These equations are equivalent to the matrix multiplication:
⎡1 2 -1⎤⎡x⎤ ⎡ 1⎤
⎢ 2 -3 1⎥⎢y⎥= ⎢-1⎥
⎣3 1 2 ⎦⎣z⎦ ⎣12⎦
which can be written AX=B where A, X, B are the matrices.
|A|=1(-3×2-1×1)-2(2×2-1×3)-1(2×1-(-3)×3)=-7-2-11=-20. Let this determinant be represented by Δ.
Now we create 3 more determinants, substituting B in the variable column for each variable in turn.
Δx=
⎢ 1 2 -1⎥
⎢-1 -3 1⎥= 1(-6-1)-2(-2-12)-1(-1+36)=-7+28-35=-14
⎢12 1 2 ⎥
x= Δx/Δ=-14/-20=0.7.
Δy=
⎢ 1 1 -1⎥
⎢ 2 -1 1⎥= 1(-2-12)-1(4-3)-1(24+3)=-14-1-27=-42
⎢ 3 12 2 ⎥
y= Δy/Δ=-42/-20=2.1.
Δz=
⎢ 1 2 1 ⎥
⎢ 2 -3 -1⎥= 1(-36+1)-2(24+3)+1(2+9)=-35-54+11=-78
⎢ 3 1 12⎥
z= Δz/Δ=-78/-20=3.9.
Therefore the solution using Cramer's Method is x=0.7, y=2.1, z=3.9.
This result satisfies the given equations:
0.7+4.2-3.9=1, 1.4-6.3+3.9=-1, 2.1+2.1+7.8=12.