Quick question about led matrices

So I have small rg led matrices from sparkfun, and I have shift registers to drive them (74hc595), three for each matrix. I don’t however have enough resistors to have one for each cathode (anode? I can’t remember). Could I just use a resistor connected to the VCC pins of the registers for the cathodes, or maybe one connected to GND of the common anode register?
Thanks.

I don't however have enough resistors to have one for each cathode (anode? I can't remember).

doesn't matter (in this case)

Could I just use a resistor connected to the VCC pins of the registers for the cathodes, or maybe one connected to GND of the common anode register?

I am pretty sure that will not work, but I have no explanation at this point

(well I keep thinking that It will pass current on the latch pin, or 2 you will be dropping Vcc, and it will only handle 0.5 or so volts above Vcc, so what happens when you shoot a 5 volt clock (or whatever) signal into it)

If i'm not mistaking you could do that on the common A/C. Problem is i think the LED brightness will dim as you light up more LEDs. Try it on a few plain LEDs and see what happens. Better to loose a few cheap LEDs rather then a matrix.

I have 8 resistors, if that makes a difference. They are 330 ohm. I think I'll try it on some other leds, as digimike suggested.

I have 8 resistors, if that makes a difference.

Yes you need 16.

Do you have any current drivers for the rows? You can not source not enough current from a shift register to light up 16 LEDs at once (one row of r and one row of g)

The sinking shift registers are OK as you only need to sink the current from one LED at a time.

Sorry, forgot to mention I have darlington sink drivers, whatever they're called. Someone earlier mentioned that it may be a problem using the resistor on one pin of the shift register, and therefore having a higher voltage on the inputs. Could I put a resistor on the darlington driver if that would work better?

It’s aright having sink drives but you also need source drives to supply the current. Like I used on this:-
http://www.thebox.myzen.co.uk/Hardware/Econo_Monome.html

You need a resistor in the path of each LED in the end (anode or cathode) that is not common. Having it in the sink lines is fine, as would be having it in the source lines, if it is not the common line. But you need one resistor to take the current from only one LED at any instant of time.

Why would I need source drives if each output of the shift register is driving 1 led max ata a time? (row/column scanning)
And I still don’t have enough resistors. So there is no safe way to use <= 8 resistors in wiring two small 8*8 led matrices?

Okay, I have another idea.
Could I power the matrices and the ICs with 3v, and use the resistors instead to connect the arduino’s outputs to the ICs?
That way, woudn’t the LEDs still get below 5v, and the ICs would get about the same voltage from the 3v power and the arduino’s outputs, avoiding a problem that someone earlier suggested?
(sorry for that long sentence)