Switch on a LED: why doesn’t this code work

The 2 rails at the top of a breadboard are all connected along the straight line. In the picture, you are connecting both ends of the resistor to the same rail, so it's as if it's not even there (all the current will bypass it and travel through the LED).

It's the equivalent circuit of having the resistor ends connected (same voltage):

          +---R---+      <----- Resistor is useless, current will bypass it
          |       |
 +--------+-------+-----+
 |                      |
(+)                     |
 V                    (LED)
(-)                     |
 |                      |
 +----------------------+

If you are trying to have an LED to let you know when your power is connected, you will either need to use a pin line outside of the voltage rails (and make sure the current passes through the resistor), or else use an LED with a built-in resistor. As your circuit is now though, you run the risk of shorting out your battery, depending on how much current the LED allows through.

If you could change your scanning code to either pull the column high, instead of low, then you don't need to do anything special to read this switch. Connect the column input to R2 and read the switch from the junction of R2 and the switch. You'd also want to enable the internal pull-up resistor on the column inputs, instead of the row inputs.

If that's not possible, then your circuit should work. Be sure you tri-state the column, rather than pulling high, otherwise you might reverse-bias the base-emitter junction too much and Q1 might not like it. You could also use a PNP for Q1, which will invert the logic (I think you want that anyway). In the PNP case you can also dispense with R2.