Witam.
Układ do rozwiązania Gaussem:
\(\displaystyle{ \left\{\begin{array}{l} 2x+3y-z+a=-3\\3x-y+2z+4a=8\\x+y+3z-2a=6\\x-2y-3z-5a=-3 \end{array}\right.}\)
Moje obliczenia:
\(\displaystyle{ \left[\begin{array}{ccccc}2&3&-1&1&|-3\\3&-1&2&4&|8\\1&1&3&-2&|6\\1&-2&-3&-5&|-3\end{array}\right]=}\)
Zamiana wiersza 1 i 3
\(\displaystyle{ =\left[\begin{array}{ccccc}1&1&3&-2&|6\\3&-1&2&4&|8\\2&3&-1&1&|-3\\1&-2&-3&-5&|-3\end{array}\right]=}\)
\(\displaystyle{ W_{2}^{`}=W _{1} \cdot (-3)+ W_{2}}\)
\(\displaystyle{ W_{3}^{`}=W _{1} \cdot (-2)+ W_{3}}\)
\(\displaystyle{ W_{4}^{`}=W _{1} \cdot (-1)+ W_{4}}\)
\(\displaystyle{ =\left[\begin{array}{ccccc}1&1&3&-2&|6\\0&-4&-7&10&|-10\\0&1&-7&5&|-14\\0&-3&-6&-3&|-9\end{array}\right]=}\)
Zamiana wiersza 2 i 3
\(\displaystyle{ =\left[\begin{array}{ccccc}1&1&3&-2&|6\\0&1&-7&5&|-14\\0&-4&-7&10&|-10\\0&-3&-6&-3&|-9\end{array}\right]=}\)
\(\displaystyle{ W_{3}^{`}=W _{2} \cdot 4+ W_{3}}\)
\(\displaystyle{ W_{4}^{`}=W _{2} \cdot 3+ W_{4}}\)
\(\displaystyle{ =\left[\begin{array}{ccccc}1&1&3&-2&|6\\0&1&-7&5&|-14\\0&-0&-35&30&|-66\\0&0&-27&12&|-51\end{array}\right]=}\)
Dobrze liczę, nie ma nigdzie błędu?
