Hi, there--

Problem:
Do you mean solve the following system of equations using Cramer's Rule?
4x+4y-5z=13
-2x+3y=-10
2x+y-2z=-5

Answer:

We have the left-hand side of the system with the variables (the "coefficient matrix") and the 
right-hand side with the answer values. Let D be the determinant of the coefficient matrix of 
the above system. 

Build a 3x3 coefficient matrix D for the left-hand side of the system:

4, 4, -5
-2, 3, 0
2, 1, -2

Build 3x1 matrix for the right-hand side of the system:

13
-10
-5

Let Dx be the determinant formed by replacing the x-column values with the answer-column 
values. 
Let Dy be the determinant formed by replacing the y-column values with the answer-column 
values.
Let Dz be the determinant formed by replacing the z-column values with the answer-column 
values.

Dx:
13, 4, -5
-10, 3, 0
-5, 1, -2 

Dy:
4, 13, -5
-2, -10, 0
2, -5, -2

Dz:
4, 4, 13
-2, 3, -10
2, 1, -5

Cramer's Rule states that

x = Dx/D
y = Dy/D
z = Dz/D

Evaluate each determinate and apply the formulas to find (x,y,z).

D:

 4, 4, -5
-2, 3,   0
 2, 1, -2

(4)(3)(-2)+(4)(0)(2)+(-5)(-2)(1)-(2)(3)(-5)-(1)(0)(4)-(-2)(-2)(4)=(-24)+(0)+(10)-(-30)-(0)-(16)=
D = 0

I'll stop here. Since D=0, we cannot use Cramer's Rule. (It would mean that we divide by zero...a definite no-no!)

You will need to go back to a different method, such as row operations.

~Mrs. Figgy