Strictly speaking, it's not the sign of the resistors that can be changed, it's the sign (direction) of the current \$I_x\$. Resistors always have positive resistance but you can choose the current in either direction as long as you are consistent.
Suppose you chose \$I_x\$ in the opposite direction. Starting from node \$f\$ and moving counterclockwise, you have a positive voltage across the \$6\Omega\$ resistor, another positive voltage across the \$3\Omega\$ resistor, and a positive voltage across the 12V supply. This gives:
\$I_{x}(6\Omega + 3\Omega) + 12 = 0\$
which results in \$I_{x} = -4/3\$A. The sign is different, indicating that the current actually flows in the opposite direction (i.e. in the original direction shown in the image).
The reason your 4A answer is wrong is that you weren't consistent with the \$I_x\$ direction for the two resistors -- you had \$I_x\$ flowing in the opposite direction through the \$3\Omega\$ resistor.
Here's the way I'd analyze this circuit:
1) By inspection, the emitter resistor is smaller than the collector resistor. So short C-E and calculate the voltage at that junction. I get about 4.13V.
2) The Base voltage is set at 6V. The voltage from above (4.13V) means that the E-B junction is forward biased. That changes things. Note that you need to do the above calculation to determine if, in fact, the junction is forward-biased.
3) The Emitter voltage is the Base voltage minus Vbe. That is: 6V - (0.7) = 5.3V. The current through the Emitter resistor is that voltage divided by the resistance: 5.3 / 3300 = ~1.61 mA.
4) Assume that the transistor is saturated. We'll do the calculations based on that assumption, then go back and check the assumption once we have some numbers to work with.
5) The collector voltage is about Vemitter plus the saturation voltage. I would normally assume saturation voltage at about 150 mV but your text says 200 mV, so use that. Ve + 0.2 = 5.5V.
6) Calculate the collector current. (10 - 5.5) / 4700 = ~0.957 mA
7) The Base current is the Emitter current minus the Collector current. 1.61 - 0.957 = ~0.649 mA
Now check to see if the transistor is in saturation. Most small-signal transistors have a Hfe of anywhere from 40 to 200. Let's use the worst-case value of 40.
If the transistor was NOT saturated, the collector current would be greater than 0.649 mA * 40 = ~25.9 mA. But we already know the collector current is about 0.957 mA. Therefore, the transistor is saturated and the above calculations hold.
Note that it took far longer to type this out than it did to calculate.
Best Answer
Nope, you're absolutely correct in both cases. A negative magnitude of a vector is the same as a positive magnitude on the vector in the opposite direction.