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Notice that the current source E1 is on a branch shared by current loop  and  . As these currents pass through E1, will face up while will face down. We know that the difference of these currents must result in a net current of 1A upwards (since the arrow points upwards). Therefore since is upwards then
Notice this is equivalent to considering down is positive in which case the net resulting current would be 1A up or -1A. In this case, choosing down as positive we'd have  which is exactly equivalent to above.
Full Solution for Those Interested:
If people are looking to practice there work with circuits, we can continue work from here and from A10 to solve for the unknown voltages and loop currents. Based on the same reasoning in A10 we can get that the Kirchhoff linear equation for loop 1 is
where recall that positive numbers mean voltage drops (across the resistors and across the current source since it points upwards while our clockwise current points downwards there) and that negative numbers mean voltage gains (across the battery since we move from the negative to positive terminal). The 7 resistor takes the difference in currents from loop 1 and loop 3 since it shares those loop currents. Similarly we can get for loop 2
where for this loop  is a voltage gain because we are traversing upward, just like the arrow. We already have the third loop from A10 as
and so the fourth and final loop is
where here crossing the battery is a voltage drop because we are going from the positive to negative terminal. So far we only have 4 equations for 6 unknowns; the last two come from the current source. From above we already have that
and through the current source E2 we have that
We can write this in a matrix problem as
We can write this is an augmented matrix as
![{\displaystyle \left[{\begin{array}{cccccc|c}11&0&-7&0&1&0&3\\0&14&0&-8&-1&0&0\\-7&0&16&0&0&1&0\\0&-8&0&18&0&-1&-5\\-1&1&0&0&0&0&1\\0&0&-1&1&0&0&2\end{array}}\right]}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/59df947790294f2f23d099ff01825de01c50d9c3)
which we can row reduce to get
![{\displaystyle \left[{\begin{array}{cccccc|c}1&0&0&0&0&0&-0.5200\\0&1&0&0&0&0&0.4800\\0&0&1&0&0&0&-1.2000\\0&0&0&1&0&0&0.8000\\0&0&0&0&1&0&0.3200\\0&0&0&0&0&1&15.5600\end{array}}\right].}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/7ce7e1767f36831924535413627336ed54635493)
Therefore we see that our solution is
Do not worry about the presence of a negative sign, a negative loop current just means that the current ( and  in this case) actually flows counter-clockwise, not clockwise like we assumed.
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