SOLUTION: Hello - this question I am stuck on. Solve x, y, z. xy - 2(y^-2) + 3zy = 8 2xy - 3(y^-2) + 2zy = 7 -xy + (y^-2) + 2zy = 4 I first made a matrix: 1 -2 3 8

Question 24539: Hello - this question I am stuck on.
Solve x, y, z.
xy - 2(y^-2) + 3zy = 8
2xy - 3(y^-2) + 2zy = 7
-xy + (y^-2) + 2zy = 4
I first made a matrix:
1 -2 3 8
2 -3 2 7
-1 1 2 4
Then reduced and got
1 0 0 5/11
0 1 0 -7/11
0 0 1 23/11
This would make:
xy = 5/11
(y^-2) = -7/11
zy = 23/11
then solve by algebra and substitution.
This this correct?
Thanks
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
EXCELLENT..YOU HAVE CORRECTLY IDENTIFIED THE PROBLEM AND METHOD OF SOLUTION..IT TAKES A GOOD EYE TO DETECT THIS AS 3 VARIABLE SIMULTANEOUS EQUATION IN XY,1/Y^2,YZ.YOUR MATRIX REDUCTION IS ALSO OK ,BUT I THINK YOU MADE SOME ARITHMAICAL ERROR IN DOING SO.THE FINAL REDUCED MATRIX IS ACTUALLY AS GIVEN BELOW.SEE MY COMMENTS.....
Solve x, y, z.
xy - 2(y^-2) + 3zy = 8
2xy - 3(y^-2) + 2zy = 7
-xy + (y^-2) + 2zy = 4
I first made a matrix: ..VERY GOOD
1 -2 3 8
2 -3 2 7
-1 1 2 4
Then reduced and got
1 0 0 5/11
0 1 0 -7/11
0 0 1 23/11
THIS SHOULD BE
1 0 0 5
0 1 0 3
0 0 1 3
YOU CAN SUBSTITUTE AND CHECK BACK.
This would make:
xy = 5.............I
(y^-2) = 3.........II
zy = 3 .............III
then solve by algebra and substitution. ...YA...CORRECT..FOR ALL THE GOOD WORK DONE BY YOU .LET ME COMPLETE THE SOLUTION
I*III....XY^2Z=15.......IV
II*IV.....XZ=45.........V
I*III*V......(X^2)(Y^2)(Z^2)=5*3*45...OR....XYZ=15 SQRT 3......VI
DIVIDING VI WITH I,III,V...WE GET
X=5SQRT 3
Y=1/SQRT 3
Z=3SQRT 3

This this correct?
Thanks

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