Question

Помогите решить алгебру,срочно

Answer 1

$\left\{\begin{array}{ccc}2x_1-x_2+2x_3=0\\4x_1+x_2+4x_3=6\\x_1+x_2+2x_3=4\end{array}\right$

$a)\ \Delta =\left|\begin{array}{ccc}2&-1&2\\4&1&4\\1&1&2\end{array}\right|=2\, (2-4)+1\, (8-4)+2(4-1)=6\ne 0\\\\\\\Delta _1=\left|\begin{array}{ccc}0&-1&2\\6&1&4\\4&1&2\end{array}\right|=1\, (12-16)+2(6-4)=0$

$\Delta _2=\left|\begin{array}{ccc}2&0&2\\4&6&4\\1&4&2\end{array}\right|=2(12-16)+2(16-6)=12\\\\\\\Delta _3=\left|\begin{array}{ccc}2&-1&0\\4&1&6\\1&1&4\end{array}\right|=2(4-6)+1\, (16-6)=6$

$x_1=\dfrac{\Delta _1}{\Delta }=\dfrac{0}{6}=0\ ,\ \ x_2=\dfrac{\Delta _2}{\Delta }=\dfrac{12}{6}=2\ ,\ \ x_3=\dfrac{\Delta _3}{\Delta }=\dfrac{6}{6}=1$

$b)\ \ A_{11}=-2\ \ ,\ \ A_{12}=-4\ \ ,\ \ A_{13}=3\\{}\ \ \ \ \ A_{21}=4\ \ ,\ \ A_{22}=2\ \ ,\ \ A_{23}=-3\\{}\ \ \ \ \ A_{31}=-6\ \ ,\ \ A_{32}=0\ \ ,\ \ A_{33}=6$

$A^{-1}=\dfrac{1}{6}\cdot \left[\begin{array}{ccc}-2&4&-6\\-4&2&0\\3&-3&6\end{array}\right]$

$X=A^{-1}\cdot B=\dfrac{1}{6}\cdot \left[\begin{array}{ccc}-2&4&-6\\-4&2&0\\3&-3&6\end{array}\right]\cdot \left[\begin{array}{l}0&6&4\end{array}\right]=\dfrac{1}{6}\cdot \left[\begin{array}{l}0&12&6\end{array}\right]=\left[\begin{array}{l}0&2&1\end{array}\right]$

$c)\ \ \left[\begin{array}{cccc}2&-1&2&\ |\ 0\\4&1&4&\ |\ 6\\1&1&2&\ |\ 4\end{array}\right]\sim \left[\begin{array}{cccc}1&1&2&\ |\ 5\\2&-1&2&\ |\ 0\\4&1&4&\ |\ 6\end{array}\right]\sim \left[\begin{array}{cccc}1&1&2&|4\\0&-3&-2&\ |\ -8\\0&-3&-4&\ |-10\end{array}\right]\sim$

$\sim \left[\begin{array}{cccc}1&1&2&\ |\ 4\\0&3&2&\ |\ 8\\0&0&2&\ |\ 2\end{array}\right]\sim \left[\begin{array}{cccc}1&1&2&\ |\ 4\\0&3&2&\ |\ 8\\0&0&1&\ |\ 1\end{array}\right]$

$\left\{\begin{array}{ccc}x_1+x_2+2x_3=4\\{}\ \ 3x_2+2x_3=8\\{}\ \ \ \ \ \ x_3=1\end{array}\right\\\\\\3x_2=8-2x_3=8-2=6\ \ ,\ \ x_2=6:3=2\\\\x_1=4-2x_3-x_2=4-2-2=0$

$Otvet:\ \ x_1=0\ ,\ x_2=2\ .\ x_3=1\ .$