# Matrix inversion with complex numbers? or faster way?

1. May 5, 2013

### asdf12312

matrix inversion with complex numbers?? or faster way?

1. The problem statement, all variables and given/known data

3. The attempt at a solution

i managed to get the answer, but it took me like 30min. to work this by hand. i probably worked it differently than my instructor's method above, but wat i did was get the coefficients of V on one side, coefficient of V(ex) on the other side.

(3/0.2 + 1/(0.1+j0.1))V = (1/(0.1+j0.1))Vex
(20-5j)V = (5-5j)Vex
V=((5-5j)/(20-5j))Vex
V=(0.343<-30.96)Vex

which in fact did equal V=[0.294− j0.176]Vex when i converted to rectangular form. the problem is, this took me way too long to work, i was wondering if there was an easier way. like perhaps matrix inversion with complex numbers? i still have no idea how that works. because i have a test on this and if i spend 30min. to simplify equation like this, and plus this isn't even the answer to the problem, it's just one equation. is there an easier way??

2. May 5, 2013

### SteamKing

Staff Emeritus
No calculator?

3. May 5, 2013

### asdf12312

i have a calculator, i used it for some of the simpler calculations actualy . just don't know how to use matrix inversion for a problem liek this.

4. May 5, 2013

### SteamKing

Staff Emeritus
If it takes you 30 min. to do the calculations in the OP with a calculator, matrix inversion will not be any quicker.

5. May 6, 2013

### asdf12312

should it not take 30min? maybe the method i used to do it was slow, i had to convert between rectangular and polar form a lot. plus i had to do multiplication with complex numbers by hand. couldnt do it with calculator.

6. May 6, 2013

### SteamKing

Staff Emeritus
For multiplication and division, work exclusively in polar form. When you obtain a final answer, you can convert it to rectangular, if you wish.

7. May 6, 2013

### The Electrician

If someone tells you how to solve the problem by matrix inversion, how would you carry out the inversion? By hand? If you have to do complex arithmetic by hand, how do you plan to invert a matrix having complex numbers among the elements of the matrix?

If you try to invert a matrix by hand, the amount of number crunching involved won't be any easier than just solving the equations directly.

I can't believe you don't have a calculator that can do complex arithmetic.

You should get a TI-86 or a HP50G. You can find a TI-86 on eBay for $20 to$30. Right now I see an HP50G for a good price:

http://www.ebay.com/itm/BEAUTIFUL-H...1039652398?pt=Calculators&hash=item232aaa322e

Both calculators can do complex arithmetic, including matrix inversion with complex entries.

Edit: You can also solve problems like this with resources available on the web, such as Wolfram Alpha.

Last edited: May 6, 2013
8. May 6, 2013

### asdf12312

I have a TI-83 plus. i can do normal matrix inversion but i don't know if i can do it with complex numbers.

9. May 6, 2013

### SteamKing

Staff Emeritus
Your TI-83 plus does complex number operations as a standard feature. It can convert between rectangular and polar form. It should not take you 30 min. to do the calculations listed in the OP. It would help you a lot to review the guidebook for this calculator. If a guidebook did not come with your calculator, you can download one from here:

10. May 6, 2013

### asdf12312

thanks very much! i actually decided to google it, found a site that had very easy to follow instructions: http://www.tc3.edu/instruct/sbrown/ti83/complx83.htm#Display

just managed to get the answer in like 10sec..i obviously was doing it wrong, haha. i didn't know my calculator could convert from rectangular to polar form or do complex number oprations, so i was doing it myself, just plugging in values into the calculator. like tan^-1(y/x) to find the angle. and yah, i don't think i need matrix inversion then.

also, i know its easier to use polar form for multiplication/division, my book just had a shortcut method that involved multiplying complex numbers this way: 1/(1-1i) = [1/(1-1i)] * [(1+1i)/(1+1i)] so it would get rid of any imaginary part in the denominator, since i^2=-1, so the answer would be (1+1i)/2 = 0.5+0.5i. this is what i meant by doing it by hand. but obviously with my calculator i can do this in like 1 sec.

Last edited: May 6, 2013
11. May 9, 2013

### asdf12312

OK, just found out i cannot input complex numbers into matrices on my ti-83 plus. i have an exam coming up pretty soon, and here is a part requiring to solve simultaneous equations from a sample problem my teacher said might show up on it:

so i was wondring, is solving simulatneous equations with complex numbers doable by hand/calculator without matrix inversion, and if so how long that would take? if it helps using my calculator i worked out the matrix elements to be:

Code (Text):

[B]V1           V2 [/B]
[0.4+0.1i     -0.2] = [4.2]
[-0.2     0.4+0.1i] = [-2.1]

Last edited: May 9, 2013
12. May 9, 2013

### SteamKing

Staff Emeritus
13. May 9, 2013

### The Electrician

You can use Cramer's rule.

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14. May 10, 2013

### asdf12312

thanks. just finished my exam, went in there hoping for at most a 2x2 complex-number matrix because thats all i knew how to calculate, turns out there was only one problem like this and with a single node circuit too. so basically just a 1x1 matrix. which was easy, just used my calculator, no detreminants at all. thanks anyway :D