Question

ПОЖАЛУЙСТА ПОМОГИТЕ РЕШИТЬ ВЫСШУЮ МАТЕМАТИКУ! ЗАРАНЕЕ СПАСИБО!

Answer 1

$1)\; \; M_{23}=\left|\begin{array}{ccc}6&2&4\\2&4&-6\\3&0&4\end{array}\right|=6(16-0)-2(8+18)+4(0-12)=-4\\\\\\2)\; \; AB=\left[\begin{array}{ccc}3&-7&2\\1&8&3\\4&-2&3\end{array}\right]\cdot \left[\begin{array}{ccc}0&5&-3\\2&4&1\\2&1&-5\end{array}\right]=\left[\begin{array}{ccc}-10&-11&-26\\22&40&-10\\2&15&-29\end{array}\right]$

$BA=\left[\begin{array}{ccc}0&5&-3\\2&4&1\\2&1&-5\end{array}\right]\cdot \left[\begin{array}{ccc}3&-7&2\\1&8&3\\4&-2&3\end{array}\right]=\left[\begin{array}{ccc}=7&46&6\\14&16&19\\-13&4&-8\end{array}\right]\\\\\\detA=\left|\begin{array}{ccc}3&-7&2\\1&-8&3\\4&-2&3\end{array}\right|=3(-24+6)+7(3-12)+2(-2+32)=-57\\\\\\A_{11}=17\; \; ,\; \; A_{12}=9\; \; ,\; \; A_{13}=30\\A_{21}=17\; \; ,\; \; A_{22}=1\; \; ,\; \; A_{23}=-22\\A_{31}=-5\; \; ,\; \; A_{32}=-7\; \; ,\; \; A_{33}=-17$

$A^{-1}=-\frac{1}{57}\cdot \left[\begin{array}{ccc}-18&17&-5\\9&1&-7\\30&-22&-17\end{array}\right] =\left[\begin{array}{ccc}6/19&-17/57&5/57\\-3/19&-1/57&7/57\\-10/19&22/57&17/57\end{array}\right]$

$3)\; \; \Delta =\left|\begin{array}{ccc}3&1&1\\-3&5&6\\1&-4&-2\end{array}\right|=3(-10+24)-(6-6)+(12-5)=49\\\\49\ne 0\; \; \to \; \; sistema\; sovmestna\\\\\\\Delta_{1}=\left|\begin{array}{ccc}-4&1&1\\36&5&6\\-19&-4&-2\end{array}\right|=-4\cdot 14-42-49=-147\\\\\\\Delta_{2}=\left|\begin{array}{ccc}3&-4&1\\-3&36&6\\1&-19&-2\end{array}\right|=3\cdot 42+4\cdot 0+21=147\\\\\\\Delta _3=\left|\begin{array}{ccc}3&1&-4\\-3&5&36\\1&-4&-19\end{array}\right|=3\cdot 49-21-4\cdot 7=98$

$x_1=\frac{-147}{49}=-3\; ,\; \; x_2= \frac{147}{49}=3\; ,\; \; x_3=\frac{98}{49}=2$