SOLUTION: could someone please help me solve this equations using Cramer's rule for three variables? I am just not getting past the point of transforming them into the three different matic

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Question 151668: could someone please help me solve this equations using Cramer's rule for three variables? I am just not getting past the point of transforming them into the three different matices.
x-2y=17
2y+z=10
-x+z=-2

Answer by Edwin McCravy(6939) About Me  (Show Source):
You can put this solution on YOUR website!
could someone please help me solve this equations using Cramer's rule for three variables? I am just not getting past the point of transforming them into the three different matices.


x-2y=17
2y+z=10
-x+z=-2

Put in +0 times every missing letter:
Put in understood 1 coefficients:

 1x - 2y + 0z = 17
 0x + 2y + 1z = 10
-1x + 0y + 1z = -2

You must memorize this:

This system:

Ax+%2B+By+%2B+Cz+=+D
Ex+%2B+Fy+%2B+Gz+=+H
Ix+%2B+Jy+%2B+Kz+=+L

has this solution:

x+=+abs%28matrix%283%2C3%2CD%2CB%2CC%2CH%2CF%2CG%2CL%2CJ%2CK%29%29%2Fabs%28matrix%283%2C3%2CA%2CB%2CC%2CE%2CF%2CG%2CI%2CJ%2CK%29%29       y+=+abs%28matrix%283%2C3%2CA%2CD%2CC%2CE%2CH%2CG%2CI%2CL%2CK%29%29%2Fabs%28matrix%283%2C3%2CA%2CB%2CC%2CE%2CF%2CG%2CI%2CJ%2CK%29%29       z+=+abs%28matrix%283%2C3%2CA%2CB%2CD%2CE%2CF%2CH%2CI%2CJ%2CL%29%29%2Fabs%28matrix%283%2C3%2CA%2CB%2CC%2CE%2CF%2CG%2CI%2CJ%2CK%29%29

The 3 denominators are all the same. They are the
coefficients of the unknowns in the same configuration
as they are in the original equations.

The column of constants is the column of numbers right of
the equal signs, namely this: matrix%283%2C1%2CD%2CH%2CL%29.

The numerator for the FIRST unknown, x, is like its denominator
except that the FIRST column matrix%283%2C1%2CA%2CE%2CI%29.is replaced by   matrix%283%2C1%2CD%2CH%2CL%29.

The numerator for the SECOND unknown, y, is like its denominator
except that the SECOND column matrix%283%2C1%2CB%2CF%2CJ%29.is replaced by   matrix%283%2C1%2CD%2CH%2CL%29.

The numerator for the THIRD unknown, x, is like its denominator
except that the THIRD column matrix%283%2C1%2CC%2CG%2CK%29.is replaced by   matrix%283%2C1%2CD%2CH%2CL%29.

-----

This system:

  1x+-+2y+%2B+0z+=+17
  0x+%2B+2y+%2B+1z+=+10
 -1x+%2B+0y+%2B+1z+=+-2

has this solution:

x+=+abs%28matrix%283%2C3%2C17%2C-2%2C0%2C10%2C2%2C1%2C-2%2C0%2C1%29%29%2Fabs%28matrix%283%2C3%2C1%2C-2%2C0%2C0%2C2%2C1%2C-1%2C0%2C1%29%29       y+=+abs%28matrix%283%2C3%2C1%2C17%2C0%2C0%2C10%2C1%2C-1%2C-2%2C1%29%29%2Fabs%28matrix%283%2C3%2C1%2C-2%2C0%2C0%2C2%2C1%2C-1%2C0%2C1%29%29       z+=+abs%28matrix%283%2C3%2C1%2C-2%2C17%2C0%2C2%2C10%2C-1%2C0%2C-2%29%29%2Fabs%28matrix%283%2C3%2C1%2C-2%2C0%2C0%2C2%2C1%2C-1%2C0%2C1%29%29

The 3 denominators are all the same. They are the
coefficients of the unknowns in the same configuration
as they are in the original equations.

The column of constants is the column of numbers right of
the equal signs, namely this: matrix%283%2C1%2C17%2C10%2C-2%29.

The numerator for the FIRST unknown, x, is like its denominator
except that the FIRST column matrix%283%2C1%2C1%2C0%2C-1%29.is replaced by   matrix%283%2C1%2C17%2C10%2C-2%29.

The numerator for the SECOND unknown, y, is like its denominator
except that the SECOND column matrix%283%2C1%2C-2%2C2%2C0%29.is replaced by   matrix%283%2C1%2C17%2C10%2C-2%29.

The numerator for the THIRD unknown, z, is like its denominator
except that the THIRD column matrix%283%2C1%2C0%2C1%2C1%29.is replaced by   matrix%283%2C1%2C17%2C10%2C-2%29.

Do you know how to evaluate the determinants? If not
post again asking how:

x=58%2F4=29%2F2       y=%28-5%29%2F4       z=50%2F4=25%2F2

Edwin