}{3} &8&5&6\\ \frac{2}{3} &4&3&1\\ \frac{1}{3} &3&2&2 \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ \frac{2}{3} &4& \frac{5}{3} & \frac{20}{3} \\ \frac{2}{3} &0&- \frac{1}{3} & \frac{5}{3} \\ \frac{1}{3} &1& \frac{1}{3} & \frac{7}{3} \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ \frac{2}{3} &4& \frac{5}{3} & \frac{20}{3} \\ \frac{2}{3} &0&- \frac{1}{3} & \frac{5}{3} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{3} & \frac{7}{3} \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ \frac{2}{3} &4& \frac{5}{3} & \frac{20}{3} \\ \frac{2}{3} &0&- \frac{1}{3} & \frac{5}{3} \\ \frac{1}{3} & \frac{1}{4} & -\frac{1}{12} & \frac{8}{12} \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ \frac{2}{3} &4& \frac{5}{3} & \frac{20}{3} \\ \frac{2}{3} &0&- \frac{1}{3} & \frac{5}{3} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)-- 5 stycznia 2010, 23:34 --\(\displaystyle{ \begin{bmatrix} 1&0&0&0 \\ \frac{2}{3}&1&0&0\\ \frac{2}{3}&0&1&0\\ \frac{1}{3}& \frac{1}{4}& \frac{1}{4}&1 \end{bmatrix} \cdot \begin{bmatrix} y_{1} \\ y_{2}\\y_{3}\\y_{4} \end{bmatrix}= \begin{bmatrix} 1 \\ 1\\2\\1 \end{bmatrix}}\)
\(\displaystyle{ \begin{cases} y_{1}=1 \\ \frac{2}{3}+y_{2}=1\\ \frac{2}{3}+y_{3}=2\\ \frac{1}{3}+ \frac{1}{4}y_{2}+ \frac{1}{4}y_{3}+y_{4}=1 \end{cases}}\)
\(\displaystyle{ \begin{cases} y_{1}=1 \\ y_{2}= \frac{1}{3} \\ y_{3}= \frac{4}{3} \\ \frac{1}{3}+ \frac{1}{12}+ \frac{1}{3}+y_{4}=1 \end{cases}}\)
\(\displaystyle{ \begin{cases} y_{1}=1 \\ y_{2}= \frac{1}{3} \\ y_{3}= \frac{4}{3} \\ \frac{3}{4} +y_{4}=1 \end{cases}}\)
\(\displaystyle{ \begin{cases} y_{1}=1 \\ y_{2}= \frac{1}{3} \\ y_{3}= \frac{4}{3} \\ y_{4}= \frac{1}{4} \end{cases}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ 0&4& \frac{5}{3}& \frac{20}{3}\\0&0& -\frac{1}{3}& \frac{5}{3}\\0&0&0& \frac{1}{4}\end{bmatrix} \cdot \begin{bmatrix} x_{1} \\ x_{2}\\x_{3}\\x_{4} \end{bmatrix}= \begin{bmatrix} 1 \\ \frac{1}{3}\\ \frac{4}{3}\\ \frac{1}{4} \end{bmatrix}}\)
\(\displaystyle{ \begin{cases} x_{4}= \frac{1}{4} \cdot 4=1 \\ x_{3}=-3 \left( \frac{4}{3}- \frac{5}{3} \right) \\ x_{2}=\frac{1}{4} \left( \frac{1}{3}- \frac{20}{3}- \frac{5}{3} \right)\\x_{1}= \frac{1}{3} \left(1+1-5+12 \right) \end{cases}}\)
\(\displaystyle{ \begin{cases} x_{4}=1 \\ x_{3}=-3 \left( -\frac{1}{3}\right) \\ x_{2}=\frac{1}{4} \left( - \frac{24}{3} \right)\\x_{1}= \frac{1}{3} \left(9 \right) \end{cases}}\)
\(\displaystyle{ \begin{cases} x_{4}=1 \\ x_{3}=1 \\ x_{2}=-2\\x_{1}=3 \end{cases}}\)
\(\displaystyle{ \begin{bmatrix}x_{1} \\ x_{2}\\x_{3}\\x_{4} \end{bmatrix}= \begin{bmatrix} 3 \\ -2\\1\\1 \end{bmatrix}}\)
a dokładniej jest opisany na ważniaku
\(\displaystyle{ \begin{bmatrix} 2&8&5&6 \\3&6&5&-1\\2&4&3&1\\1&3&2&2 \end{bmatrix} \begin{bmatrix} 1 \\2\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\2&8&5&6\\2&4&3&1\\1&3&2&2 \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ \frac{2}{3} &8&5&6\\ \frac{2}{3} &4&3&1\\ \frac{1}{3} &3&2&2 \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ \frac{2}{3} &4& \frac{5}{3} & \frac{20}{3} \\ \frac{2}{3} &0&- \frac{1}{3} & \frac{5}{3} \\ \frac{1}{3} &1& \frac{1}{3} & \frac{7}{3} \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ \frac{2}{3} &4& \frac{5}{3} & \frac{20}{3} \\ \frac{2}{3} &0&- \frac{1}{3} & \frac{5}{3} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{3} & \frac{7}{3} \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ \frac{2}{3} &4& \frac{5}{3} & \frac{20}{3} \\ \frac{2}{3} &0&- \frac{1}{3} & \frac{5}{3} \\ \frac{1}{3} & \frac{1}{4} & -\frac{1}{12} & \frac{8}{12} \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ \frac{2}{3} &4& \frac{5}{3} & \frac{20}{3} \\ \frac{2}{3} &0&- \frac{1}{3} & \frac{5}{3} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} \end{bmatrix} \begin{bmatrix} 2 \\1\\3\\4 \end{bmatrix}}\)-- 5 stycznia 2010, 23:34 --\(\displaystyle{ \begin{bmatrix} 1&0&0&0 \\ \frac{2}{3}&1&0&0\\ \frac{2}{3}&0&1&0\\ \frac{1}{3}& \frac{1}{4}& \frac{1}{4}&1 \end{bmatrix} \cdot \begin{bmatrix} y_{1} \\ y_{2}\\y_{3}\\y_{4} \end{bmatrix}= \begin{bmatrix} 1 \\ 1\\2\\1 \end{bmatrix}}\)
\(\displaystyle{ \begin{cases} y_{1}=1 \\ \frac{2}{3}+y_{2}=1\\ \frac{2}{3}+y_{3}=2\\ \frac{1}{3}+ \frac{1}{4}y_{2}+ \frac{1}{4}y_{3}+y_{4}=1 \end{cases}}\)
\(\displaystyle{ \begin{cases} y_{1}=1 \\ y_{2}= \frac{1}{3} \\ y_{3}= \frac{4}{3} \\ \frac{1}{3}+ \frac{1}{12}+ \frac{1}{3}+y_{4}=1 \end{cases}}\)
\(\displaystyle{ \begin{cases} y_{1}=1 \\ y_{2}= \frac{1}{3} \\ y_{3}= \frac{4}{3} \\ \frac{3}{4} +y_{4}=1 \end{cases}}\)
\(\displaystyle{ \begin{cases} y_{1}=1 \\ y_{2}= \frac{1}{3} \\ y_{3}= \frac{4}{3} \\ y_{4}= \frac{1}{4} \end{cases}}\)
\(\displaystyle{ \begin{bmatrix} 3&6&5&-1 \\ 0&4& \frac{5}{3}& \frac{20}{3}\\0&0& -\frac{1}{3}& \frac{5}{3}\\0&0&0& \frac{1}{4}\end{bmatrix} \cdot \begin{bmatrix} x_{1} \\ x_{2}\\x_{3}\\x_{4} \end{bmatrix}= \begin{bmatrix} 1 \\ \frac{1}{3}\\ \frac{4}{3}\\ \frac{1}{4} \end{bmatrix}}\)
\(\displaystyle{ \begin{cases} x_{4}= \frac{1}{4} \cdot 4=1 \\ x_{3}=-3 \left( \frac{4}{3}- \frac{5}{3} \right) \\ x_{2}=\frac{1}{4} \left( \frac{1}{3}- \frac{20}{3}- \frac{5}{3} \right)\\x_{1}= \frac{1}{3} \left(1+1-5+12 \right) \end{cases}}\)
\(\displaystyle{ \begin{cases} x_{4}=1 \\ x_{3}=-3 \left( -\frac{1}{3}\right) \\ x_{2}=\frac{1}{4} \left( - \frac{24}{3} \right)\\x_{1}= \frac{1}{3} \left(9 \right) \end{cases}}\)
\(\displaystyle{ \begin{cases} x_{4}=1 \\ x_{3}=1 \\ x_{2}=-2\\x_{1}=3 \end{cases}}\)
\(\displaystyle{ \begin{bmatrix}x_{1} \\ x_{2}\\x_{3}\\x_{4} \end{bmatrix}= \begin{bmatrix} 3 \\ -2\\1\\1 \end{bmatrix}}\)