Question

Метот крамера. x+3y-2z=5. 2x-5y+6z=1. 3x+4y-5z=2. нужно

Answer 1

$\begin{equation*} \begin{cases} x+3y-2z=5, \\ 2x-5y+6z=1, \\ 3x+4y-5z=2. \end{cases} \end{equation*} \\\\\\\\ \Delta= \left|\begin{array}{ccc}1&3&-2\\2&-5&6\\3&4&-5\end{array}\right|=1\cdot(-5) \cdot(-5)+(-2)\cdot4\cdot2+3\cdot3\cdot6-\\\\-(-2)\cdot(-5)\cdot3-1\cdot4\cdot6-(-5)\cdot2\cdot3=25-16+54-30-24+\\+30=39;$

$\Delta_1=\left|\begin{array}{ccc}5&3&-2\\1&-5&6\\2&4&-5\end{array}\right|=5\cdot(-5) \cdot(-5)+(-2)\cdot4\cdot1+3\cdot2\cdot6-\\\\-(-2)\cdot(-5)\cdot2-5\cdot4\cdot6-(-5)\cdot1\cdot3=125-8+36-20-120+\\+15=28;$

$\Delta_2=\left|\begin{array}{ccc}1&5&-2\\2&1&6\\3&2&-5\end{array}\right|=1\cdot1 \cdot(-5)+(-2)\cdot2\cdot2+3\cdot5\cdot6-\\\\-(-2)\cdot1\cdot3-1\cdot2\cdot6-(-5)\cdot2\cdot5=-5-8+90+6-12+\\+50=121;$

$\Delta_3=\left|\begin{array}{ccc}1&3&5\\2&-5&1\\3&4&2\end{array}\right|=1\cdot2 \cdot(-5)+5\cdot2\cdot4+3\cdot1\cdot3-\\\\-5\cdot(-5)\cdot3-1\cdot4\cdot1-2\cdot3\cdot2=-10+40+9+75-4-\\-12=98;$

$x=\frac{\Delta_1}{\Delta}=\frac{28}{39},\ \ y=\frac{\Delta_2}{\Delta}=\frac{121}{39}=3\frac{4}{39},\ \ z=\frac{\Delta_3}{\Delta}=\frac{98}{39}=2\frac{20}{39}.$

$OTBET:\ \ x=\frac{28}{39},\ \ y=3\frac{4}{39},\ \ z=2\frac{20}{39}.$