Question

Необходимо решить задачу по высшей математике

Answer 1

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

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

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

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