Question

Помогите решить матрицы
M=3
N=7

Answer 1

$a); ; A=left(begin{array}{cc}-4&3\-7&10end{array}right)\\\detA=-4cdot 10+7cdot 3=-19\\A_{11}=10; ; ; ,; ; ; A_{12}=7; ; ; ,; ; ; A_{21}=-3; ; ; ,; ; ; A_{22}=-4\\\A^{-1}=-frac{1}{19}left(begin{array}{cc}10&-3\7&-4end{array}right)=left(begin{array}{cc}-frac{10}{19}&frac{3}{19}\-frac{7}{19}&frac{4}{19}end{array}right)\\\Acdot A^{-1}=left(begin{array}{cc}-4&3\-7&10end{array}right)cdot left(begin{array}{cc}-frac{10}{19}&frac{3}{19}\-frac{7}{19}&frac{4}{19}end{array}right)=left(begin{array}{cc}1&0\0&1end{array}right)=E$

$A=left(begin{array}{ccc}3&7&10\7&-4&3\2&1&2end{array}right)\\\detA=3(-8-3)-7(14-6)+10(7+8)=61ne 0\\\A_{11}=-11; ; ,; ; ; A_{12}=-8; ; ,; ; ; A_{13}=15\\A_{21}=-4; ; ,; ; ; A_{22}=-14; ; ,; ; A_{23}=11\\A_{31}=61; ; ,; ; A_{32}=61; ; ,; ; A_{33}=-61$

$A^{-1}=frac{1}{61}cdot left(begin{array}{ccc}-11&-4&61\-8&-14&61\15&11&-61end{array}right)\\\Acdot A^{-1}=frac{1}{61}cdot left(begin{array}{ccc}3&7&10\7&-4&3\2&1&2end{array}right)cdot left(begin{array}{ccc}-11&-4&61\-8&-14&61\15&11&-61end{array}right)=\\\=left(begin{array}{ccc}1&0&0\0&1&0\0&0&1end{array}right)=E$