Question

Solve equations by Cramer's rule
$x+y+z=2,x-2y+z=3,2x-y-3z=-1$

Answer 1

$x+y+z=2$,$x-2y+z=3$,$2x-y-3z=-1$

$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& -2& 1\\ 2& -1& -3\end{array}\right|$

$=1(6+1)-1(-3-2)+1(-1+4)$

$=7+5+3=15$

${\Delta }_x=\left|\begin{array}{ccc}2& 1& 1\\ 3& -2& 1\\ -1& -1& -3\end{array}\right|$

$=2(6+1)-1(-9+1)+1(-3-2)$

$=14+8-5=17$

$x=\frac{{\Delta }_x}{\Delta }=\frac{17}{15}$

${\Delta }_y=\left|\begin{array}{ccc}1& 2& 1\\ 1& 3& 1\\ 2& -1& -3\end{array}\right|$

$=1(-9+1)-2(-3-2)+1(-1-6)$

$=-8+10-7=-5$

$y=\frac{{\Delta }_y}{\Delta }=\frac{-5}{15}=\frac{-1}{3}$

${\Delta }_z=\left|\begin{array}{ccc}1& 1& 2\\ 1& -2& 3\\ 2& -1& -1\end{array}\right|$

$=1(2+3)-1(-1-6)+2(-1+4)$

$=5+7+6=18$

$z=\frac{{\Delta }_z}{\Delta }=\frac{18}{15}=\frac{6}{5}$

$\therefore x=\frac{17}{15},y=\frac{-1}{3}$ and $z=\frac{6}{5}$