Question

Solve using Cramer's Rule:

2x-y+2z=4
3x+2y+3z=16
x+3y+2z=12

• A. $x=\frac{-4}{3},y=0,z=\frac{20}{3}$
• B. $x=9,y=5,z=\frac{20}{3}$
  
• C. $x=\frac{20}{3},y=0,z=\frac{-4}{3}$
  
• D. $x=\frac{24}{7},y=\frac{20}{7},z=0$

Answer 1

The correct option is D $x=\frac{24}{7},y=\frac{20}{7},z=0$

Given equation are

$2x-y+2z=4$

$3x+2y+3z=16$

$x+3y+2z=12$

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

$=2(4-9)+1(6-3)+2(9-2)$

$=-10+3+14=7$

${\Delta }_x=\left|\begin{array}{ccc}4& -1& 2\\ 16& 2& 3\\ 12& 3& 2\end{array}\right|$

$=4(4-9)+1(32-36)+2(48-24)$

$=-20-4+48=24$

${\Delta }_y=\left|\begin{array}{ccc}2& 4& 2\\ 3& 16& 3\\ 1& 12& 2\end{array}\right|$

$=2(32-36)-4(6-3)+2(36-16)$

$=-8-12+40=20$

${\Delta }_z=\left|\begin{array}{ccc}2& -1& 4\\ 3& 2& 16\\ 1& 3& 12\end{array}\right|$

$=2(24-36)+1(36-16)+4(9-2)$

$=-48+20+28=0$

$\therefore x=\frac{{\Delta }_x}{\Delta },y=\frac{{\Delta }_y}{\Delta },z=\frac{{\Delta }_z}{\Delta }$

$⟹x=\frac{24}{7},y=\frac{20}{7},z=0$