solve the system of equations: 2x-y+z=-2 x+2y-z=6 3x+y-2z=0 (By Determininates only) D[ ] Dx[ ] Dy[ ]?

Question

solve the system of equations: 2x-y+z=-2 x+2y-z=6 3x+y-2z=0 (By Determininates only) D[ ] Dx[ ] Dy[ ]

Answer 1

2x - y + z = -2
x + 2y - z = 6
3x + y - 2z = 0

Set up as matrices:

2 -1 1 : -2
1 2 -1 : 6
3 1 -2 : 0

The use row manipulation till you get to row eschelon form:

STEP 1 - add row 2 to row 1:

3 1 0 : 4
1 2 -1 : 6
3 1 -2 : 0

STEP 2 - subtract row 1 from row 3:

3 1 0 : 4
1 2 -1 : 6
0 0 -2 : -4

STEP 3 - divide row 3 by -2:

3 1 0 : 4
1 2 -1 : 6
0 0 1 : 2

STEP 4 - add row 3 to row 2:

3 1 0 : 4
1 2 0 : 8
0 0 1 : 2

STEP 5 - subtract 3 times row 2 from row 1:

0 -5 0 : -20
1 2 0 : 8
0 0 1 : 2

STEP 6 - divide row 1 by -5:

0 1 0 : 4
1 2 0 : 8
0 0 1 : 2

STEP 7 - subtract 2 times row 1 from row 2:

0 1 0 : 4
1 0 0 : 0
0 0 1 : 2

STEP 8 - flip rows 1 and 2:

1 0 0 : 0
0 1 0 : 4
0 0 1 : 2

So your answers are:
x = 0
y = 4
z = 2

Hope that helped!

~ ♥ ~

Answer 2

2x - y + z = -2
x + 2y - z = 6
3x + y - 2z = 0

3x + y - 2z = 0
y = -3x + 2z

2x - (-3x + 2z) + z = -2
x + 2(-3x + 2z) - z = 6

2x + 3x - 2z + z = -2
x - 6x + 4z - z = 6

5x - z = -2
-5x + 3z = 6

2z = 4
z = 2

5x - 2 = -2
5x = 0
x = 0

y = -3x + 2z
y = -3(0) + 2(2)
y = 0 + 4
y = 4

x = 0
y = 4
z = 2

By going to www.quickmath.com, clicking on Determinants, typing in your problem, it states that your problem must be in a square form.

You have a 4 by 3 matrix.

Answer 3

d[-8]dx[4]dy[-2] i don't know