# Need help on matrices using cramer's rule

I am learning how to solve a matrice using cramer's rule, and not sure if this is the correct answer.

Solve the following systems of equations
x - y + 3z = 8
3x + y - 2z = -2
2x + 4y + z = 0
so I figured out the solution is x = 1, y = -1, z = 2

but is this equation consider a
dependent equation that all solutions that satisfy x - y + 3z = 8 ??

thanks

dextercioby
Homework Helper
qdv said:
I am learning how to solve a matrice using cramer's rule, and not sure if this is the correct answer.

Solve the following systems of equations
x - y + 3z = 8
3x + y - 2z = -2
2x + 4y + z = 0
so I figured out the solution is x = 1, y = -1, z = 2

thanks
It's okay.The solution satisfies all equations and it's unique,therefore...Congratulations!! Daniel.

Online calculator

I solve it using a online calculator and I got

Cramer rule's solver step by step
Coeficients Matrix
1 -1 3 8
3 1 -2 -2
2 4 1 0
Δ = determinant1 -1 3
3 1 -2
2 4 1
Δ sub x = det8 -1 3
-2 1 -2
0 4 1
Δ sub y = det1 8 3
3 -2 -2
2 0 1
Δ sub z = det1 -1 8
3 1 -2
2 4 0
Δ = det1 -1 3
3 1 -2
2 4 1
1 -1 3
3 1 -2

[(1) (1) (1) + (3) (4) (3) + (2) (-1) (-2)] - [(3) (-1) (1) + (1) (4) (-2) + (2) (1) (3)]
(1) + (36) + (4)- (-3) + (-8) + (6)
( 41) - ( -5)
Δ = 46

Δx = det8 -1 3
-2 1 -2
0 4 1
8 -1 3
-2 1 -2

[(8) (1) (1) + (-2) (4) (3) + (0) (-1) (-2)] - [(-2) (-1) (1) + (8) (4) (-2) + (0) (1) (3)]
(8) + (-24) + (0)- (2) + (-64) + (0)
( -16) - ( -62)
Δx = 46

Δy = det1 8 3
3 -2 -2
2 0 1
1 8 3
3 -2 -2

[(1) (-2) (1) + (3) (0) (3) + (2) (8) (-2)] - [(3) (8) (1) + (1) (0) (-2) + (2) (-2) (3)]
(-2) + (0) + (-32)- (24) + (0) + (-12)
( -34) - ( 12)
Δy = -46

Δz = det1 -1 8
3 1 -2
2 4 0
1 -1 8
3 1 -2

[(1) (1) (0) + (3) (4) (8) + (2) (-1) (-2)] - [(3) (-1) (0) + (1) (4) (-2) + (2) (1) (8)]
(0) + (96) + (4)- (0) + (-8) + (16)
( 100) - ( 8)
Δz = 92

x =46/46

y =-46/46

z =-46/46

x =1

y =-1

z =2

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www.algebrasolver.totalh.com

Last edited:
HallsofIvy
Homework Helper
I am learning how to solve a matrice using cramer's rule, and not sure if this is the correct answer.

Solve the following systems of equations
x - y + 3z = 8
3x + y - 2z = -2
2x + 4y + z = 0
so I figured out the solution is x = 1, y = -1, z = 2

but is this equation consider a
dependent equation that all solutions that satisfy x - y + 3z = 8 ??

thanks
What do you mean by "this equation"? It's not at all clear what your question is. Yes, as dextercioby said, and you could easily have checked, x= 1, y= -1, z= 2 satisfies the three equations and, since the determinant of coefficients is not 0, is the only solution to that system of equations.