Question

Необходимо решить задание по высшей математике

Answer 1

1. 
$3\cdot \left(\begin{array}{ccc}2&1&3\\1&3&4\\2&1&1\end{array}\right) + \left(\begin{array}{ccc}1&-1&0\\1&2&-1\\-1&1&2\end{array}\right) \cdot \left(\begin{array}{ccc}1&0&1\\0&1&0\\1&0&1\end{array}\right) =\\\\\\= \left(\begin{array}{ccc}6&3&9\\3&9&12\\6&3&3\end{array}\right) + \left(\begin{array}{ccc}1&-1&1\\0&2&0\\1&1&1\end{array}\right)= \left(\begin{array}{ccc}7&2&10\\3&11&12\\7&4&4\end{array}\right)$

2.
$\left|\begin{array}{cc}1&2\\4&7\end{array}\right|=1\cdot 7 - 4\cdot 2=-1.$

$\left|\begin{array}{cc}2&3\\12&x\end{array}\right|=2\cdot x - 12\cdot 3=2\cdot x - 36=0 \, \Rightarrow \, x = 18.$

$\left|\begin{array}{cc}1&2\\4&x-5\end{array}\right|=1\cdot (x-5) - 4\cdot 2 =x-5-8=x-13 \ \textless \ 0\, \Rightarrow \, x \ \textless \ 13.$

3.
$\left|\begin{array}{ccc}1&2&3\\2&3&1\\1&3&2\end{array}\right|= 1\cdot \left|\begin{array}{cc}3&1\\3&2\end{array}\right| - 2\cdot \left|\begin{array}{cc}2&1\\1&2\end{array}\right| + 3\cdot \left|\begin{array}{cc}2&3\\1&3\end{array}\right| = 3 - 6 + 9 =6.$

$\left|\begin{array}{ccc}1&1&2\\2&3&1\\3&2&x\end{array}\right|= 1\cdot \left|\begin{array}{cc}3&1\\2&x\end{array}\right| - 1\cdot \left|\begin{array}{cc}2&1\\3&x\end{array}\right| + 2\cdot \left|\begin{array}{cc}2&3\\3&2\end{array}\right| = \\\\\\ = 3x-2-(2x-3) -10=x-9=0 \, \Rightarrow\,x = 9.$

$\left|\begin{array}{ccc}1&2&4\\3&x&5\\3&1&2\end{array}\right|= 1\cdot \left|\begin{array}{cc}x&5\\1&2\end{array}\right| - 2\cdot \left|\begin{array}{cc}3&5\\3&2\end{array}\right| + 4\cdot \left|\begin{array}{cc}3&x\\3&1\end{array}\right| = \\\\\\ = 2x-5-2(6-15)+4(3-3x) = -10x+25\ \textless \ 0 \Rightarrow x \ \textgreater \ 2.5.$