Question

Să se determine dacă poate fi aplicată metoda lui Cramer și să se rezolve în C sistemul de ecuații:​

Answer 1

$\displaystyle a). \left\{\begin{array}{ccc}-x+8y+3z=2\\2x+4y-z=1\\2x-z=1\end{array}\right~~~~~~~~~~~~~~~~~~~~~~~A=\left(\begin{array}{ccc}-1&8&3\\2&4&-1\\2&0&-1\end{array}\right)\\ \\ \Delta=det~A=\left|\begin{array}{ccc}-1&8&3\\2&4&-1\\2&0&-1\end{array}\right|=-1\cdot4\cdot(-1)+3\cdot2\cdot0+8 \cdot (-1)\cdot2-\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-3\cdot4\cdot2-8\cdot2\cdot(-1)-(-1)\cdot(-1)\cdot 0=\\ \\ =4+0-16-24+16-0=-20\\ \\ \Delta=det~A=-20\ne0$

$\displaystyle\Delta_x=\left|\begin{array}{ccc}2&8&3\\1&4&-1\\1&0&-1\end{array}\right|=2\cdot4\cdot(-1)+3\cdot1\cdot0+8 \cdot (-1) \cdot 1-3 \cdot 4 \cdot 1-\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-8 \cdot 1 \cdot (-1)-2 \cdot (-1) \cdot 0=-8-8-12+8=-20\\ \\ \Delta_x=-20\\ \\ \Delta_y=\left|\begin{array}{ccc}-1&2&3\\2&1&-1\\2&1&-1\end{array}\right|=-1\cdot1\cdot(-1)+3\cdot2\cdot1+2\cdot(-1)\cdot2-3\cdot1\cdot2- \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-2 \cdot 2 \cdot (-1)-(-1) \cdot (-1) \cdot 1=\\ \\ =1+6-4-6+4-1=0\\ \\ \Delta_y=0$

$\displaystyle\Delta_z=\left|\begin{array}{ccc}-1&8&2\\2&4&1\\2&0&1\end{array}\right|=-1\cdot4\cdot1+2 \cdot 2 \cdot 0+8 \cdot 1\cdot2-2\cdot4\cdot2-8\cdot2\cdot1-\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-(-1) \cdot1\cdot0=-4+0+16-16-16=-20\\ \\ \Delta_z=-20\\ \\ x=\frac{\Delta_x}{\Delta} =\frac{-20}{-20} =1~~~~~~~~~~~~~~~~~~~~~~~x=1\\ \\ y=\frac{\Delta_y}{\Delta}=\frac{0}{-20} =0~~~~~~~~~~~~~~~~~~~~~~~y=0\\ \\ z=\frac{\Delta_z}{\Delta}=\frac{-20}{-20} =1~~~~~~~~~~~~~~~~~~~~~~~z=1$

$\displaystyle b).\left\{\begin{array}{ccc}x+4y+9z=16\\2x+2y+2z=2\\x+2y+3z=4\end{array}\right~~~~~~~~~~~~~~~~~~~~~~~~~~A=\left(\begin{array}{ccc}1&4&9\\2&2&2\\1&2&3\end{array}\right) \\ \\ \Delta=det~A=\left|\begin{array}{ccc}1&4&9\\2&2&2\\1&2&3\end{array}\right|=1 \cdot 2 \cdot 3+9\cdot2\cdot2+4\cdot2\cdot1-9\cdot2\cdot1-\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-4\cdot2\cdot3-1\cdot2\cdot2=\\ \\ =6+36+8-18-24-4=4\\ \\ \Delta=det~A=4\neq 0$

$\displaystyle \Delta_x=\left|\begin{array}{ccc}16&4&9\\2&2&2\\4&2&3\end{array}\right|=16\cdot2\cdot3+9\cdot2\cdot2+4\cdot2\cdot4-9\cdot2\cdot4-4\cdot2\cdot3-\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-16\cdot2\cdot2=96+36+32-72-24-64=4 \\ \\ \Delta_x=4\\ \\ \Delta_y=\left|\begin{array}{ccc}1&16&9\\2&2&2\\1&4&3\end{array}\right|=1\cdot2\cdot3+9\cdot2\cdot4+16\cdot2\cdot1-9\cdot2\cdot1-16\cdot2\cdot3-\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-1\cdot2\cdot4=6+72+32-18-96-8=-12\\ \\ \Delta_y=-12$

$\displaystyle \Delta_z=\left|\begin{array}{ccc}1&4&16\\2&2&2\\1&2&4\end{array}\right|=1\cdot2\cdot4+16\cdot2\cdot2+4\cdot2\cdot1-16\cdot2\cdot1-4\cdot2\cdot4-\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-1\cdot2\cdot2=8+64+8-32-32-4=12\\ \\ \Delta_z=12\\ \\ x=\frac{\Delta_x}{\Delta} =\frac{4}{4} =1~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=1\\ \\ y=\frac{\Delta_y}{\Delta} =\frac{-12}{4} =-3~~~~~~~~~~~~~~~~~~~~~~y=-3\\ \\ z=\frac{\Delta_z}{\Delta} =\frac{12}{4} =3~~~~~~~~~~~~~~~~~~~~~~~~~~~z=3$

$\displaystyle c). \left\{\begin{array}{ccc}x+4z=-7\\-2x+y+3z=-7\\x+2y-2z=7\end{array}\right~~~~~~~~~~~~~~~~~~~~~~A=\left(\begin{array}{ccc}1&0&4\\-2&1&3\\1&2&-2\end{array}\right)\\ \\ \Delta=det~A=\left|\begin{array}{ccc}1&0&4\\-2&1&3\\1&2&-2\end{array}\right|=1\cdot1\cdot(-2)+4\cdot(-2)\cdot2+0\cdot3\cdot1-\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-4\cdot1\cdot1~-0\cdot(-2)\cdot(-2)-1\cdot3\cdot2=\\ \\ =-2-16-4-6=-28\\ \\ \Delta=det~A=-28\ne 0$

$\displaystyle\Delta_x=\left|\begin{array}{ccc}-7&0&4\\-7&1&3\\7&2&-2\end{array}\right|=-7\cdot1\cdot(-2)+4\cdot(-7)\cdot2+0\cdot3\cdot7-4\cdot1\cdot7-\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-0\cdot(-7)\cdot(-2)-(-7)\cdot3\cdot2=\\\\ =14-56-28+42=-28\\\\\Delta_x=-28$

$\displaystyle \Delta_y=\left|\begin{array}{ccc}1&-7&4\\-2&-7&3\\1&7&-2\end{array}\right|=1\cdot(-7)\cdot(-2)+4\cdot(-2)\cdot7+(-7)\cdot3\cdot1-\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-4\cdot(-7)\cdot1~-(-7)\cdot(-2)\cdot(-2)-1\cdot3\cdot7=\\\\=14-56-21+28+28-21=-28\\ \\ \Delta_y=-28$

$\displaystyle \Delta_z=\left|\begin{array}{ccc}1&0&-7\\-2&1&-7\\1&2&7\end{array}\right|=1\cdot1\cdot7+(-7)\cdot(-2)\cdot2+0\cdot(-7)\cdot1-\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-(-7)\cdot1\cdot1- 0\cdot(-2)\cdot7-1\cdot(-7)\cdot 2=\\ \\ =7+14+7+14=56\\ \\ \Delta_z=56\\ \\ x=\frac{\Delta_x}{\Delta} =\frac{-28}{-28} =1~~~~~~~~~~~~~~~~~~~~~~~~~~x=1\\ \\ y=\frac{\Delta_y}{\Delta} =\frac{-28}{-28} =1~~~~~~~~~~~~~~~~~~~~~~~~~~y=1\\ \\ z=\frac{\Delta_z}{\Delta} =\frac{56}{-28} =-2~~~~~~~~~~~~~~~~~~~~~~~~z=-2$