Question

Задание 2, под номером 17) помогите

Answer 1

$A=\left(\begin{array}{ccc}3&-5&-4\\1&-1&5\\4&-3&-6\end{array}\right) \\\\\\detA=3\cdot (6+15)+5(-6-20)-4(-3+4)=-71\ne 0\\\\A_{11}=21\; ,\; \; A_{12}=26\; ,\; \; A_{13}=1\\\\A_{21}=-18\; ,\; \; A_{22}=-2\; ,\; \; A_{23}=-11\\\\A_{31}=-29\; ,\; \; A_{32}=-19\; ,\; \; A_{33}=2\\\\\\A^{-1}=-\frac{1}{71}\cdot \left(\begin{array}{ccc}21&-18&-29\\26&-2&-19\\1&-11&2\end{array}\right)$


$A^{-1}\cdot A=-\frac{1}{71}\cdot \left(\begin{array}{ccc}21&-18&-29\\26&-2&-19\\1&-11&2\end{array}\right)\cdot \left(\begin{array}{ccc}3&-5&-4\\1&-1&5\\4&-3&-6\end{array}\right)=$

$=-\frac{1}{71}\cdot \left(\begin{array}{ccc}-71&0&0\\0&-71&0\\0&0&-71\end{array}\right) = \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)$

Answer 2

=3*(-1)*(-6)+(-5)*5*4+(-4)*1*(-3)-((-4)*(-1)*4+3*5*(-3)+(-5)*1*(-6))=18-100+12-(16-45+30)=30-100-16+45-30=75-146-71