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Hi

how to solve the system

(12+10i)x + (-2-10i)y + (-10+0i)z = (190+110i)
(-2-10i)x + (2+0i)y +(0+10i)z = (0-220i)
(-10+oi)x + (0+10i)y +(10-10i)z = (-190+110i)

Thanks
Hi!, fcgaino:

You can get complex results by using the reduced row echelon (RREF) method, of solving, a linear system.

Best Regards.
informach
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...
Other approach:

If you want find out if your system of equations has independent equations (or how many are dependent) then you can use rank of matrix (which is built in HP Prime too). In your example RANK(M1) is 2, so there are only two independent equations in your system of equations (I assume a31 = (-10+oi) is typo and it should be (-10+0i)).
At HP 48, in Solve linear system, I found the following values

x = 10.09<10.6 degrees
y = 2.45<-55.3 degrees
z = 11.31<179.2 degrees

but, in the HP Prime, i don't find these values.

Thanks
Are you sure the coefficient in the system you gave are correct? Because there is no one single point solution to the system you wrote (it is an entire complex plan of infinite number of solutions). The HP 48 gave a point answer so I do not think you are solving the same system there (some one who have HP 48 can answer this better). In any case the Prime is solving in correctly by using pivot() that I wrote above.
You can also use solve() in CAS:

Code:
`solve([(12+10i)*x + (-2-10i)*y + (-10+0i)*z = (190+110i), (-2-10i)*x + (2+0i)*y +(0+10i)*z = (0-220i), (-10+0i)*x + (0+10i)*y +(10-10i)*z = (-190+110i)],[x,y,z])`

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(07-04-2015 12:50 AM)fcgaino Wrote: [ -> ]At HP 48, in Solve linear system, I found the following values

x = 10.09<10.6 degrees
y = 2.45<-55.3 degrees
z = 11.31<179.2 degrees

but, in the HP Prime, i don't find these values.

Thanks

From memory (I don't have an HP48 with me here) the HP48 and other calculators in that series provide the solution that minimises the length of the solution vector when solving an underdetermined set of equations such as this. Your equations have infinitely many solutions so there is no reason for the Prime to return one particular solution. The failure of the HP48 to point out the existence of other solutions in this case is not helpful; the Prime's behaviour is more correct.

Nigel (UK)
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