Question

Найти определитель по методу Крамера

Answer 1

$\Delta =\left|\begin{array}{ccc}5&-3&4\\2&-1&-1\\1&-2&1\end{array}\right|=5(-1-2)+3(2+1)+4(-4+1)=-18\\\\\\\Delta _{x}=\left|\begin{array}{ccc}6&-3&4\\0&-1&-1\\0&-2&1\end{array}\right|=6(-1-2)=-18\\\\\\\Delta _{y}=\left|\begin{array}{ccc}5&6&4\\2&0&-1\\1&0&1\end{array}\right|=-6(2+1)=18\\\\\\\Delta _{z}=\left|\begin{array}{ccc}5&-3&6\\2&-1&0\\1&-2&0\end{array}\right|=6(-4+1)=-18$

$x=\frac{\Delta _{x}}{\Delta }=\frac{-18}{-18}=1\; ,\; \; y=\frac{\Delta _{y}}{\Delta }=\frac{18}{-18}=-1\; \; ,\; \; z=\frac{\Delta _{z}}{\Delta }=\frac{-18}{-18}=1\\\\Otvet:\; \; (1;-1;1)\; .$