equations with complex results

[Anders]

Hi,

Normally, you can use RREF. Or you can simply use matrix inversion putting the left hand side coefficients in a 3x3 matrix (just call it M1 for simplicity) and the right hand side values into a vector (say M2) and then simply do M1^-1 * M2 and then you have (x y z) in the resulting vector. You can do this directly in home view using [] function.

However, in this specific case the matrix is singular because it turns out that 1 equation is redundant (you can quickly see this by taking the determinate of M1 i.e. DET(M1) which is equal to 0 which means it is singular). You can also see this by doing Gaussian elimination row by row. If you do Gaussian elimination you can in the final step eliminate the last equation all together and the result is actually the entire space that satisfies equation 1 and 2 which is a complex plan in the 3 dimensional complex space:
(12+10i)x + (-2-10i)y + (-10+0i)z = (190+110i)
(-2-10i)x + (2+0i)y +(0+10i)z = (0-220i)

Gaussian elimination is done using the pivot() function and in this case you may want to put the entire system into a 3x4 matrix where the last column is the right hand side in the equation system.

Then pivot(M1,1,1) and the pivot(result of pivot(M1,1,1) matrix, 2, 2) and you get:
[
1640+960i 0 -1640-960i ¦ 33,160+22,160i
0 120-20i -120+20i ¦ 1,480 -520i
0 0 0 ¦ 0
]
This can be simplified further with divisions and displayed in more elegant matrix form with some more linear algebra...