Question

2x1-x2+3x3=-4, x1+3x2-x3=11, x1-2x2+2x3=-7
Нужно решить методом Гаусса, и если можно Крамером. (Хз, правильно ли, что я написал)​

Answer 1

Метод Гаусса

$\displaystyle\begin{cases}2x_1-x_2+3x_3=-4\\x_1+3x_2-x_3=11\\x_!-2x_2+2x_3=-7\end{cases}$

$\displaystyle\left[\begin{array}{ccccccc}2&-1&3&|&-4&|&\downarrow\\1&3&-1&|&11&|&\uparrow\\1&-2&2&|&-7&|&\end{array}\right]\to\left[\begin{array}{ccccccc}1&3&-1&|&11&|&\downarrow\\2&-1&3&|&-4&|&*(-2)\\1&-2&2&|&-7&|&*(-1)\end{array}\right]\to\\\to\left[\begin{array}{ccccccc}1&3&-1&|&11&|&\\0&-7&5&|&-26&|&:(-7)\\0&-5&3&|&-18&|&\end{array}\right]\to\\\to\left[\begin{array}{ccccccc}1&3&-1&|&11&|&\\0&1&-\frac{5}{7}&|&\frac{26}{7}&|&\downarrow\\0&-5&3&|&-18&|&*5\end{array}\right]\to$

$\displaystyle\left[\begin{array}{ccccccc}1&3&-1&|&11&|&\\0&1&-\frac{5}{7}&|&\frac{26}{7}&|&\\0&0&-\frac{4}{7}&|&\frac{4}{7}&|&*(-\frac{7}{4})\end{array}\right]\to\\\to\left[\begin{array}{ccccc}1&3&-1&|&11\\0&1&-\frac{5}{7}&|&\frac{26}{7}\\0&0&1&|&-1\end{array}\right]$

$\displaystyle\begin{cases}x+3y-z=11\to x=1\\y-\frac{5}{7}z=\frac{26}{7}\to y=3\\z=-1\end{cases}\\\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}1\\3\\-1\end{array}\right]$

Метод Крамера:

$\displaystyle\mathcal4=\left[\begin{array}{ccc}2&-1&3\\1&3&-1\\1&-2&2\end{array}\right] =12+1-6-9+2-4=-4\\\mathcal4_x=\left[\begin{array}{ccc}-4&-1&3\\11&3&-1\\-7&-2&2\end{array}\right] =-24-7-66+63+22+8=-4\\\mathcal4_y=\left[\begin{array}{ccc}2&-4&3\\1&11&-1\\1&-7&2\end{array}\right] =44+4-21-33+8-14=-12\\\mathcal4_z=\left[\begin{array}{ccc}2&-1&-4\\1&3&11\\1&-2&-7\end{array}\right] =-42-11+8+12-7+44=4$

$\displaystyle x=\frac{\mathcal4_x}{\mathcal4}=1\\y=\frac{\mathcal4_y}{\mathcal4}=3\\z=\frac{\mathcal4_z}{\mathcal4}=-1\\\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}1\\3\\-1\end{array}\right]$