Необходимо решить задачу по высшей математике

Question 1

Question 2

1.
$12\cdot \left ( \begin{matrix} 1&2&1\\ 3&4&5\\ 6&7&8 \end{matrix}\right )- \left ( \begin{matrix} 1&2&1\\ 3&1&3\\ 2&1&2 \end{matrix}\right )\cdot\left ( \begin{matrix} 1&3&5\\ 1&2&1\\ 2&1&2 \end{matrix}\right ) =\\ =\left ( \begin{matrix} 12&24&12\\ 36&48&60\\ 72&84&96 \end{matrix}\right )- \left ( \begin{matrix} 5&8&9\\ 10&14&22\\ 7&10&15 \end{matrix}\right )= \left ( \begin{matrix} 7&16&3\\ 26&34&38\\ 65&74&81 \end{matrix}\right )$

2.
$\left | \begin{matrix} 1&2\\ 4&6 \end{matrix}\right |=1*6-4*2=6-8=-2\\\\ \left | \begin{matrix} x&2\\ 3&6 \end{matrix}\right |=6x-6=0 \Rightarrow x=1\\\\ \left | \begin{matrix} 1&x\\ 6&2 \end{matrix}\right |=2-6x\ \textless \ 0 \Rightarrow x\ \textgreater \ \frac{1}{3}$

3.
$\left | \begin{matrix} 1&2&4\\ 6&8&10\\ 3&5&1 \end{matrix}\right |= 1\cdot\left | \begin{matrix} 8&10\\ 5&1 \end{matrix}\right |- 2\cdot\left | \begin{matrix} 6&10\\ 3&1 \end{matrix}\right |+ 4\cdot\left | \begin{matrix} 6&8\\ 3&5 \end{matrix}\right |=\\ =1*(8-50)-2*(6-30)+4*(30-24)=\\=-42+48+24=30$

$\left | \begin{matrix} 1&2&3\\ 4&x&1\\ 1&2&1 \end{matrix}\right |= 1\cdot\left | \begin{matrix} x&1\\ 2&1 \end{matrix}\right |- 2\cdot\left | \begin{matrix} 4&1\\ 1&1 \end{matrix}\right |+ 3\cdot\left | \begin{matrix} 4&x\\ 1&2 \end{matrix}\right |=\\ =1*(x-2)-2*(4-1)+3*(8-x)=\\=x-2-6+24-3x=16-2x=0\\ x=8$

$\left | \begin{matrix} 2&1&x\\ 1&2&0\\ 3&2&1 \end{matrix}\right |= 2\cdot\left | \begin{matrix} 2&0\\ 2&1 \end{matrix}\right |- 1\cdot\left | \begin{matrix} 1&0\\ 3&1 \end{matrix}\right |+ x\cdot\left | \begin{matrix} 1&2\\ 3&2 \end{matrix}\right |=\\ =2*(2)-1*(1)+x*(2-6)=\\=1-4x\ \textless \ 0\\ x\ \textgreater \ \frac{1}{4}$