[VA] SRC #015b - HP-15C & clones: COMPLEX Matrix Inverse up to 8x8

[Gerson W. Barbosa]

(09-29-2023 01:20 PM)Werner Wrote:  - to solve (A+iB)*(X+iY) = C+iD, press (GSB) C and (GSB) D

Thank you both, Werner and Valentín!
The matrix capabilities in the HP-15C were nice, but the methods in the manual for solving complex linear systems were somewhat complicated and not easy to remember. I would need to carry the manual along to use it for that task. Doing it programmatically is a great idea, I wonder why no one thought of this back then – at least I don’t remember of any. These would have been handy when I got my 15C a few years later.

Here’s a real-world example to illustrate the usage of your latest method. Perhaps not the best example as the main matrix has only real elements:

Three-phase voltage source (Wye), unbalanced resistive load (Δ), line voltage = 220V.

R₁I₁ + 220√3/3∠120° = 220√3/3

R₂I₂ + 220√3/3 = 220√3/3∠-120°

R₃I₃ + 220√3/3∠-120° = 220√3/3∠120°

R₁ = R₂ = 12.1 Ω; R₃ = 24.2 Ω


A =
| 12.1 0 0 |
| 0 12.1 0 |
| 0 0 24.2 |

B =
| 0 0 0 |
| 0 0 0 |
| 0 0 0 |

C =
| 190.5255888 |
|-190.5255888 |
| 0.000000000 |

D =
|-110 |
|-110 |
| 220 |

GSB C GSB D ->

X =
| I₁ |
| I₂ |
| I₃ |

=

C =
| 15.74591643 |
|-15.74591643 |
| 0.000000000 |

D =
|-9.090909091 |
|-9.090909091 |
| 9.090909091 |

or

I₁ = 18.18181818∠-30°

I₂ = 18.18181828∠-150°

I₃ = 9.090909091∠90°

These are the load currents. The source currents are I₁ - I₂, I₂ - I₃ and I₃ - I₁.

Edited to add three missing negative signs (corrections in red).