50 ! разрешите систему уравнений: а) методом крамера б) матричным методом в) методом гаусса 4х + у-3z = 9 х + у -z = -2 8х + 3y-6z = 12

Ответ:

GoRussia228

08.10.2020 04:25

$a)\; \; \Delta = \left|\begin{array}{lll}4&1&-3\\1&1&-1\\8&3&-6\end{array}\right|=4(-6+3)-(-6-8)-3(3-8)=1\\\\\\\Delta _{x}= \left|\begin{array}{ccc}9&1&-3\\-2&1&-1\\12&3&-6\end{array}\right|=9(-6+3)-(12+12)-3(-6-12)=3\\\\\\\Delta _{y}= \left|\begin{array}{lll}4&9&-3\\1&-2&-1\\8&12&-6\end{array}\right|=4(12+12)-9(-6+8)-3(12+16)=-6\\\\\\\Delta _{z}= \left|\begin{array}{lll}4&1&9\\1&1&-2\\8&3&12\end{array}\right|=4(12+6)-(12+16)+9(3-8)=-1\\\\\\x= \frac{\Delta _{x}}{\Delta }=\frac{3}{1}=3\; ,\; y=\frac{\Delta _{y}}{\Delta }=\frac{-6}{1}=-6\; ,\; z=\frac{\Delta _{z}}{\Delta }=\frac{-1}{1}=-1$

$b)\; \; detA= \left|\begin{array}{lll}4&1&-3\\1&1&-1\\8&3&-6\end{array}\right|=1\; \; \; \; \; (detA=\Delta )\\\\A_{11}=-3\; \; \; A_{12}=-2\; \; \; A_{13}=-5\\A_{21}=-3\; \; \; A_{22}=0\; \; \; \; A_{23}=-4\\A_{31}=2\; \; \; \; \; A_{32}=1\; \; \; \; A_{33}=3$

$A^{-1}= \left(\begin{array}{lll}-3&-3&2\\-2&0&1\\-5&-4&3\end{array}\right)$

$X=A^{-1}\cdot B=$

$=\left(\begin{array}{lll}-3&-3&2\\-2&0&1\\-5&-4&3\end{array}\right)\cdot \left(\begin{array}{l}9\\-2\\12\end{array}\right)=\left(\begin{array}{c}3\\-6\\-1\end{array}\right)$

$c)\; \; \left(\begin{array}{llll}4&1&-3&|9\\1&1&-1&|-2\\8&3&-6&|12\end{array}\right)\sim \left(\begin{array}{llll}1&1&-1&|-2\\0&-3&\, 1&|17\\0&1&0&|-6\end{array}\right)\sim \\\\\\\sim \left(\begin{array}{llll}1&1&-1&|-2\\0&1&\; 0&|-6\\0&0&\; 1&|-1\end{array}\right)\; \; \to \\\\\\z=-1\\y=-6\\x+y-z=-2\; \; \to \; \; x-6+1=-2\; ,\; \; x=3\\\\Otvet:\; \; x=3\; ,\; y=-6\; ,\; z=-1\; .$