матрични уравнения-ПОМОГНЕТЕ !

[tutmanika]

Да се реши матри4но уравнението:
1зад.
(2 -1 -1)
(3 4 -2).X=(4)
(3 -2 4) (11)
(11)
2.зад (1 2 3 4) (5)
(2 1 2 3) (1)
(3 2 1 2) .X=(1)
(4 3 2 1) (-5)
3.зад.
(2-1 3) (6 4 4)
X (3 2 2)=(0 0 0)
(1 3 -1) (18 12 12)
Извинявам се 4е я написах така но не знам как се ползва програмата.Ако може с метога на Гаус.. предварително благодаря

[Реклама]

[ferry2]

Тук е обяснено подробно как се решават

http://www.math10.com/forumbg/viewtopic.php?t=7892
http://www.math10.com/forumbg/viewtopic.php?t=7957

[tutmanika]

с метода на Гаус ми трябват тия уравнения

[nikko1]

Само за яснота, за да не се занимавам излишно. Това ли са задачите?
1 зад. [tex]\left(\begin{array}{rrr}2&-1&-1\\3&4&-2\\3&-2&4\end{array}\right).X=\left(\begin{array}{c} 4\\11\\11\end{array}\right).[/tex]

2 зад. [tex]\left(\begin{array}{rrrr}1&2&3&4\\2&1&2&3\\3&2&1&2\\4&3&2&1\end{array}\right).X=\left(\begin{array}{c} 5\\1\\1\\-5\end{array}\right).[/tex]

3 зад. [tex]\left(\begin{array}{rrrr}2&-1&3\\3&2&2\\1&3&-1\end{array}\right).X=\left(\begin{array}{rrrr}6&4&4\\0&0&0\\18&12&12\end{array}\right)[/tex]

[tutmanika]

да това са.

[nikko1]

1 зад. Работим по метод на Гаус-Жордан. Написваме матрица във вида [tex]\left(A|E\right)[/tex], като трябва да правим елементарни преобразувания с редовете, за да получим отпред единичната матрица.
[tex]\left(\begin{array}{rrr|rrr}2&-1&-1&1&0&0\\3&4&-2&0&1&0\\3&-2&4&0&0&1\end{array}\right)\longright^{-R_2+R_2}\left(\begin{array}{rrr|rrr}1&5&-1&-1&1&0\\3&4&-2&0&1&0\\3&-2&4&0&0&1\end{array}\right)\longright^{R_2-3R_1}_{R_3-3R_1}\left(\begin{array}{rrr|rrr}1&5&-1&-1&1&0\\0&-11&1&3&-2&0\\0&-17&7&3&-3&1\end{array}\right)\longright^{3R_2-2R_3}\left(\begin{array}{rrr|rrr}1&5&-1&-1&1&0\\0&1&-11&3&0&-2\\0&-17&7&3&-3&1\end{array}\right)\longright^{R_1-5R_2}_{R_3+17R_2}[/tex]

[tex]\rightarrow\left(\begin{array}{rrr|rrr}1&0&54&-16&1&10\\0&1&-11&3&0&-2\\0&0&-180&54&-3&-33\end{array}\right)\longright^{-\frac{1}{180}R_3}\left(\begin{array}{rrr|rrr}1&0&54&-16&1&10\\0&1&-11&3&0&-2\\0&0&1&-3/10&1/60&11/60\end{array}\right)\longright^{R_1-54R_3}_{R_2+11R_3}\left(\begin{array}{rrr|rrr}1&0&0&1/5&1/10&1/10\\0&1&0&-3/10&11/60&1/60\\0&0&1&-3/10&1/60&11/60\end{array}\right).[/tex]

[tex]A^{-1}=\left(\begin{array}{rrr}\frac{1}{5}&\frac{1}{10}&\frac{1}{10}\\-\frac{3}{10}&\frac{11}{60}&\frac{1}{60}\\-\frac{3}{10}&\frac{1}{60}& \frac{11}{60}\end{array}\right)[/tex]
[tex]A.X=b\Leftrightarrow X=\left(\begin{array}{rrr}\frac{1}{5}&\frac{1}{10}&\frac{1}{10}\\-\frac{3}{10}&\frac{11}{60}&\frac{1}{60}\\-\frac{3}{10}&\frac{1}{60}& \frac{11}{60}\end{array}\right).\left(\begin{array}{r}4\\11\\11\end{array}\right)=\left(\begin{array}{r}3\\1\\1\end{array}\right).[/tex]

Другите по-нататък.

[nikko1]

[tex]\left(\begin{array}{rrrr|rrrr}1&2&3&4&1&0&0&0\\2&1&2&3&0&1&0&0\\3&2&1&2&0&0&1&0\\4&3&2&1&0&0&0&1\end{array}\right)\longright^{\begin{matrix}R_2-2R_1\\R_3-3R_1\end{matrix}}_{R_4-4R_1}\left(\begin{array}{rrrr|rrrr}1&2&3&4&1&0&0&0\\0&-3&-4&-5&-2&1&0&0\\0&-4&-8&-10&-3&0&1&0\\0&-5&-10&-15&-4&0&0&1\end{array}\right)\longright^{-R_2\\-\frac{2}{3}R_3}_{-\frac{1}{5}R_4}\left(\begin{array}{rrrr|rrrr}1&2&3&4&1&0&0&0\\0&3&4&5&2&-1&0&0\\0&2&4&5&3/2&0&-1/2&0\\0&1&2&3&4/5&0&0&-1/5\end{array}\right)\longright^{R_2\leftrightarrow R_4}[/tex]

[tex]\rightarrow\left(\begin{array}{cccc|cccc}1&2&3&4&1&0&0&0\\0&1&2&3&4/5&0&0&-1/5\\0&2&4&5&3/2&0&-1/2&0\\0&3&4&5&2&-1&0&0\end{array}\right)\longright^{\begin{matrix}R_1-2R_2\\R_3-2R-2\end{matrix}}_{R_4-3R_2}\left(\begin{array}{cccc|cccc}1&0&-1&-2&-3/5&0&0&2/5\\0&1&2&3&4/5&0&0&-1/5\\0&0&0&-1&-1/10&0&-1/2&2/5\\0&0&-2&-4&-2/5&-1&0&3/5\end{array}\right)\longright^{R_3\leftrightarrow R_4}\left(\begin{array}{cccc|cccc}1&0&-1&-2&-3/5&0&0&2/5\\0&1&2&3&4/5&0&0&-1/5\\0&0&-2&-4&-2/5&-1&0&3/5\\0&0&0&-1&-1/10&0&-1/2&2/5\end{array}\right)\rightarrow[/tex]

[tex]\longright^{-\frac{1}{2}R_3}\left(\begin{array}{cccc|cccc}1&0&-1&-2&-3/5&0&0&2/5\\0&1&2&3&4/5&0&0&-1/5\\0&0&1&2&1/5&1/2&0&-3/10\\0&0&0&-1&-1/10&0&-1/2&2/5\end{array}\right)\longright^{R_1+R_3}_{R_2-2R_3}\left(\begin{array}{cccc|cccc}1&0&0&0&-2/5&1/2&0&1/10\\0&1&0&-1&2/5&-1&0&2/5\\0&0&1&2&1/5&1/2&0&-3/10\\0&0&0&-1&-1/10&0&-1/2&2/5\end{array}\right)\longright^{-R_4}\left(\begin{array}{cccc|cccc}1&0&0&0&-2/5&1/2&0&1/10\\0&1&0&-1&2/5&-1&0&2/5\\0&0&1&2&1/5&1/2&0&-3/10\\0&0&0&1&1/10&0&1/2&-2/5\end{array}\right)\rightarrow[/tex]

[tex]\longright^{R_2+R_4}_{R_3-2R_4}\left(\begin{array}{cccc|cccc}1&0&0&0&-2/5&1/2&0&1/10\\0&1&0&0&1/2&-1&1/2&0\\0&0&1&0&0&1/2&-1&1/2\\0&0&0&1&1/10&0&1/2&-2/5\end{array}\right).[/tex]

[tex]A^{-1}=\left(\begin{array}{cccc}-2/5&1/2&0&1/10\\1/2&-1&1/2&0\\0&1/2&-1&1/2\\1/10&0&1/2&-2/5\end{array}\right).[/tex]

[tex]A.X=b\Leftrightarrow X=A^{-1}.b=\left(\begin{array}{cccc}-2/5&1/2&0&1/10\\1/2&-1&1/2&0\\0&1/2&-1&1/2\\1/10&0&1/2&-2/5\end{array}\right).\left(\begin{array}{r}5\\1\\1\\-5\end{array}\right)=\left(\begin{array}{r}-2\\2\\-3\\3\end{array}\right).[/tex]


За да решиш трета задача [tex]\underbrace{\left(\begin{array}{rrrr}2&-1&3\\3&2&2\\1&3&-1\end{array}\right)}_{A}.X=\underbrace{\left(\begin{array}{rrrr}6&4&4\\0&0&0\\18&12&12\end{array}\right)}_{B}[/tex]

този път пишеш [tex](A|B)[/tex] ... [tex](E|X)[/tex] и намираш X.