Question

Необходимо решить задание по высшей математике.

Необходимо с последовательным выполнением шагов, а не просто ответ

Answer 1

$1)\; \; 10\cdot \left(\begin{array}{ccc}1&2&4\\3&2&1\\2&1&3\end{array}\right) -\left(\begin{array}{ccc}1&1&2\\2&1&2\\1&2&1\end{array}\right)\cdot \left(\begin{array}{ccc}1&1&1\\2&2&2\\3&2&1\end{array}\right)=\\\\\\= \left(\begin{array}{ccc}10&20&40\\30&20&10\\20&10&30\end{array}\right)- \left(\begin{array}{ccc}9&7&5\\10&8&6\\8&7&6\end{array}\right)= \left(\begin{array}{ccc}1&13&35\\20&12&4\\12&3&24\end{array}\right)$

$2)\; \; \left|\begin{array}{cc}1&2\\3&4\end{array}\right|=1\cdot 4-3\cdot 2=4-6=-2\\\\b)\; \; \left|\begin{array}{cc}1&6\\8&x\end{array}\right|\ \textless \ 0\; ,\; \; x-48\ \textless \ 0\; ,\; \; x\ \textless \ 48\\\\c)\; \; \left|\begin{array}{cc}1&4\\10&x\end{array}\right|\ \textless \ 0\; ,\; \; x-40\ \textless \ 0\; ,\; \; x\ \textless \ 40\\\\\\3)\; \; \left|\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right|=\left|\begin{array}{cc}5&6\\8&9\end{array}\right|-2\cdot \left|\begin{array}{cc}4&6\\7&9\end{array}\right|+3\cdot \left|\begin{array}{cc}4&5\\7&8\end{array}\right|=\\\\=-3-2\cdot -2\cdot (-6)+3\cdot (-3)=0$

$b)\; \; \left|\begin{array}{ccc}1&3&5\\7&x&9\\3&2&1\end{array}\right|=0\\\\\\-3\cdot \left|\begin{array}{cc}7&9\\3&1\end{array}\right|+x\cdot \left|\begin{array}{cc}1&5\\3&1\end{array}\right|-2\cdot \left|\begin{array}{cc}1&5\\7&9\end{array}\right|=\\\\=-3\cdot (-20)+x\cdot (-14)-2\cdot (-26)=112-14x=0\\\\14x=112\; ,\; \; x=8$

$c)\; \; \left|\begin{array}{ccc}1&2&3\\4&5&x\\3&2&1\end{array}\right|\ \textless \ 0\\\\\\3\cdot \left|\begin{array}{cc}4&5\\3&2\end{array}\right|-x\cdot \left|\begin{array}{cc}1&2\\3&2\end{array}\right|+\left|\begin{array}{cc}1&2\\4&5\end{array}\right|\ \textless \ 0\\\\\\3\cdot (-7)-x\cdot (-4)+(-3)\ \textless \ 0\\\\4x-24\ \textless \ 0\\\\4x\ \textless \ 24\; ,\; \; x\ \textless \ 6$