Математика: Решите матрчное уравнение и распишите решение

Решите матрчное уравнение и распишите решение

Ответы

dogsno 12.02.2021 19:05

$x = \left(\begin{array}{c}-2\\1\\-1\end{array}\right)$

Пошаговое объяснение:

$\left(\begin{array}{cc}2&-1\\4&5\\0&2\end{array}\right) *\left(\begin{array}{c}1\\3\end{array}\right) + \left(\begin{array}{ccc}-1&2&4\\1&0&-1\\2&-1&3\end{array}\right)*x = \left(\begin{array}{c}-1\\18\\-2\end{array}\right)\\\\\\\left(\begin{array}{c}2\cdot1+(-1)\cdot3\\4\cdot1+5\cdot3\\0\cdot1+2\cdot3\end{array}\right) + \left(\begin{array}{ccc}-1&2&4\\1&0&-1\\2&-1&3\end{array}\right)*x = \left(\begin{array}{c}-1\\18\\-2\end{array}\right)$

$\left(\begin{array}{c}-1\\19\\6\end{array}\right) + \left(\begin{array}{ccc}-1&2&4\\1&0&-1\\2&-1&3\end{array}\right)*x = \left(\begin{array}{c}-1\\18\\-2\end{array}\right)\\\\\\\left(\begin{array}{ccc}-1&2&4\\1&0&-1\\2&-1&3\end{array}\right)*x = \left(\begin{array}{c}-1\\18\\-2\end{array}\right)-\left(\begin{array}{c}-1\\19\\6\end{array}\right)$

$\left(\begin{array}{ccc}-1&2&4\\1&0&-1\\2&-1&3\end{array}\right)*x = \left(\begin{array}{c}0\\-1\\-8\end{array}\right)\\\\\\x = \left(\begin{array}{ccc}-1&2&4\\1&0&-1\\2&-1&3\end{array}\right)^{-1}*\left(\begin{array}{c}0\\-1\\-8\end{array}\right)$

Найдем обратную матрицу методом алгебраических дополнений (метод Крамера):

$A = \left(\begin{array}{ccc}-1&2&4\\1&0&-1\\2&-1&3\end{array}\right)\\\\\\detA = \left|\begin{array}{ccc}-1&2&4\\1&0&-1\\2&-1&3\end{array}\right| = \left|\begin{array}{ccc}-1&2&4\\0&2&3\\0&3&11\end{array}\right| = \left|\begin{array}{ccc}-1&2&4\\0&2&3\\0&0&\frac{13}{2} \end{array}\right| = -1\cdot2\cdot\frac{13}{2} =-13\\\\\\C_{11} = (-1)^{1+1}\left|\begin{array}{cc}-0&-1\\-1&3 \end{array}\right| = 0\cdot3-(-1)\cdot(-1) = -1\\\\$

$C_{12} = -1\cdot\left|\begin{array}{cc}1&-1\\2&3 \end{array}\right| = -1\cdot3+2\cdot(-1) = -5\\\\C_{13} = 1\cdot\left|\begin{array}{cc}1&0\\2&-1 \end{array}\right| = 1\cdot(-1)-0\cdot2 = -1\\\\C_{21} = -1\cdot\left|\begin{array}{cc}2&4\\-1&3 \end{array}\right| = -2\cdot3+(-1)\cdot4 = -10\\\\C_{22} = 1\cdot\left|\begin{array}{cc}-1&4\\2&3 \end{array}\right| = (-1)\cdot3-4\cdot2 = -11\\\\C_{23} = -1\cdot\left|\begin{array}{cc}-1&2\\2&-1 \end{array}\right| = -(-1)\cdot(-1)+2\cdot2 = 3\\\\$

$C_{31} = 1\cdot\left|\begin{array}{cc}2&4\\0&-1 \end{array}\right| = 2\cdot(-1)-0\cdot4 = -2\\\\C_{32} = -1\cdot\left|\begin{array}{cc}-1&4\\1&-1 \end{array}\right| = -(-1)\cdot(-1)+1\cdot4 = 3\\\\C_{33} = 1\cdot\left|\begin{array}{cc}-1&2\\1&0 \end{array}\right| = 0\cdot(-1)-1\cdot2 = -2\\\\\\A^{-1} = \frac{1}{detA}\left(\begin{array}{ccc}C_{11}&C_{12}&C_{13}\\C_{21}&C_{22}&C_{23}\\C_{31}&C_{32}&C_{33}\end{array}\right)^T =$

$= \frac{1}{-13}\left(\begin{array}{ccc}-1&-5&-1\\-10&-11&3\\-2&3&-2\end{array}\right)^T =\frac{1}{13}\left(\begin{array}{ccc}1&10&2\\5&11&-3\\1&-3&2\end{array}\right)$

Тогда:

$x = \frac{1}{13} * \left(\begin{array}{ccc}1&10&2\\5&11&-3\\1&-3&2\end{array}\right)*\left(\begin{array}{c}0\\-1\\-8\end{array}\right) = \frac{1}{13} *\left(\begin{array}{c}1\cdot0+10\cdot(-1)+2\cdot(-8)\\5\cdot0+11\cdot(-1)+(-3)\cdot(-8)\\1\cdot0+(-3)\cdot(-1)+2\cdot(-8)\end{array}\right) =\\\\\\= \frac{1}{13}*\left(\begin{array}{c}-26\\13\\-13\end{array}\right) = \left(\begin{array}{c}-2\\1\\-1\end{array}\right)\\\\\\x = \left(\begin{array}{c}-2\\1\\-1\end{array}\right)$