Question

Решите пожалуйста систему линейных уравнений методом Крамера.
2x-y+3z=3
X+2y+z=2
X-3y+4z=-1

Answer 1

$\displaystyle\begin{cases}2x-y+3z=3\\x+2y+z=2\\x-3y+4z=-1\end{cases}\\\mathcal4=\left[\begin{array}{ccc}2&-1&3\\1&2&1\\1&-3&4\end{array}\right] =16-1-9-6+4+6=10\\\mathcal4_x=\left[\begin{array}{ccc}3&-1&3\\2&2&1\\-1&-3&4\end{array}\right]=24+1-18+6+8+9=30\\\mathcal4_y=\left[\begin{array}{ccc}2&3&3\\1&2&1\\1&-1&4\end{array}\right] =16+3-3-6-12+2=0\\\mathcal4_z=\left[\begin{array}{ccc}2&-1&3\\1&2&2\\1&-3&-1\end{array}\right] =-4-2-9-6-1+12=-10$

$\displaystyle x=\frac{\mathcal4_x}{\mathcal4}=\frac{30}{10}=3\\y=\frac{\mathcal4_y}{\mathcal4}=\frac{0}{10}=0\\z=\frac{\mathcal4_z}{\mathcal4}=-\frac{10}{10}=-1$