Question

Решить матричное уравнение

Answer 1

$X\cdot A=B\ \ \to \ \ \ X\cdot \underbrace {A\cdot A^{-1}}_{E}=B\cdot A^{-1}\ \ ,\ \ \ X=B\cdot A^{-1}\\\\\\A=\left(\begin{array}{ccc}-1&-4&3\\-2&1&-5\\0&-1&2\end{array}\right)\ \ \ B=\left(\begin{array}{ccc}9&9&6\\10&2&18\\11&6&15\end{array}\right)\ \ \ E=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)$

$detA=\left|\begin{array}{ccc}-1&-4&3\\-2&\ 1&-5\\\ 0&-1&2\end{array}\right|=-1(2-5)+4(-4-0)+3(2-0)=-7\ne 0\\\\\\A_{11}=-3\ ,\ \ A_{12}=4\ \ ,\ \ A_{13}=2\\\\A_{21}=5\ \ ,\ \ A_{22}=-2\ \ ,\ \ A_{23}=-1\\\\A_{31}=17\ \ ,\ \ A_{32}=-11\ \ ,\ \ A_{33}=-9\\\\\\A^{-1}=-\dfrac{1}{7}\cdot \left(\begin{array}{ccc}-3&5&17\\4&-2&-11\\2&-1&-9\end{array}\right)$

$X=B\cdot A^{-1}=-\dfrac{1}{7}\cdot \left(\begin{array}{ccc}9&9&6\\10&2&18\\11&6&15\end{array}\right)\cdot \left(\begin{array}{ccc}-3&5&17\\4&-2&-11\\2&-1&-9\end{array}\right)=\\\\\\=-\dfrac{1}{7}\cdot \left(\begin{array}{ccc}21&21&0\\14&28&-14\\21&28&-14\end{array}\right)=\left(\begin{array}{ccc}-3&-3&0\\-2&-4&2\\-3&-4&2\end{array}\right)$