Question

I am desperate i need help PLEASE!!
Solve the following system of equations.

3x – 2y + z = –6
x – 3y + z = –3
2x + y – 2z = –7

Answer 1

Can anyone help me??

Answer 2

wait

Answer 3

Do you know how to get the determinate of a matrix? If so, Cramer's rule is the easiest.
x=Dx/D
y=Dy/D
z=Dz/D
where D is thee determinate of the matrix
Dx is the determinate of the matrix where the x values are substituted by the solutions,
Dy is the determinate of the matrix where the y values are substituted by the solutions, and
Dz is the determinate of the matrix where the z values are substituted by the solutions.

Answer 4

are you set? stgreen posted the answer.

elimination is how a computer would solve this. (Cramer's rule is computationally very expensive)

But doing elimination by hand can be tedious and error prone, because you end up with fractions.

Answer 5

Here is how I would do it
x – 3y + z = –3
3x – 2y + z = –6
2x + y – 2z = –7
noticed I switched the order of the equations (that is allowed) so the one with a coeff on x is 1 (it makes things a little easier)

write down the coeffs

1 -3 1 | -3
3 -2 1 | -6
2 1 -2 | -7

Answer 6

i got the answe as -2,1,2?? is that not right??

Answer 7

now "eliminate" the values below the 1 in the first column.
you do this for the 2nd row by: 2nd row - 3* 1st row

3 -2 1 | -6
-3 - -9 -3 | --9
______________________
0 +7 -2 | +3 < -- this is your new 2nd row


to eliminate the 1st entry in the bottom column: do 3rd row - 2*1st row
2 1 -2 | -7
-2 +6 -2 | +6
___________________
0 7 -4 | -1 <-- this is your new 3rd row


so far we have
1 -3 1 | -3
0 +7 -2 | +3
0 7 -4 | -1

now use the 2nd row to eliminate the numbers below the 7 in the 3rd row

you do this by 3rd row - 1*2nd row
0 7 -4 | -1
-0 -7 +2 | -3
_________________
0 0 -2 | -4 <--- your new 3rd row


you now have
1 -3 1 | -3
0 +7 -2 | +3
0 0 -2 | -4

if we put the letters back in, you can solve. start at the bottom and work up