Metoda Gaussa - źle mi wychodzi...
: 10 lis 2012, o 22:54
\(\displaystyle{ \begin{vmatrix} 2&-1&1& |2\\3&2&2&|-2\\1&-2&1&|1\end{vmatrix}}\) W1<>W3
\(\displaystyle{ \begin{vmatrix} 1&-2&1& |1\\3&2&2&|-2\\2&-1&1&|2\end{vmatrix}}\)W3-2W1, W2-3W1
\(\displaystyle{ \begin{vmatrix} 1&-2&1& |1\\0&12&-6&|-5\\0&-4&-1&|0\end{vmatrix}}\) W2<>W3
\(\displaystyle{ \begin{vmatrix} 1&-2&1& |1\\0&-4&-1&|0\\0&12&-6&|-5\end{vmatrix}}\) 3W2+W3
\(\displaystyle{ \begin{vmatrix} 1&-2&1& |1\\0&-4&-1&|0\\0&0&-9&|-5\end{vmatrix}}\)
i z tego wychodzi
\(\displaystyle{ -9z=-5 \\
z= \frac{5}{2} \\ \\
-4y-\frac{5}{2} =0 \\ \\
y=-\frac{5}{36} \\ \\
x+\frac{5}{18} +\frac{5}{9} =1 \\ \\
x=\frac{1}{6}}\)
Co robię źle? W odpowiedziach jest: \(\displaystyle{ x=2, \ y=-1, \ z=-3}\)
\(\displaystyle{ \begin{vmatrix} 1&-2&1& |1\\3&2&2&|-2\\2&-1&1&|2\end{vmatrix}}\)W3-2W1, W2-3W1
\(\displaystyle{ \begin{vmatrix} 1&-2&1& |1\\0&12&-6&|-5\\0&-4&-1&|0\end{vmatrix}}\) W2<>W3
\(\displaystyle{ \begin{vmatrix} 1&-2&1& |1\\0&-4&-1&|0\\0&12&-6&|-5\end{vmatrix}}\) 3W2+W3
\(\displaystyle{ \begin{vmatrix} 1&-2&1& |1\\0&-4&-1&|0\\0&0&-9&|-5\end{vmatrix}}\)
i z tego wychodzi
\(\displaystyle{ -9z=-5 \\
z= \frac{5}{2} \\ \\
-4y-\frac{5}{2} =0 \\ \\
y=-\frac{5}{36} \\ \\
x+\frac{5}{18} +\frac{5}{9} =1 \\ \\
x=\frac{1}{6}}\)
Co robię źle? W odpowiedziach jest: \(\displaystyle{ x=2, \ y=-1, \ z=-3}\)