solve the system of equations x-y+2z=1 2y-3z=1 3x-2y+4z=2 - Maths - Determinants

Question

solve the system of equations
x-y+2z=1
2y-3z=1
3x-2y+4z=2

Gursheen kaur. · Accepted Answer

The given system of equations is x-y+2z=1 2y-3z=1 3x-2y+4z=2 By Cramer's Rule,

D=1(8-6)-(-1)(0+9)+2(0-6) D=2+9-12 D=-1 Now,

$D_{1}$ =1(8-6)-(-1)(4+6)+2(-2-4)=2+10-12=0

$D_{2}$ =1(4+6)-1(0+9)+2(0-3)=10-9-6=-5

$D_{3}$ =1(4+2)-(-1)(0-3)+1(0-6)=6-3-6=-3 So, by Cramer's rule, $x = \frac{D_{1}}{D} = \frac{0}{- 1} = 0 y = \frac{D_{2}}{D} = \frac{- 5}{- 1} = 5 z = \frac{D_{3}}{D} = \frac{- 3}{- 1} = 3 x = 0, y = 5, z = 3$

solve the system of equationsx-y+2z=1 2y-3z=1 3x-2y+4z=2

solve the system of equations
x-y+2z=1

2y-3z=1

3x-2y+4z=2