Question

x＋2y＋3z=5

sistem 3x＋2y＋z=3

2x＋3y＋z=4



de rezolvat în 3 metode

Answer 1

x=5-2y-3z inlocuind in celelalte doua ecuatii avem
3(5-2y-3z)+2y+z = 3  =>15-6y-9z+2y+z=3  => 4y+8z = 12 => y+2z=3
2(5-2y-3z)+3y+z = 4  =>10-4y-6z+3y+z=4  =>  y +5z =  6  scadem cele doua relatii si obtinem: 3z = 3  => Z=1
y+2z=3, Z=1 => Y=1
x+2y+3z=5, Y=1 ,Z=1  =>  X=0

Metota II

prima relatie se inmulteste convenabil cu 3 si avem 3x+6y+9z=15 si 3x+2y+z=3 se scad cele doua relatii si se obtine 4y+8z=12 => y+2z=3
 si apoi cu 2 si avem 2x+4y+6z=10 si 2x+3y+z=4 se scad una din alta si avem y+5z=6

Metoda III se aduna toate relatiile 6x+7y+6z=12 si se folosesc si celelalte relatii si se fac inlocuiri

Answer 2

$\displaystyle 1.~~~~ \left\{\begin{array}{ccc}x+2y+3z=5\\3x+2y+z=3\\2x+3y+z=4\end{array}~~~~~~~~~~~~~~~~~~A= \left(\begin{array}{ccc}1&2&3\\3&2&1\\2&3&1\end{array}\right)\right}\\ \\ \\\Delta=det~A=\left|\begin{array}{ccc}1&2&3\\3&2&1\\2&3&1\end{array}\right|=1\cdot2\cdot1+3\cdot3\cdot3+2\cdot1\cdot2-3\cdot2\cdot2-\\\\ \\-2\cdot3\cdot1-1\cdot1\cdot3=2+27+4-12-6-3=12\\ \\ ~~~~~~~~~~~~~~\Delta=det~A=12\ne 0$

$\displaystyle\Delta_x=\left|\begin{array}{ccc}5&2&3\\3&2&1\\4&3&1\end{array}\right|=5\cdot2\cdot1+3\cdot3\cdot3+2\cdot1\cdot4-3\cdot2\cdot4-2\cdot3\cdot1-\\\\\\-5\cdot1\cdot3=10+27+8-24-6-15=0\\ \\ ~~~~~~~~~~~~~~~~~\Delta_x=0$

$\displaystyle \Delta_y=\left|\begin{array}{ccc}1&5&3\\3&3&1\\2&4&1\end{array}\right|=1\cdot3\cdot1+3\cdot3\cdot4+5\cdot1\cdot2-3\cdot3\cdot2-5\cdot3\cdot1-\\ \\ \\ -1\cdot1\cdot4=3+36+10-18-15-4=12\\ \\ ~~~~~~~~~~~~~~\Delta_y=12$

$\displaystyle \Delta_z=\left|\begin{array}{ccc}1&2&5\\3&2&3\\2&3&4\end{array}\right|=1\cdot2\cdot4+5\cdot3\cdot3+2\cdot3\cdot2-5\cdot2\cdot2-2\cdot3\cdot4-\\\\\\-1\cdot3\cdot3=8+45+12-20-24-9=12\\\\~~~~~~~~~~~~~~~~~\Delta_z=12\\\\\\x=\frac{\Delta_x}{\Delta}= \frac{0}{12}=0~~~~~~~~~~~~~~~~~~~~~x=0~\\\\\\y=\frac{\Delta_y}{\Delta} = \frac{12}{12} =1~~~~~~~~~~~~~~~~~~~~~y=1\\\\\\z=\frac{\Delta_z}{\Delta}= \frac{12}{12}=1~~~~~~~~~~~~~~~~~~~~~z=1$

$\displaystyle -------------------------------\\ \\ 2.~~~~\left\{\begin{array}{ccc}x+2y+3z=5\\3x+2y+z=3\\2x+3y+z=4\end{array}\right}\\ \\ \\ A= \left(\begin{array}{ccc}1&2&3\\3&2&1\\2&3&1\end{array}\right)~~~~~~~~~~X= \left(\begin{array}{ccc}x\\y\\z\end{array}\right)~~~~~~~~~~~~B= \left(\begin{array}{ccc}5\\3\\4\end{array}\right)$

$\displaystyle AX=B\Rightarrow X=A^{-1}B~~~~~~~~~~~~~~~~~~~~A^{-1}= \frac{1}{det~A}\cdot A^* \\\\det~A=\left|\begin{array}{ccc}1&2&3\\3&2&1\\2&3&1\end{array}\right|=1\cdot2\cdot1+3\cdot3\cdot3+2\cdot1\cdot2-3\cdot2\cdot2-\\\\ \\-2\cdot3\cdot1-1\cdot1\cdot3=2+27+4-12-6-3=12 \\ \\ ~~~~~~~~~~~~~~~det~A=12\ne 0 \\ \\ \\ A= \left(\begin{array}{ccc}1&2&3\\3&2&1\\2&3&1\end{array}\right)\Rightarrow A^T= \left(\begin{array}{ccc}1&3&2\\2&2&3\\3&1&1\end{array}\right)$

$\displaystyle \delta_{11}=(-1)^{1+1} \cdot \left|\begin{array}{ccc}2&3\\1&1\end{array}\right|=1\cdot(-1)=-1\\ \\ \\ \delta_{12}=(-1)^{1+2}\cdot \left|\begin{array}{ccc}2&3\\3&1\end{array}\right|=(-1)\cdot(-7)=7\\\\\\\delta_{13}=(-1)^{1+3}\cdot \left|\begin{array}{ccc}2&2\\3&1\end{array}\right|=1\cdot(-4)=-4\\\\\\\delta_{21}=(-1)^{2+1}\cdot \left|\begin{array}{ccc}3&2\\1&1\end{array}\right|=(-1)\cdot 1=-1\\ \\ \\ \delta_{22}=(-1)^{2+2}\cdot \left|\begin{array}{ccc}1&2\\3&1\end{array}\right|=1\cdot(-5)=-5$

$\displaystyle \delta_{23}=(-1)^{2+3}\cdot \left|\begin{array}{ccc}1&3\\3&1\end{array}\right|=(-1)\cdot(-8)=8\\ \\ \\ \delta_{31}=(-1)^{3+1}\cdot \left|\begin{array}{ccc}3&2\\2&3\end{array}\right|=1\cdot5=5\\ \\ \\ \delta_{32}=(-1)^{3+2}\cdot \left|\begin{array}{ccc}1&2\\2&3\end{array}\right|=(-1)\cdot (-1)=1\\ \\ \\ \delta_{33}=(-1)^{3+3}\cdot \left|\begin{array}{ccc}1&3\\2&2\end{array}\right|=1\cdot(-4)=-4$

$\displaystyle A^*= \left(\begin{array}{ccc}-1&7&-4\\-1&-5&8\\5&1&-4\end{array}\right)\\ \\ \\ A^{-1}= \frac{1}{12} \left(\begin{array}{ccc}-1&7&-4\\-1&-5&8\\5&1&-4\end{array}\right)$

$\displaystyle X=\frac{1}{12} \left(\begin{array}{ccc}-1&7&-4\\-1&-5&8\\5&1&-4\end{array}\right)\left(\begin{array}{ccc}5\\3\\4\end{array}\right)\\\\\\ X=\frac{1}{12}\left(\begin{array}{ccc}(-1)\cdot 5+7\cdot3+(-4)\cdot4\\(-1)\cdot5+(-5)\cdot3+8\cdot4\\5\cdot5+1\cdot3+(-4)\cdot4\end{array}\right) \\ \\ \\ X= \frac{1}{12} \left(\begin{array}{ccc}(-5)+21-16\\(-5)-15+32\\25+3-16\end{array}\right)\\\\\\X=\frac{1}{12} \left(\begin{array}{ccc}0\\12\\12\end{array}\right)$

$\displaystyle X=\left(\begin{array}{ccc} \frac{1}{12}\cdot0 \\ \\\frac{1}{12} \cdot12\\\\\frac{1}{12}\cdot12\end{array}\right)\\\\\\X=\left(\begin{array}{ccc}0\\1\\1\end{array}\right)$

$\displaystyle -------------------------------\\ \\ 3.~~~~\left\{\begin{array}{ccc}x+2y+3z=5\\3x+2y+z=3\\2x+3y+z=4\end{array}\right}\\ \\ \\ \left\{\begin{array}{ccc}x=-2y-3z+5\\3x+2y+z=3\\2x+3y+z=4\end{array}\right \Rightarrow \left\{\begin{array}{ccc}x=-2y-3z+5\\3(-2y-3z+5)+2y+z=3\\2(-2y-3z+5)+3y+z=4\end{array}\right \Rightarrow$

$\displaystyle\Rightarrow\left\{\begin{array}{ccc}x=-2y-3z+5\\-6y-9z+15+2y+z=3\\-4y-6z+10+3y+z=4\end{array}\right \Rightarrow\left\{\begin{array}{ccc}-4y-8z=-12~/:(-4)\\-y-5z=-6\end{array}\right \Rightarrow\\\\ \\\Rightarrow\left \{ {{y+2z=3} \atop {-y-5z=-6}} \right.~~~\Rightarrow\left \{ {{z=1}\atop {-y-5\cdot1=-6}}\right.\Rightarrow\left\{{{z=1}\atop{-y=-6+5}} \right.\Rightarrow\\~~~~~~-------\\~~~~~~~~~/~-3z=-3 \\ ~~~~~~~~~~~~~~~~~z=1$

$\displaystyle\Rightarrow \left\{{{z=1}\atop {y=1}}\right.\Rightarrow \left\{\begin{array}{ccc}z=1\\y=1\\x=(-2)\cdot1-3\cdot1+5\end{array}\right\Rightarrow\left\{\begin{array}{ccc}x=0\\y=1\\z=1\end{array}\right$